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Appendix A

Exercises

We end this book by giving a list of exercises and providing full solutions forall of them. Many of these exercises regard properties of point-line geometriesthat were already mentioned (without proofs) in Chapter 2. The list ofproblems is however not restricted to only those properties. Many exercisesalso concern the construction of certain models of point-line geometries orthe uniqueness of certain point-line geometries with small parameters.

A.1 List of problems

(1) Show that every ordinary n-gon with n ∈ N \ {0, 1, 2} is self-dual.

(2) Show that every near-pencil is self-dual.

(3) Let n1, n2 ∈ N\{0, 1}. Determine all hyperplanes of the (n1×n2)-grid.

(4) Show that the dual grids are precisely the complete bipartite graphswhose two parts have size at least 2.

(5) Show that if S is a nondegenerate projective plane, then there exists apossibly infinite cardinal number s ≥ 2 such that every point is incidentwith precisely s+1 lines and every line is incident with precisely s+1points.

(6) Show that the degenerate projective planes are precisely the near-pencils.

© Springer International Publishing AG 2016B. De Bruyn, An Introduction to Incidence Geometry, Frontiers in Mathematics,DOI 10.1007/978-3-319-43811-5

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Chapter A - Exercises

(7) Show that Properties (�) and (�′) of Section 2.2 are equivalent.

(8) Prove that a projective space of dimension at least 2 is irreducibleif and only if for every three noncollinear points x1, x2 and x3, the

point-line geometry ˜〈x1, x2, x3〉 induced on the subspace 〈x1, x2, x3〉 isan irreducible projective plane.

(9) Let V be a right vector space of dimension n ≥ 1 over a skew field F.If W is a subspace of V , then show that the set SW of all points ofthe form 〈w〉 with w ∈ W \ {o} is a subspace of PG(V ) (regarded as apoint-line geometry). Show also that every subspace of PG(V ) can beobtained in this way.

(10) Suppose V is a right vector space over a skew field F with dim(V ) ≥ 1.Show that the generating index of PG(V ) is equal to dim(V ) and thatthe dimension of the projective space PG(V ) is equal to dim(V )− 1.

(11) Show that a Desarguesian projective space is always irreducible.

(12) Let S be a (possibly reducible) projective space. Define the followingrelation R on the point set P of S. If x1 and x2 are two points of S, then(x1, x2) ∈ R if and only if either x1 = x2 or x1 �= x2 and the unique linethrough x1 and x2 is thick. Prove that R is an equivalence relation. LetXi, i ∈ I, denote the equivalence classes of R for some suitable indexset I. Prove that Xi is a subspace, that Xi is an irreducible projectivespace and that S is the direct sum of the Xi’s.

(13) Prove that there exist up to isomorphism unique projective planes oforder 2 and 3.

(14) Let S be an irreducible projective space of dimension d ≥ 2 and leti ∈ {0, 1, . . . , d − 1}. Show that the Grassmannian Gr(S, i) of thei-dimensional subspaces of S is a partial linear space.

(15) Prove that if S is an irreducible projective plane, then Gr(S, 1) is iso-morphic to the point-line dual of S. Prove that if S ∼= PG(d,F) forsome d ≥ 2 and some skew field F, then Gr(S, d − 1) ∼= PG(d,Fo).

(16) Show on basis of the axioms of an affine plane that the parallelismrelation is an equivalence relation on the set of lines.

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Section A.1 - List of problems

(17) Let V be a right vector space of dimension n ≥ 2 over a skew field F,and denote by AG(n,F) the affine space associated with V . If W is asubspace of V and w ∈ V , then show that SW,w = {w+ w′ | w′ ∈ W} isa subspace of AG(n,F) (regarded as a point-line geometry). If |F| ≥ 3,then also show that every subspace of AG(n,F) can be obtained in thisway.

(18) Let V be a right vector space of dimension n ≥ 2 over a skew field F ofsize at least three, and denote by AG(n,F) the affine space associatedwith V . Show that the generating index of AG(n,F) is equal to n+ 1.

(19) Prove that every net in which there are precisely two parallel classes oflines is a grid.

(20) Prove that the affine planes are precisely the nets that are also linearspaces.

(21) Prove that the finite affine planes are precisely the finite nets that havean order (s, t) for which t = s+ 1.

(22) Suppose that in the definition of dual net, one replaces the condition“there are at least two equivalence classes and each equivalence classcontains at least two points” by the weaker condition “there are atleast two equivalence classes”. Then show that the only extra point-line geometries (besides the dual nets) that arise are the lines and theK1,a’s with a ≥ 2.

(23) Let S = (P,L, I) be an affine plane and let L′ be a parallel class oflines of S. Let S ′ be the point-line geometry (P,L \ L′, I′), where I′ isthe restriction of I to P × (L \ L′). Then prove that S ′ is a dual net.

(24) Prove that those nets that are also dual nets are precisely the affineplanes in which one parallel class of lines has been removed.

(25) Prove that if S is a net, then there exists a constant t such that everypoint of S is incident with precisely t+ 1 lines. Prove also that if S isa dual net, then there exists a constant s such that every line of S isincident with precisely s+ 1 points.

(26) Let Σ be a 3-dimensional projective space and let L be a line of Σ.Then let S be the point-line geometry whose points are the lines of Σ

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disjoint from L and whose lines are the planes of Σ not containing L,with incidence being containment. Then show that S is a net.

(27) Let X be a set of 23 + 1 = 9 points in PG(2, 4) intersecting each linein either 1 or 3 points. Then show that X is a Hermitian curve.

(28) Prove that there exists up to isomorphism a unique Steiner quadruplesystem of type S(3, 4, 8).

(29) Prove that there exists up to isomorphism a unique inversive plane oftype S(3, 4, 10).

(30) Prove that the Witt designs S(5, 6, 12) and S(5, 8, 24) are not extend-able.

(31) If v is the number of points of a Steiner triple system, then prove thatv is congruent to either 1 or 3 modulo 6. If v is the number of pointsof a Steiner quadruple system, then prove that v is congruent to either2 or 4 modulo 6.

(32) Let q be a prime power and n, d nonnegative integers satisfying n ≥d ≥ 1. Let S be the point-line geometry whose points are the points ofthe n-dimensional projective space PG(n, q) and whose lines are the d-dimensional subspaces of PG(n, q), where incidence is the one inducedby PG(n, q). Prove that S is a 2-(v, k, λ)-design, and determine theparameters v, k and λ.

(33) Let q be a prime power and n, d nonnegative integers satisfying n ≥ 2and d ≥ 1. Let S be the point-line geometry whose points are the pointsof the n-dimensional affine space AG(n, q) and whose lines are the d-dimensional subspaces of AG(n, q), where incidence is the one inducedby AG(n, q). Prove that S is a 2-(v, k, λ)-design, and determine theparameters v, k and λ.

(34) Let n and d be nonnegative integers satisfying n ≥ d ≥ 2. Let S be thepoint-line geometry whose points are the points of the n-dimensionalaffine space AG(n, 2) and whose lines are the d-dimensional subspacesof AG(n, 2), where incidence is the one induced by AG(n, 2). Provethat S is a 3-(v, k, λ)-design, and determine the parameters v, k and λ.

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(35) Suppose H is a hyperoval of a projective plane π of even order n. LetS be the point-line geometry whose points are the lines of π disjointfrom H and whose lines are the points of π not contained in H, withincidence being reverse containment. Prove that S is a Steiner systemand determine its parameters.

(36) Prove that the point-line dual SD of a generalized quadrangle S (asdefined in Section 2.7) is again a generalized quadrangle.

(37) Prove that there exists up to isomorphism a unique generalized quad-rangle of order 2.

(38) Let H be a hyperoval of a hyperplane PG(2, q) of the 3-dimensionalprojective space PG(3, q), q even. Let S be the point-line geometrywhose points are the points of PG(3, q) \ PG(2, q) and whose linesare those lines of PG(3, q) not contained in PG(2, q) that contain aunique point of H, with incidence being containment. Show that S isa generalized quadrangle of order (q − 1, q + 1).

(39) Let x be a point of the polar space W (3, q). Let L1 denote the set oflines of W (3, q) not containing x and let L2 denote the set of hyperboliclines of W (3, q) containing x. Let S be the point-line geometry whosepoints are the points of W (3, q) noncollinear with x and whose linesare the elements of L1 ∪ L2, with incidence being containment. Showthat S is a generalized quadrangle of order (q − 1, q + 1).

(40) Let PG(d, q), d ∈ {2, 3}, be a hyperplane of PG(d+ 1, q) and let O bea set of points of PG(d, q) which is an oval if d = 2 and an ovoid ifd = 3. Let Td(O) be the point-line geometry whose points are of oneof the following types:

(i) the points of PG(d+ 1, q) not contained in PG(d, q);

(ii) the hyperplanes of PG(d+ 1, q) containing a unique point of O;

(iii) a new symbol (∞).

The lines of Td(O) are of two types:

(a) the lines of PG(d+1, q) containing a point of O and not containedin PG(d, q);

(b) the points of O.

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A point of Type (i) is incident with no line of Type (b) and with alllines of Type (a) it is incident with in PG(d + 1, q). A point of Type(ii) is incident with those lines of Type (a) and (b) which are containedin it. The point (∞) is incident with no line of Type (a) and with alllines of Type (b). Prove that Td(O) is a generalized quadrangle anddetermine its order.

(41) Let H be a hyperoval of a plane PG(2, 2h) of the projective spacePG(3, 2h), h ∈ N \ {0}, and let x, y be two distinct points of H. Thenthe following point-line geometry S(H, {x, y}) can be defined. Thepoints of S(H, {x, y}) are of three types:

(i) the points of PG(3, 2h) not contained in PG(2, 2h);

(ii) the planes through x not containing y;

(iii) the planes through y not containing x.

The lines of S(H, {x, y}) are those lines of PG(3, 2h) that are notcontained in PG(2, 2h) and meet H \ {x, y}. A point and a line ofS(H, {x, y}) are incident if and only if they are incident as objectsof the projective space PG(3, 2h). Then show that S(H, {x, y}) is ageneralized quadrangle of order (2h + 1, 2h − 1).

(42) Let Π = (P,Σ) be a Veldkamp-Tits polar space of rank n ≥ 1 and leti ∈ {0, 1, . . . , n− 2}. Show that Gr(Π, i) is a partial linear space.

(43) Put X := {1, 2, 3, 4, 5, 6, 7, 8}. Let S = (P,L, I) be the point-linegeometry whose points are the partitions of X in four subsets of size2 and whose lines are the partitions of X in two subsets of size 2 andone subset of size 4. A point is incident with a line if and only if thepartition corresponding to the point is a refinement of the partitioncorresponding to the line. Prove that S is a near hexagon and thatevery line of S is incident with precisely three points.

(44) Prove that the near 3-gons are precisely the linear spaces that are notlines.

(45) Let S be a partial geometry with parameters (s, t, α) where α �= s+ 1.Prove that the collinearity graph Γ of S is a strongly regular graph anddetermine its parameters.

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(46) Show that every partial geometry S with parameters (s, t, 1) wheres, t ∈ N \ {0} is a generalized quadrangle.

(47) Show that the finite linear spaces of order (s, t) where s, t ≥ 1 areprecisely the partial geometries with parameters (s, t, s+ 1).

(48) Let s, t ∈ N\{0}. Prove (on basis of the axioms of a dual net) that thepartial geometries with parameters (s, t, s) are precisely the dual netsof order (s, t).

(49) Let s, t ∈ N\{0}. Prove (on basis of the axioms of a net) that the partialgeometries with parameters (s, t, t) are precisely the nets of order (s, t).

(50) Let S be a partial quadrangle with parameters (s, t, μ). Prove that thecollinearity graph Γ of S is a strongly regular graph and determine itsparameters.

(51) Prove that every partial quadrangle is either a generalized quadrangleor a near 5-gon.

(52) Prove that the partial quadrangles with parameters (s, t, μ) = (1, t, μ)are precisely the strongly regular graphs whose parameters (v, k, λ, μ)satisfy λ = 0, k = t + 1 and μ > 0.

(53) Let S be a semipartial geometry with parameters (s, t, α, μ) where α �=s + 1. Prove that the collinearity graph Γ of S is a strongly regulargraph and determine its parameters.

(54) Show that every partial geometry with parameters (s, t, α) is a semi-partial geometry with parameters (s, t, α, μ) = (s, t, α, (t+ 1)α). Showalso that if S is a semipartial geometry with parameters (s, t, α, μ) thatis not a linear space, then μ ≤ (t + 1)α with equality if and only if Sis a partial geometry.

(55) Show that if S is a semipartial geometry with parameters (s, t, α, μ),then α ≤ s+ 1 with equality if and only if S is a linear space.

(56) Show that the copolar spaces of order (1, t) are precisely those regulargraphs of valency t+ 1 that contain no triangles.

(57) Show that dual nets are examples of copolar spaces.

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(58) Show that the disjoint union of copolar spaces is again a copolar space.

(59) A maximal arc of degree d in a projective plane π of order q is anonempty set of points intersecting each line in either 0 or d points.Suppose K is a maximal arc of degree d ∈ {2, 3, . . . , q − 1} of a pro-jective plane π of order q. Prove that d | q. Let S be the point-linegeometry whose points are the points of π \ K and whose lines are thelines of π intersecting K in precisely d points, where incidence is de-rived from π. Prove that S is a partial geometry and determine itsparameters. When is S a generalized quadrangle?

(60) Let K be a maximal arc of degree d ≥ 2 in a hyperplane PG(2, q) ofthe 3-dimensional projective space PG(3, q). Let S be the point-linegeometry whose points are the points of PG(3, q) \PG(2, q) and whoselines are those lines of PG(3, q) not contained in PG(2, q) that containa unique point of K, where incidence is derived from PG(3, q). Provethat S is a partial geometry and determine its parameters. When is Sa generalized quadrangle?

(61) Let Q be a nonsingular elliptic quadric of a hyperplane PG(3, q) ofthe 4-dimensional projective space PG(4, q). Let S be the point-linegeometry whose points are the points of PG(4, q) \PG(3, q) and whoselines are those lines of PG(4, q) not contained in PG(3, q) that containa unique point of Q, where incidence is derived from PG(4, q). Provethat S is a partial quadrangle.

(62) Prove that the geometry of the hyperbolic lines of W (2n−1, q), n ≥ 2,is a copolar space as well as a semipartial geometry. What are theparameters of this semipartial geometry?

(63) Let S be the point-line geometry whose points and lines are the linesand planes of the projective space PG(n, q), n ≥ 3, where incidence iscontainment. Prove that S is a semipartial geometry and determine itsparameters. When is S a partial geometry?

(64) Let U be a nonsingular Hermitian curve of a hyperplane PG(2, q2) ofPG(3, q2). Let S be the point-line geometry whose points are the pointsof PG(3, q2)\PG(2, q2) and whose lines are those lines of PG(3, q2) notcontained in PG(2, q2) that contain a unique point of U , where incidence

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is derived from PG(3, q2). Prove that S is a semipartial geometry anddetermine its parameters.

(65) Let PG(2, q2) be embedded as a hyperplane in the projective spacePG(3, q2) and let PG(2, q) be a Baer subplane of PG(2, q2). Let S be thepoint-line geometry whose points are the points of PG(3, q2)\PG(2, q2)and whose lines are those lines of PG(3, q2) not contained in PG(2, q2)that contain a unique point of PG(2, q), with incidence being the onederived from PG(3, q2). Prove that S is a semipartial geometry anddetermine its parameters.

(66) Let Q be a nonsingular elliptic quadric of PG(5, 2). Let S be thepoint-line geometry whose points and lines are those points and linesof PG(5, 2) that lie in PG(5, 2)\Q, with incidence being the one derivedfrom PG(5, 2). Prove that S is a semipartial geometry and determineits parameters.

(67) Let PG(n− 2, q) be an (n− 2)-dimensional subspace of the projectivespace PG(n, q), n ≥ 3. Let S be the point-line geometry whose pointsare the lines of PG(n, q) disjoint from PG(n−2, q) and whose lines arethe planes of PG(n, q) intersecting PG(n − 2, q) in a singleton, withincidence being inclusion. Prove that S is a semipartial geometry anddetermine its parameters.

(68) Prove that the generalized Moore geometries of diameter 2 are preciselythe partial quadrangles.

(69) Prove that every generalized Moore geometry of diameter d is either anear 2d-gon or a near (2d+ 1)-gon.

(70) Show that if Γ is a regular graph of valency k and diameter d, then thenumber of vertices of Γ is bounded above by 1 + k + k(k − 1) + · · ·+k(k − 1)d−1, with equality if and only if Γ is a Moore graph.

(71) Show that there exists up to isomorphism a unique Moore graph ofdiameter 2 and valency 3.

(72) Prove that a Moore graph of diameter 2 is a partial quadrangle as wellas a copolar space.

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(73) Let Γ be a Moore graph of diameter 2. For every vertex x of Γ, let x⊥

denote the set of vertices of Γ adjacent to x (not including x). Let S bethe point-line geometry whose points are the vertices of Γ and whoselines are the sets x⊥ where x is some vertex of Γ, with incidence beingcontainment. Prove that S is a semi-partial geometry and determineits parameters.

(74) Prove that the point-line dual of the circular space on four vertices isisomorphic to the partial linear space obtained from the Fano plane byremoving one point and all three lines incident with that point.

(75) Show that the affine plane AG(2, 3) and the point-line dual of the cir-cular space on 4 vertices are examples of Fischer spaces.

(76) Prove that if S is a Fischer space, then the partial linear space S ′

constructed in Section 2.20 satisfies the properties (F1’) and (F2’).Prove also the converse, namely that if S ′ is a partial linear spacesatisfying (F1’) and (F2’), then the point-line geometry S derived fromS ′ as described in Section 2.20 is a Fischer space.

(77) Prove that the point-line geometry SO constructed in Section 2.21 isan inversive plane.

(78) Show that the finite inversive planes are precisely the Steiner systemsof type S(3, n+ 1, n2 + 1) for some n ∈ N \ {0, 1}.

(79) Prove that the point-line geometry SX constructed in Section 2.22 is aLaguerre plane.

(80) Prove that the point-line geometry SQ constructed in Section 2.23 is aMinkowski plane.

A.2 Solutions of the problems

Solution of Problem 1. Let S = (P,L, I) be an ordinary n-gon with pointset P = {1, 2, . . . , n} and line set L = {{1, 2}, {2, 3}, . . . , {n, 1}}, whereincidence is containment. Then the permutation of P ∪ L defined by

1 �→ {n, 1}, 2 �→ {1, 2}, 3 �→ {2, 3}, . . . , n �→ {n− 1, n},

{1, 2} �→ 1, {2, 3} �→ 2, {3, 4} �→ 3, . . . , {n, 1} �→ n

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Section A.2 - Solutions of the problems

induces an isomorphism between S and its point-line dual SD. �

Solution of Problem 2. Suppose S = (P,L, I) is a near-pencil. Then thereis a line L which is incident with all but one of the points. We denote theunique point non-incident with L by p. The permutation of P ∪ L definedby p �→ L, L �→ p, x �→ xp (x IL), xp �→ x (x IL) defines an isomorphismbetween S and SD. �

Solution of Problem 3. Let S be an (n1 × n2)-grid. Then the line set Lof S can be partitioned into two subsets L1 and L2 such that two lines ofthe same subset Li, i ∈ {1, 2}, are disjoint. For every point x of S and everyi ∈ {1, 2}, let Li(x) denote the unique line of Li containing x.

Suppose H is a hyperplane of S not containing any line. Then everyelement of L1 ∪ L2 contains a unique point of H , implying that for everyline L1 ∈ L1, there exists a unique line θ(L1) ∈ L2 such that L1 ∩ θ(L1)is a singleton contained in H . The map θ : L1 → L2 must be a bijection,showing that n1 = n2. Conversely, with every bijection θ : L1 → L2, therecorresponds a unique hyperplane of S not containing any lines, namely thehyperplane

⋃L1∈L1

(L1 ∩ θ(L1)). So, if n1 and n2 are equal, say to n, thenthe total number of hyperplanes of S not containing a line is equal to thenumber of bijections θ : L1 → L2, i.e. equal to n!.

Suppose H is a hyperplane of S containing a line L. Then without loss ofgenerality, we may suppose that L ∈ L1. Since there are lines of S disjointfrom L containing points of H , there must exist a point x of H not containedin L. Since the line K := L2(x) contains two points of H , namely x andK ∩ L, it must be contained in H . We now show that H = K ∪ L. Supposethat this is not the case. Then there exists a point y ∈ H \ (K ∪ L). Thenthe line M := L1(y) contains two distinct points of H , namely y and M ∩K,implying that M is contained in H . Now, every line N of L2 contains twopoints of S, namely N ∩ L and N ∩M , and hence is contained in H . Sincethe lines of L2 cover all points of S, the hyperplane H would contain allpoints of S, an obvious contradiction. So, H is equal to K ∪ L, i.e. of theform u⊥ for some point u of S. Clearly, K ∪L itself is a hyperplane as everyline meets it in either a point or a line. The total number of hyperplanes ofS containing a line is thus equal to the number of points of S, i.e. equal ton1 · n2. �

Solution of Problem 4. Suppose S is a grid whose points are the elementsof {1, 2, . . . , n1} × {1, 2, . . . , n2} and whose lines are all the sets of the form

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Li := {1, 2, . . . , n1} × {i} for some i ∈ {1, 2, . . . , n2} and all the sets of theform Mj := {j} × {1, 2, . . . , n2} for some j ∈ {1, 2, . . . , n1}, where incidenceis containment.

If i1, i2 ∈ {1, 2, . . . , n2} with i1 �= i2, then Li1 and Li2 are disjoint. Ifj1, j2 ∈ {1, 2, . . . , n1} with j1 �= j2, then Mj1 and Mj2 are disjoint. If i ∈{1, 2, . . . , n2} and j ∈ {1, 2, . . . , n1}, then Li and Mj meet in the point (j, i).

So, the point-line dual SD of S is a complete bipartite graph whose twoparts have sizes n1 ≥ 2 and n2 ≥ 2. �

Solution of Problem 5. For every point x of S, let Nx denote the totalnumber of lines of S incident with x, and for every line L of S, let ML denotethe total number of points incident with L. If (x, L) is an anti-flag of S, thenevery line through x meets L showing that Nx = ML.

Now, let x1, x2, x3, x4 be a collection of four points of S, no three ofwhich are collinear. Put s := Nx4 − 1. Since x4x1, x4x2 and x4x3 arethree distinct lines through x4, we have s ≥ 2. By the foregoing, we havethat Mx1x2 = Mx1x3 = Mx2x3 = Nx4 = s + 1. For every point x of S,there exists a line L ∈ {x1x2, x1x3, x2x3} not incident with x, and henceNx = ML = Nx4 = s + 1. For every line K of S, there exists a point y notincident with K and hence MK = Ny = s+ 1. �

Solution of Problem 6. Every near-pencil is a projective plane since thereare three noncollinear points and every two distinct lines meet in a uniquepoint. In a near-pencil, it is impossible to find a collection of four points, nothree of which are contained in a line. Indeed, there is a line containing allbut one of the points. So, every near-pencil is a degenerate projective plane.

Conversely, suppose that S is a degenerate projective plane, i.e. a pro-jective plane not having a collection of four points no three of which are ona line. Let L be one of the lines containing the largest amount of points.Suppose there are two distinct points x1, x2 not contained on L. The linex1x2 meets L in a point x3 and contains therefore at least three points. So,also L contains at least 3 points. If x4 and x5 are two points of L distinctfrom x3, then {x1, x2, x4, x5} is a collection of four points, no three of whichare collinear. As this is impossible, there can be at most one point outsideL. This implies that S is a near-pencil. �

Solution of Problem 7. We show that Property (∗′) implies Property (∗).Suppose L1 and L2 are two distinct lines through a point z. For i ∈ {1, 2},

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let xi and yi be two distinct points of Li \{z}. Since ˜〈z, x1, x2〉 is a projectiveplane, the lines x1x2 and y1y2 meet in a point.

We show that Property (∗) also implies Property (∗′). So, suppose S is alinear space satisfying Property (∗).

Suppose x1, x2, x3 are three noncollinear points and x′3 ∈ x1x3 \ {x1, x3}.

Let L denote the set of lines through x1 meeting x2x3, and let L′ denote theset of lines through x1 meeting x2x

′3. We show that L = L′. By symmetry,

it suffices to show that L ⊆ L′. Let L be an arbitrary line of L. If L ∈{x1x2, x1x3}, then L meets x2x

′3. Suppose therefore that L �∈ {x1x2, x1x3}

and denote by x4 the unique intersection point of the lines L and x2x3. ByProperty (∗), the lines x2x

′3 and x1x4 indeed meet in a point.

Suppose x1, x2, x3 are three noncollinear points. Let L2 and L3 be twodistinct lines through x1 meeting x2x3, let x

′2 ∈ L2 \{x1} and x′

3 ∈ L3 \{x1}.By applying the previous paragraph at most two times, we see that the setL of lines through x1 meeting x2x3 coincides with the set L′ of lines throughx1 meeting x′

2x′3. Now, put X :=

⋃L∈L L. Since L = L′, we see that the

line x′2x

′3 is contained in X. So, X must be a subspace, necessarily equal to

〈x1, x2, x3〉.Suppose x1, x2, x3 are three noncollinear points of S. Since S is a linear

space, ˜〈x1, x2, x3〉 is also a linear space. In order to prove that ˜〈x1, x2, x3〉is a projective plane, we must prove that any two distinct lines K and L of

˜〈x1, x2, x3〉 meet in a point. We may suppose that one of these lines, say L,does not contain x1. Let x′

2 and x′3 be two distinct points of L and let L′

denote the set of lines through x1 meeting x′2x

′3. Then 〈x1, x2, x3〉 =

⋃L′∈L′ L′

by the above. So, if x1 ∈ K, then K and L meet. If x1 �∈ K, then K and Lmeet by Property (∗).

So, any two distinct lines of ˜〈x1, x2, x3〉meet and ˜〈x1, x2, x3〉 is a projectiveplane. �

Solution of Problem 8. Suppose S is an irreducible projective space, and

denote by x1, x2, x3 three noncollinear points. Then ˜〈x1, x2, x3〉 is a projectiveplane for which each line is incident with at least three points. So, ˜〈x1, x2, x3〉cannot be a near-pencil and must be an irreducible projective plane.

Conversely, suppose S is a projective space of dimension at least 2 having

the property that for every three noncollinear points x1, x2 and x3, ˜〈x1, x2, x3〉is an irreducible projective plane. Since the dimension of S is at least 2, any

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line L of S is contained in such an irreducible projective plane and musttherefore contain at least three points. So, S is irreducible. �

Solution of Problem 9. If v1 and v2 are two linearly independent vectorsof V , and if p1 = 〈v1〉 and p2 = 〈v2〉 are the two corresponding points ofPG(V ), then the points of the line p1p2 are precisely the points of the form〈v1λ1+ v2λ2〉, where (λ1, λ2) ∈ (F×F)\{(0, 0)}. So, if W is a subspace of V ,then the set SW is a subspace of PG(V ) (regarded as a point-line geometry).

Conversely, suppose that S is a subspace of PG(V ), denote by W ′ the setof all nonzero vectors w ∈ V for which 〈w〉 ∈ S, and put W := W ′ ∪ {o}.Then for all w ∈ W and all λ ∈ F, we have wλ ∈ W . If w1 and w2 are twolinearly independent vectors of W , then also w1 + w2 ∈ W since 〈w1+ w2〉 isa point contained on the line connecting the points 〈w1〉 ∈ S and 〈w2〉 ∈ S.So, W is a subspace of V and S = SW . �

Solution of Problem 10. If pi = 〈wi〉, i ∈ {1, 2, . . . , k}, is a collec-tion of points of PG(V ), then by the previous exercise the smallest sub-space of PG(V ) containing p1, p2, . . . , pk coincides with SW , where W =〈w1, w2, . . . , wk〉. Now, SW is the whole point set of PG(V ) if and only ifW = V . So, the smallest size of a generating set of PG(V ) is equal todim(V ). The dimension of the projective space PG(V ) is one less than thegenerating index, i.e. equal to dim(V )− 1. �

Solution of Problem 11. Suppose V is a right vector space of dimensionat least 2 over a skew field F. If p1 = 〈v1〉 and p2 = 〈v2〉 are two distinctpoints of PG(V ), then the points of the line p1p2 are the points 〈v1〉 and〈v1λ + v2〉 where λ ∈ F. So, each line of PG(V ) contains |F|+ 1 ≥ 3 points,implying that PG(V ) is irreducible. �

Solution of Problem 12. We prove that R is an equivalence relation.Clearly, R is reflexive and symmetric. We should also prove that R is tran-sitive. So, let x1, x2 and x3 be three points for which (x1, x2) ∈ R and(x2, x3) ∈ R. We need to prove that (x1, x3) ∈ R. This would be trivial ifnot all of x1, x2, x3 would be distinct. So, we suppose that x1, x2 and x3 aremutually distinct. Consider the lines L1 = x1x2 and L2 = x2x3. If L1 = L2,then clearly (x1, x3) ∈ R and we are done. So, suppose L1 �= L2. Since(x1, x2) ∈ R, there exists a point u on L1 distinct from x1 and x2. Since(x2, x3) ∈ R, there exists a point v on L2 distinct from x2 and x3. As the

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lines uv and x1x3 meet in a point w, the line x1x3 contains at least threepoints and we have (x1, x3) ∈ R.

Denote by Xi, i ∈ I, the equivalence classes of R for some suitable indexset I. We show that each Xi is a subspace. Let a and b be two distinct pointsof Xi, denote by L the unique line containing these points and let c be anarbitrary point of L distinct from a and b. Since (a, b) ∈ R, the line ab = accontains at least three points, implying that (a, c) ∈ R, i.e. c ∈ Xi. So, Xi

is indeed a subspace.Suppose x1, x2 and x3 are three noncollinear points contained in the same

Xi. Since the lines x1x2 and x1x3 are thick, the projective plane ˜〈x1, x2, x3〉only contains thick lines and hence all its points are contained in Xi, provingthat Xi is an irreducible projective space. It is also clear that S is the directsum of the Xi’s. �

Solution of Problem 13. If L is a line of a projective plane of order n ≥ 2,then by removing L and all points incident with L, we obtain an affine planeof order n. There is a canonical way to reconstruct the projective plane fromthis affine plane (the points of L correspond to the parallel classes of lines ofthe affine plane). In view of this, its suffices to prove that there exist uniqueaffine planes of order 2 and 3 (up to isomorphism). This is certainly truefor affine planes of order 2 since any such plane is a complete graph on 4vertices.

We now show the uniqueness of the affine plane of order 3. SupposeA is an affine plane of order 3. The full subgeometry of A defined bytwo parallel classes of lines is a (3 × 3)-grid. So, we can label the pointsin such a way that {xij | 1 ≤ i, j ≤ 3} is the point set of A and that{x11, x12, x13}, {x21, x22, x23}, {x31, x32, x33}, {x11, x21, x31}, {x12, x22, x32},{x13, x23, x33} are lines. The remaining lines of A are now uniquely deter-mined: {x11, x22, x33}, {x12, x23, x31}, {x13, x21, x32}, {x31, x22, x13},{x21, x12, x33}, {x23, x32, x11}. �

Solution of Problem 14. Let S be an irreducible projective space ofdimension d ≥ 2 and let i ∈ {0, 1, . . . , d − 1}. Suppose α1 and α2 are twodistinct i-dimensional subspaces of S. If π1, π

′1 are two (i − 1)-dimensional

subspaces of S and π2, π′2 are two (i + 1)-dimensional subspaces of S such

that π1 ≤ α1, α2 ≤ π2 and π′1 ≤ α1, α2 ≤ π′

2, then as α1 �= α2, we necessarilyhave π1 = α1 ∩ α2 = π′

1 and π2 = 〈α1, α2〉 = π′2. So, there is at most one line

of Gr(S, i) containing α1 and α2. �

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Solution of Problem 15. Suppose S is an irreducible projective plane.Then the points of Gr(S, 1) are the lines of S and the lines of Gr(S, 1) areall collections of lines through a given point. So, we see that Gr(S, 1) isisomorphic to the dual plane of S.

Let V be a right vector space of dimension d+1 ≥ 3 over a skew field F, andsuppose S = PG(d,F) is the associated projective space. Let {e1, e2, . . . , ed+1}be a basis of V , and denote by V ′ the dual space of V . Then V ′ is a leftvector space over F with basis {f1, f2, . . . , fd+1} such that (a1f1+a2f2+ · · ·+ad+1fd+1)(e1b1 + e2b2 + · · · + ed+1bd+1) = a1b1 + a2b2 + · · · + ad+1bd+1. Wedenote by PGl(V

′) the corresponding projective space. Now, if we definev ∗ k := kv for all v ∈ V ′ and all k ∈ F, then V ′ can be regarded as a rightvector space over F

◦. If we denote by PG(d,F◦) the corresponding projec-tive space, then we have PGl(V

′) ∼= PG(d,F◦). So, it suffices to prove thatGr(S, d− 1) ∼= PGl(V

′).For every f ∈ V ′, the set of all vectors v ∈ V for which f(v) = 0 is

a hyperplane Wf of V . In this way, we obtain a bijective correspondencebetween the 1-spaces 〈f〉 of V ′ and the hyperplanes Wf of V , i.e. a bijectivecorrespondence between the points of PGl(V

′) and the hyperplanes of PG(V )(which are precisely the points of Gr(S, d − 1)).

For any two linear independent vectors f1 and f2 of V ′, the set of allvectors v ∈ V for which f1(v) = f2(v) = 0 is a subspace Wf1,f2 of co-dimension 2 of V . In this way, we obtain a bijective correspondence betweenthe subspaces 〈f1, f2〉 of dimension 2 of V ′ and the subspaces Wf1,f2 of co-dimension 2 of V , i.e. a bijective correspondence between the lines of PGl(V

′)and the co-dimension 2 subspaces of PG(V ). Note that there also exists anatural bijective correspondence between the co-dimension 2 subspaces ofPG(V ) and the lines of Gr(S, d− 1).

The just-defined bijective correspondences define an isomorphism be-tween PGl(V

′) and Gr(S, d− 1). �

Solution of Problem 16. Clearly, the parallelism relation is reflexive andsymmetric. We now also show that it is transitive.

Let L1, L2 and L3 be three lines such that L1 is parallel with L2 andL2 is parallel with L3. We need to show that L1 and L3 are parallel. Thisis obviously the case if two of the lines L1, L2, L3 are equal. So, we maysuppose that L1, L2, L3 are mutually distinct. If L1 and L3 were not parallelthen they would meet in a unique point p. As L1 and L2 are disjoint, we

318


have p �∈ L2. But then through p, there would go two lines L1 and L3 thatare parallel with L2, an obvious contradiction. �

Solution of Problem 17. By the definition of the lines of AG(n,F), thefollowing holds.

(A) If w and w′ are two distinct vectors of V , then the unique line ofAG(n,F) containing them consists of all vectors of the formwλ+ w′(1− λ) where λ ∈ F.

Since translations of AG(n,F) map lines to lines, we see that the followingmust hold.

(B) If S is a subspace of AG(n,F) (regarded as a point-line geometry) andw ∈ V , then the set {w + w′ | w′ ∈ S} is also a subspace of AG(n,F).

From (A) and (B), it follows that for every subspace W of V and everyw ∈ W , the set SW,w is a subspace of AG(n,F) (regarded as a point-linegeometry).

Suppose now that |F| ≥ 3 and that S is a subspace of AG(n,F) (regardedas a point-line geometry). Our intention is to show that S = SW,w for acertain subspace W of V . By (B), we may assume that o ∈ S. The subspaceW of V then necessarily consists of all vectors w contained in S. So, we mustshow the following:

(i) for all w ∈ S and all λ ∈ F, we have wλ ∈ S;

(ii) for all w1, w2 ∈ S, we have w1 + w2 ∈ S.

Obviously, (i) holds since wλ lies on the line through o and w. We now showthe validity of (ii). Since |F| ≥ 3, there exists a λ ∈ F such that λ+ 1 �= 0 �=λ+2. Suppose w1, w2 ∈ S. Then (w1 + w2(λ+1)) · (λ+ 2)−1 ∈ S and hencew1+w2(λ+1) ∈ S. It follows that (w1+w2(λ+1)+w1λ) ·(λ+1)−1 = w1+w2

then also belongs to S. �

Solution of Problem 18. If w1, w2, . . . , wk is a collection of vectors of V ,then by the previous exercise, the smallest subspace of AG(n,F) containingthese vectors coincides with

wk + 〈w1 − wk, w2 − wk, . . . , wk−1 − wk〉

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Now, wk+〈w1−wk, w2−wk, . . . , wk−1−wk〉 = V if and only if 〈w1−wk, w2−wk, . . . , wk−1 − wk〉 = V . This implies that the generating index of AG(n,F)must be equal to n + 1. (Besides the zero-vector, for instance, we need totake a basis of V to obtain a generating set of AG(n,F)). �

Solution of Problem 19. Let {Li | i ∈ I} and {Mj | j ∈ J} denote thetwo parallel classes of lines, where I and J are suitable index sets. We maysuppose that Li �= Li′ for all i, i′ ∈ I with i �= i′ and Mj �= Mj′ for allj, j′ ∈ J with j �= j′. Every line Li, i ∈ I, intersects every line Mj , j ∈ J , ina unique point, which we will denote by xij . Since every point is containedin a unique line of each parallel class, the point set of the net is equal to{xij | i ∈ I, j ∈ J}. It is now clear that the points and lines form a grid. �

Solution of Problem 20. We already know that affine planes are nets andlinear spaces. Conversely, suppose that S = (P,L, I) is a linear space that isalso a net.

Let C1 and C2 denote two equivalence classes of the parallelism relation.Let x be a point of S and let Li, i ∈ {1, 2}, the unique line of Ci through x.For every i ∈ {1, 2}, let yi ∈ Li \ {x}. Then x, y1, y2 are three noncollinearpoints.

For every anti-flag (x, L), there exists a unique line through x parallelwith L, i.e. a unique line through x not meeting L.

We conclude that S is an affine plane. �

Solution of Problem 21. Every affine plane of order n is a net of order(n − 1, n). Conversely, suppose that S = (P,L, I) is a finite net of order(s, t) with t = s+1 and let (x, L) be an anti-flag of S. Through x, there is aunique line disjoint from L and t = s+ 1 lines meeting L. So, x is collinearwith every point of L and S is a linear space. By the previous exercise, wethen know that S is an affine plane. �

Solution of Problem 22. If each equivalence class contains a unique point,then the fact that each line contains a unique point of each equivalence classimplies that the geometry must be a single line. So, we may assume thatthere is an equivalence class C1 of size 1 and an equivalence class C2 of sizea �= 1. If there was another equivalence class C3, then the lines through apoint x ∈ C3 are precisely the lines xy1 where y1 ∈ C1 and also the lines xy2where y2 ∈ C2. So, such a point y would be incident with precisely |C1| = 1lines and also with precisely |C2| = a �= 1 lines, an obvious contradiction. So,

320


C1 and C2 are the only equivalence classes and the geometry is isomorphicto K1,a. �

Solution of Problem 23. Define the following relation T on P: if x1, x2 ∈P, then (x1, x2) ∈ T if and only if x1 = x2 or (x1 �= x2 and x1x2 ∈ L′). ThenT is an equivalence relation. Two distinct points are equivalent if and only ifthey are noncollinear in S ′. Obviously, each line of S ′ contains a unique pointof each equivalence class. Since |L′| ≥ 2, there are at least two equivalenceclasses, and since |L| ≥ 2 for every L ∈ L′, each equivalence class containsat least two points. �

Solution of Problem 24. If S is a point-line geometry that arises from anaffine plane by removing one of its parallel classes of lines, then S is a netand a dual net (the latter by the previous exercise).

Conversely, suppose that S = (P,L, I) is a net that is also a dual net.Let T denote the transversality relation of S, and let C denote the set of allequivalence classes of the equivalence relation T . Then S ′ = (P,L ∪ C, I) isa linear space that satisfies all axioms of a net. Hence, S ′ is an affine plane.In this affine plane, C is a parallel class of lines. Hence, S is an affine planein which one parallel class of lines has been removed. �

Solution of Problem 25. Suppose S is a net and denote by t + 1 ≥ 2the number of equivalence classes of the parallelism relation. Since a pointx is incident with precisely one line of each equivalence class, every point xis incident with precisely t+1 lines. The second claim is just the dual of thefirst claim. �

Solution of Problem 26. Let A denote the set of planes of Σ meeting L ina singleton. For every α ∈ A, let xα denote the unique point in the singletonα ∩ L. For every point x ∈ L, let Ax �= ∅ denote the set of all planes α forwhich x = xα. We define an equivalence relation R on the line set A of Sby calling two elements α1, α2 ∈ A equivalent whenever xα1 = xα2 . Since|L| ≥ 2, there are at least two equivalence classes. If L is a point of S andAy with y ∈ L is an equivalence class of R, then 〈L, y〉 is the unique line ofS incident with L and belonging to Ay. So, every point of S is incident withprecisely one line of each equivalence class of R.

Let α1, α2 ∈ A with α1 �= α2. Then α1 ∩ α2 is a line. This line is disjointfrom L if and only if xα1 �= xα2 . Hence, every two distinct lines of S meet ifand only if they belong to distinct equivalence classes of the relation R.

We conclude that S is a net. �321


Solution of Problem 27. Suppose X is a set of 9 points in PG(2, 4)intersecting each line in either 1 or 3 points. Lines intersecting in 1 pointare called tangent lines, while lines intersecting in precisely three points arecalled secant lines. Each point x ∈ X is incident with precisely 9−1

2= 4

secant lines and a unique tangent line. For every point y �∈ X , we denoteby αy the number of tangent lines incident with y and by βy the number ofsecant lines incident with y. From αy + βy = 5 and αy + 3βy = 9, it thenfollows that αy = 3 and βy = 2.

Now, choose an arbitrary secant line L. We denote by x1, x2, x3 the threepoints of X incident with L, and by x4, x5 the two other points. The pointxi, i ∈ {4, 5}, is incident with a unique secant line Mi distinct from L. Wedenote by z the unique intersection point of the lines M4 and M5. If z ∈ X,then all five lines through z (namely M4, M5, zx1, zx2, zx3) would be secantlines, which is impossible. So, z �∈ X. Clearly,

X =((zx4 ∪ zx5) \ {z, x4, x5}

)∪ {x1, x2, x3}.

If we choose a reference system in PG(2, 4) such that

z = (1, 0, 0), x1 = (0, 1, 0), x2 = (0, 0, 1), x3 = (0, 1, 1),

thenX consists of all points (X0, X1, X2) of PG(2, 4) that satisfy the equationX3

0 +X1X22 +X2X

21 = 0. So, X is a Hermitian curve of PG(2, 4). �

Solution of Problem 28. Suppose D is a Steiner quadruple system of typeS(3, 4, 8).

We show that no two blocks B1 and B2 can intersect in a single pointx. If this were the case, then the derived design Dx would have two disjointlines, which is impossible as Dx is an S(2, 3, 7), i.e. a projective plane oforder 2.

We show that if B is a block, then the complement of B is also a block.If x1, x2 and x3 are three points outside B, then the unique block contain-ing {x1, x2, x3} cannot meet B by the previous paragraph, and hence mustcoincide with the complement of B.

Denote the point set of D by {x1, x2, . . . , x8} and suppose {x1, x2, x3, x4}and {x5, x6, x7, x8} are two blocks of D. Then we know that every otherblock intersects each of {x1, x2, x3, x4}, {x5, x6, x7, x8} in precisely two points.Without loss of generality, we may suppose that we have labeled the pointsin such a way that {x1, x2, x5, x6} and {x1, x3, x5, x7} are blocks. (Indeed,

322


once we know that {x1, x2, x5, x6} is a block, we also know that the uniqueblock containing {x1, x3, x5} cannot contain x6, since there is only one blockthrough {x1, x5, x6}.) The unique block containing {x1, x4, x5} cannot con-tain x6 nor x7, and so must coincide with {x1, x4, x5, x8}. The remainingblocks through {x1, x2}, {x1, x3} and {x1, x4} are now uniquely determined:{x1, x2, x7, x8}, {x1, x3, x6, x8}, {x1, x4, x6, x7}. The blocks not containingx1 are now also uniquely determined: they arise as complements of blockscontaining x1. So, we have found all blocks:

{x1, x2, x3, x4}, {x5, x6, x7, x8}, {x1, x2, x5, x6}, {x1, x2, x7, x8},

{x1, x3, x5, x7}, {x1, x3, x6, x8}, {x1, x4, x5, x8}, {x1, x4, x6, x7},{x3, x4, x7, x8}, {x3, x4, x5, x6}, {x2, x4, x6, x8}, {x2, x4, x5, x7},

{x2, x3, x6, x7}, {x2, x3, x5, x8}.It is straightforward to verify that the set {x1, x2, . . . , x8} together with these14 blocks form a Steiner system of type S(3, 4, 8). �

Solution of Problem 29. Suppose A is an affine plane of order 3 defined ona given set X of 9 points. Then we show that the line set of A is uniquely de-termined as soon as all lines through two distinct points are known. SupposeA has point set {x11, x12, . . . , x33} and the following lines:

• {x11, x12, x13}, {x21, x22, x23}, {x31, x32, x33},

• {x11, x21, x31}, {x12, x22, x32}, {x13, x23, x33},

• {x11, x22, x33}, {x12, x23, x31}, {x21, x32, x13},

• {x31, x22, x13}, {x21, x12, x33}, {x23, x32, x11}.

Suppose A′ is another affine plane on the set X for which {x11, x12, x13},{x11, x22, x33}, {x11, x21, x31}, {x11, x23, x32}, {x31, x22, x13}, {x31, x32, x33}and {x31, x12, x23} are lines. As {x11, x12, x13} and {x31, x32, x33} are two dis-joint lines of A′, also {x21, x22, x23} should be a line of A′. As {x31, x22, x13}and {x11, x23, x32} are two disjoint lines of A′, also {x21, x12, x33} is a line ofA′. The remaining line of A′ through x21 should then be {x21, x32, x13}. Westill need two additional lines of A′ that cover the set {x12, x22, x32, x13, x23,x33}. As each of the pairs {x12, x13}, {x12, x23}, {x12, x33} is already con-tained in a line, we know that {x12, x22, x32} and {x13, x23, x33} are the re-maining two lines.

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We show that if X is a set of four points in AG(2, 3) no three of whichare collinear, then the points of X can be labeled in such a way that X ={x1, x2, x3, x4}, x1x2‖x3x4 and x1x3‖x2x4. Indeed, if X = {a, b, c, d} withab‖cd, then one readily verifies that either ac‖bd or ad‖bc (but not both). IfX = {a, b, c, d} with ab ∩ cd �= ∅, then one readily verifies that ac‖bd andad‖bc.

With what we have derived above, we are now able to show that thereexists up to isomorphism at most one and hence precisely one inversive planeof type S(3, 4, 10), namely the classical inversive plane with these parameters.We label one of the vertices of S(3, 4, 10) by ∞ and the other points by1, 2, . . . , 9. Since the internal structure at the point ∞ is an affine plane oforder 3, we may without loss of generality suppose that the following setsare blocks:

• {∞, 1, 2, 3}, {∞, 4, 5, 6}, {∞, 7, 8, 9},

• {∞, 1, 4, 7}, {∞, 2, 5, 8}, {∞, 3, 6, 9},

• {∞, 1, 5, 9}, {∞, 2, 6, 7}, {∞, 3, 4, 8},

• {∞, 3, 5, 7}, {∞, 2, 4, 9}, {∞, 1, 6, 8}.

We need to determine the blocks not containing ∞. By the previous para-graph, we may without loss of generality suppose that {1, 2, 4, 5} is a block.We now show that the remaining blocks of S(3, 4, 10) are uniquely deter-mined. In view of the first paragraph and the fact that all derived designsare affine planes of order 3, it suffices to show that all 12 blocks through thepoint 1 are uniquely determined. At this stage, we already know five suchblocks:

{∞, 1, 2, 3}, {∞, 1, 4, 7}, {∞, 1, 5, 9}, {∞, 1, 6, 8}, {1, 2, 4, 5}.

Since {∞, 1, 2, 3}, {1, 2, 4, 5}, {∞, 1, 6, 8} and {∞, 2, 6, 7} are blocks, theunique block through {1, 2, 6} must contain the point 9. So, {1, 2, 6, 9} is ablock. The remaining block through {1, 2} is then the block {1, 2, 7, 8}. In asimilar way one shows that {1, 4, 8, 9} is the unique block containing {1, 4, 8}and that the remaining block through {1, 4} is equal to {1, 3, 4, 6}.

We already know three blocks through {1, 6}. The remaining block shouldbe {1, 5, 6, 7}. Similarly, we already know three blocks through {1, 8}. Theremaining block should be {1, 3, 5, 8}. We are still missing one block. Thisshould be the block {1, 3, 7, 9}. �

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Solution of Problem 30. If S(5, 6, 12) would be extendable, then theextension would be a Steiner system of type S(6, 7, 13). The total numberof blocks in such a Steiner system would be equal to

(136

)/(76

), which is not

integral since 7 is not a divisor of(136

).

If S(5, 8, 24) would be extendable, then the extension would be a Steinersystem of type S(6, 9, 25). The total number of blocks in an S(6, 9, 25) wouldbe equal to

(256

)/(96

), which is not integral. �

Solution of Problem 31. Suppose S is a Steiner triple system on v points.Then the total number of blocks of S through a given point is equal to v−1

2

and so v is odd. The total number of blocks of S is equal to v(v−1)6

and so vcannot be congruent to 5 modulo 6. It follows that v is congruent to 1 or 3modulo 6.

Suppose S is a Steiner quadruple system on v points. Since the deriveddesign of a Steiner quadruple system is a Steiner triple system, v − 1 mustbe congruent to 1 or 3 modulo 6, i.e. v is congruent to 2 or 4 modulo 6. �

Solution of Problem 32. The point-line geometry S has v = qn+1−1q−1

points

and k = qd+1−1q−1

points on each line. The total number of lines through two

distinct points is equal to the total number of (d− 2)-dimensional subspaces

of a PG(n− 2, q), i.e. equal to λ =

[n− 1d− 1

]

q

= (qn−1−1)(qn−1−q)···(qn−1−qd−2)(qd−1−1)(qd−1−q)···(qd−1−qd−2)

.

�

Solution of Problem 33. The point-line geometry S has v = qn pointsand every line is incident with precisely k = qd points. The total num-ber of lines through two distinct points is equal to the number of (d − 2)-

dimensional subspaces of a PG(n − 2, q), i.e. equal to λ =

[n− 1d− 1

]

q

=

(qn−1−1)(qn−1−q)···(qn−1−qd−2)(qd−1−1)(qd−1−q)···(qd−1−qd−2)

. �

Solution of Problem 34. The point-line geometry S has v = 2n pointsand every line is incident with precisely k = 2d points. The total numberof lines through three distinct points is equal to the number of (d − 3)-

dimensional subspaces of a PG(n − 3, 2), i.e. equal to λ =

[n− 2d− 2

]

2

=

(2n−2−1)(2n−2−2)···(2n−1−2d−3)(2d−2−1)(2d−2−2)···(2d−2−2d−3)

. �

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Solution of Problem 35. The total number of lines intersecting H inprecisely 2 points is equal to

(n+22

)= (n+2)(n+1)

2and hence the total number

of lines disjoint from H is equal to n2 + n+1− 12(n+ 2)(n+1) = 1

2(n2 − n).

Through a point outside H, there are n+22

lines intersecting H in precisely 2points and n + 1 − n+2

2= n

2lines disjoint from H. Every two lines disjoint

from H meet in a point outside H. We conclude that S is a Steiner systemS(2, n

2, n2−n

2). �

Solution of Problem 36. Let L1 and L2 be two disjoint lines of S. Let x1

and x′1 be two distinct points of L1 and let x2 denote the unique point of L2

collinear with x1. Then x′1 and x2 are two noncollinear points of S, i.e. two

lines of SD that have no points in common.Let (L, x) be an anti-flag of SD. Then (x, L) is a flag of S and so there

exists a unique point x′ on L collinear with x, i.e. there exists a unique lineM through x meeting L (namely M = xx′). Any point of SD incident withx is a line of S through x. Such a point of SD is collinear in SD with L ifand only if the corresponding line of S meets L. It follows that there is aunique such point of SD, namely the line M of S. �

Solution of Problem 37. Suppose S is a generalized quadrangle of order2. We prove that every two disjoint lines L1 and L2 are contained in a unique(3×3)-subgrid. Let K1, K2 and K3 denote the three lines meeting L1 and L2

(through every point of L1, there is precisely one such line). Since S has notriangles, these three lines are mutually disjoint. Let xi, i ∈ {1, 2, 3}, denotethe unique point of Ki which is incident with neither of L1, L2. The point x1

is collinear with a unique point of K2 which necessarily equals x2 since S hasno triangles. Similarly, x1 ∼ x3 and x2 ∼ x3. Since x1, x2 and x3 are threemutually collinear points, they form a line which we denote by L3. Clearly,the lines L1, L2, L3, K1, K2, K3 are the six lines of a (3× 3)-subgrid G.

Now, let G denote an arbitrary (3×3)-grid of S. For every point x of S not

contained in G, x⊥∩G is an ovoid of G, i.e. a set of points of G meeting eachline of G in a singleton. Conversely, if {y1, y2, y3} is an ovoid of G, then thereexists a unique point x of S not contained in G for which x⊥∩G = {y1, y2, y3}.Indeed, such a point necessarily lies on the unique line Ly1 through y1 notcontained in G and hence must coincide with the unique point x′ of Ly1

collinear with y2. Since (x′)⊥ ∩G is an ovoid of G containing y1 and y2, thisovoid must be equal to {y1, y2, y3}. So, there exists a bijective correspondencebetween the points of S not contained in G and the ovoids of G.

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The generalized quadrangle S has 15 lines. Six of these lines are containedin G and the remaining nine lines meet G in a unique point. Indeed, througheach point of G there is a unique line which is not contained in G. Supposethat {x, y1, y2} is one of the nine lines of S intersecting G in some point x.Since S has no triangles, the two ovoids y⊥1 ∩G and y⊥2 ∩G of G intersect inthe singleton {x}.

From the above discussion, one now sees that S is isomorphic to thepoint-line geometry S ′ which is defined as follows.The points of S ′ are of two types:

• the points of a given (3× 3)-grid G∗;

• the ovoids of G∗.

The lines of S ′ are of two types:

• the lines of G∗;

• all the sets {x,O1, O2} where x is a point of G∗ and O1, O2 are the twoovoids of G∗ through x.

Incidence is containment. Since S ∼= S ′, there exists up to isomorphism aunique generalized quadrangle of order 2. This generalized quadrangle isisomorphic to W (2) and to Q(4, 2). �

Solution of Problem 38. Every line of S is incident with precisely s+1 :=q ≥ 2 points and every point of S is incident with precisely t + 1 := |H| =q + 2 ≥ 4 lines.

Suppose (x, L) is an anti-flag of S and let z denote the unique point ofH contained in L. The plane 〈x, L〉 intersects PG(2, q) in a line M and wedenote by z′ the unique point of M ∩ H distinct from z. If y is a point ofL \ {z}, then y is collinear with x in S if and only if xy ∩PG(2, q) is a pointof H. Since xy ∩PG(2, q) ⊆ 〈x, L〉 ∩PG(2, q) = M and y �= z, we know thatx is collinear with y if and only if xy ∩ PG(2, q) = {z′}. So, the point inxz′∩L is the unique point of L collinear with x in S. It follows that throughthe point x there is a unique line meeting L and t ≥ 1 lines disjoint from L.

We conclude that S is a generalized quadrangle of order (q − 1, q + 1). �

Solution of Problem 39. Let P denote the set of points of W (3, q) non-collinear with x. Every L ∈ L1 contains a unique point collinear with x in

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W (3, q) and hence precisely q points of P. Every L ∈ L2 contains q pointsdistinct from x and all these points belong to P. We conclude that everyline of S is incident with precisely s+ 1 := q points.

Every point of S is incident with q+1 totally isotropic lines ofW (3, q), i.e.with q + 1 lines of S belonging to L1, and with one hyperbolic line throughx, i.e. with one line of S belonging to L2. We conclude that every point ofS is incident with precisely t+ 1 := q + 2 lines of S.

Now, denote by ζ the symplectic polarity of PG(3, q) defining W (3, q).Note that the hyperbolic lines of W (3, q) through x define a partition of P.Let (y, L) denote an anti-flag of S.

Suppose L is a hyperbolic line through x. The unique hyperbolic linethrough x and y has besides x no points in common with L. So, every lineof S through y meeting L must belong to L1. The plane yζ intersects L ina unique point z �= x and yz ∈ L1. The point z is the unique point of Lcollinear with x in the geometry S.

Suppose L ∈ L1 such that x belongs to the plane 〈y, L〉. Let x′ denotethe unique point of L collinear with x. The totally isotropic lines of W (3, q)contained in 〈y, L〉 are precisely the lines of 〈y, L〉 through x′. So, yx′ is theunique totally isotropic line through y meeting L and xx′ is also a totallyisotropic line. So x′ �∈ P and no line of L1 through y has a point in commonwith L in the point-line geometry S. The unique hyperbolic line through xand y meets L in a point y′ and this point is therefore the unique point of Lcollinear with y.

Suppose L ∈ L1 such that x does not belong to the plane 〈y, L〉. Then theunique line of L2 through y has no point in common with L in the geometryS. In the plane yζ the only totally isotropic lines of W (3, q) are the lines of yζ

through y. So, L is not contained in yζ and yζ ∩ L is a singleton {y′}. If y′xwould be a totally isotropic line ofW (3, q), then the plane (y′)ζ would containthe lines y′x, L and y′y, which is impossible since x �∈ 〈L, y〉. So, y′ ∈ P andy′ is the unique point of L that is collinear with y in the geometry S.

So, for every anti-flag (y, L) there exists a unique line through y meetingL and t ≥ 1 lines disjoint from L.

We conclude that S is a generalized quadrangle of order (q − 1, q + 1). �

Solution of Problem 40. For every point x of O, let πx denote the uniquehyperplane of PG(d, q) through x such that a line L of PG(d, q) through xis contained in πx if and only if L ∩ O = {x}. Through each point x ∈ O,

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there are qd−1 lines of PG(d, q) not contained in πx and each of these linescontains besides x one extra point of O. We conclude that |O| = qd−1 + 1.

From the description of the incidences, we can draw the following conclu-sions:

(1) Two distinct points x and y of Type (i) are collinear if and only if theunique line through them contains a point of O. If this is the case, thenxy is the unique line of Td(O) incident with these points.

(2) A point x of Type (i) is collinear with a point π of Type (ii) if and onlyif x ∈ π. If this is the case, then there is a unique line of Td(O) incidentwith these points, namely the line xo where o is the unique point of Ocontained in π.

(3) A point x of Type (i) is never collinear with the point (∞) of Type (iii).

(4) Two distinct points π1 and π2 of Type (ii) are collinear if and only ifthe singletons {o1} = π1 ∩O and {o2} = π2 ∩O coincide. If this is thecase, then the line o1 = o2 is the unique line of Td(O) incident withthese points.

(5) A point π of Type (ii) is always collinear with the point (∞) of Type(iii). If o denotes the unique point of O inside π, then o is the uniqueline of Td(O) incident with these points.

So, Td(O) is a partial linear space.Every point x of Type (i) is incident with 0 lines of Type (b) and with

|O| = qd−1+1 lines of Type (a), namely the |O| lines through x containing apoint of O. If π is a hyperplane of PG(d+1, q) containing a unique point x ofO, then there are qd−1 lines of π through x not contained in πx, showing thatthe point π of Type (ii) is incident with a unique line of Type (b) (namely x)and qd−1 lines of Type (a). The point (∞) of Type (iii) is incident with noline of Type (a) and with all |O| = qd−1 + 1 lines of Type (b). We concludethat every point of Td(O) is incident with precisely t + 1 := qd−1 + 1 lines.

Suppose L is a line of PG(d + 1, q) containing a point x ∈ O and notcontained in PG(d, q). Then L is incident in Td(O) with q points of Type(i) and a unique point of Type (ii), namely the hyperplane 〈L, πx〉. Supposeu is a point of O. Then through πu, there are q hyperplanes distinct fromPG(d, q), showing that the line u of Type (b) is incident with precisely q+1points, namely 0 points of Type (i), q points of Type (ii) and the point (∞)

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of Type (iii). We conclude that every line of Td(O) is incident with preciselys+ 1 := q + 1 points.

We now consider seven possibilities for an anti-flag.

• Suppose x is a point of Type (i) and L is a line of Type (a) suchthat 〈x, L〉 ∩PG(d, q)∩O is a singleton. Put L∩O = {u}. By (1)–(5)above, 〈L, πu〉 is the unique point of Td(O) incident with L and collinearwith x.

• Suppose x is a point of Type (i) and L is a line of Type (a) such that|〈x, L〉 ∩ PG(d, q) ∩ O| = 2. Put 〈x, L〉 ∩ PG(d, q) ∩ O = {u, v}, whereL∩O = {u}. By (1)–(5) above, the unique point in xv∩L is the uniquepoint of Td(O) incident with L and collinear with x.

• Suppose x is a point of Type (i) and y ∈ O is a line of Type (b). By(1)–(5) above, 〈x, πy〉 is the unique point of Td(O) incident with y andcollinear with x.

• Suppose π is a hyperplane of PG(d+ 1, q) containing a unique point xof O, and L is a line through x not contained in PG(d, q) ∪ π. By (1)–(5) above, 〈L, πx〉 is the unique point of Td(O) incident with L andcollinear with π.

• Suppose π is a hyperplane of PG(d + 1, q) containing a unique pointx of O, and L is a line of PG(d + 1, q) not contained in PG(d, q) thatcontains a point y �= x of O. By (1)–(5) above, the unique point inthe intersection π∩L is the unique point of Td(O) incident with L andcollinear with π.

• Suppose π is a hyperplane of PG(d+ 1, q) containing a unique point xof O, and y ∈ O \ {x} is a line of Type (b). By (1)–(5) above, (∞) isthe unique point of Td(O) incident with y and collinear with π.

• Let x be the point (∞) and L a line of PG(d + 1, q) not contained inPG(d, q) that contains a point y of O. By (1)–(5) above, 〈L, πy〉 is theunique point of Td(O) incident with L and collinear with x = (∞).

We conclude that for every anti-flag (x, L), there exists a unique line throughx meeting L and t = qd−1 ≥ 1 lines through x disjoint from L.

We conclude that S is a generalized quadrangle of order (q, qd−1). �

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Solution of Problem 41. For every plane π through x not containing y,the intersection π ∩ PG(2, 2h) is a line containing two points x and zπ ofH. Similarly, for every plane π through y not containing x, the intersectionπ ∩ PG(2, 2h) is a line containing two points y and zπ of H.

From the description of the incidences, we can draw the following conclu-sions:

(1) Two distinct points u and v of Type (i) are collinear if and only if theunique line through them contains a point of H \ {x, y}. If this is thecase, then uv is the unique line of S(H, {x, y}) incident with u and v.

(2) A point u of Type (i) and a point π of Type (ii) are collinear if and onlyif u ∈ π. If this is the case, then uzπ is the unique line of S(H, {x, y})incident with these points.

(3) A point u of Type (i) and a point π of Type (iii) are collinear if and onlyif u ∈ π. If this is the case, then uzπ is the unique line of S(H, {x, y})incident with these points.

(4) Two distinct points of Type (ii) are never collinear.

(5) Two distinct points of Type (iii) are never collinear.

(6) A point π1 of Type (ii) is collinear with a point of Type (iii) if andonly if zπ1 = zπ2. If this is the case, then π1 ∩ π2 is the unique line ofS(H, {x, y}) incident with these points.

We conclude that S(H, {x, y}) is a partial linear space.Let x be point of Type (i). Then x is incident with |H \ {x, y}| = 2h

lines of S(H, {x, y}), namely the 2h lines of PG(3, 2h) through x containinga point of H\{x, y}. Let π be a point of Type (ii) or (iii). Then there are 2h

lines of π through zπ distinct from π ∩ PG(2, 2h), showing that π is incidentwith 2h lines of S(H, {x, y}). We conclude that every point of S(H, {x, y})is incident with precisely t+ 1 := 2h lines.

Let L be a line of S(H, {x, y}). Then L is incident with 2h points ofType (i) (namely the 2h points in L \ PG(2, 2h)), a unique point of Type(ii) (namely 〈L, x〉) and a unique point of Type (iii) (namely 〈L, y〉). So,every line of S(H, {x, y}) is incident with precisely s+ 1 := 2h + 2 points ofS(H, {x, y}).

We now consider four possibilities for an anti-flag of S(H, {x, y}).

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• Suppose u is a point of Type (i) and L is a line of S(H, {x, y}) suchthat u �∈ L and the plane 〈u, L〉 does not contain x nor y. The plane〈u, L〉 intersects PG(2, 2h) in a line containing two points v1 and v2 ofH. One of these points, say v1, is contained in L. By (1)–(6) above,the unique point in uv2∩L is the unique point of S(H, {x, y}) incidentwith L and collinear with u.

• Suppose u is a point of Type (i) and L is a line of S(H, {x, y}) suchthat u �∈ L and the plane 〈u, L〉 contains one of x, y. By (1)–(6) above,the plane 〈u, L〉 is the unique point of S(H, {x, y}) incident with L andcollinear with u.

• Suppose π is a point of Type (ii) or (iii) and L is a line of S(H, {x, y})through the point zπ but not contained in π. Let π′ be the uniquepoint of Type (ii) or (iii) which is incident with L and whose type isdistinct from the type of π. By (1)–(6) above, π′ is the unique point ofS(H, {x, y}) incident with L and collinear with π.

• Suppose π is a point of Type (ii) or (iii) and L is a line of S(H, {x, y})not containing the point zπ. By (1)–(6) above, the unique point in theintersection L ∩ π is the unique point of S(H, {x, y}) incident with Land collinear with π.

We conclude that for every anti-flag (x, L), there exists a unique line throughx meeting L and t = 2h − 1 ≥ 1 lines through x disjoint from L.

We conclude that S is a generalized quadrangle of order (2h + 1, 2h − 1).�

Solution of Problem 42. Let Π = (P,Σ) be a Veldkamp-Tits polar spaceof rank n ≥ 1 and let i ∈ {0, 1, . . . , n−2}. Suppose α1 and α2 are two distincti-dimensional singular subspaces of Π. Suppose also that π1, π

′1 are two (i−1)-

dimensional singular subspaces of Π, and π2, π′2 are two (i + 1)-dimensional

singular subspaces of Π such that π1 ≤ α1, α2 ⊆ π2 and π′1 ≤ α1, α2 ⊆ π′

2. By(VT2), α1 ∩ α2 is a singular subspace of dimension at most i − 1, implyingthat π1 = α1 ∩ α2 = π′

1. If π2 �= π′2, then by (VT2), π2 ∩ π′

2 is a singularsubspace of dimension at most i, implying that α1 = π2∩π′

2 = α2. As this isimpossible, we should have π2 = π′

2. So, there is at most one line of Gr(Π, i)containing α1 and α2. �

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Solution of Problem 43. We show that every line L of S is incident withprecisely three points. Without loss of generality, we may suppose that L ={{1, 2}, {3, 4}, {5, 6, 7, 8}}. Then {{1, 2}, {3, 4}, {5, 6}, {7, 8}}, {{1, 2}, {3, 4},{5, 7}, {6, 8}} and {{1, 2}, {3, 4}, {5, 8}, {6, 7}} are the points of S incidentwith L.

Let x1 = {P1, Q1, R1, S1} and x2 = {P2, Q2, R2, S2} be two points of S,and let Γ be the graph on the vertex set {1, 2, . . . , 8} having {P1, Q1, R1, S1}∪{P2, Q2, R2, S2} as edge set. The vertices x1 and x2 are at distance 1 if andonly if |x1 ∩ x2| = 2. We distinguish the following cases:

(1) If x1 = x2, then d(x1, x2) = 0.

(2) If |x1 ∩ x2| = 2, then d(x1, x2) = 1.

(3) If |x1 ∩ x2| = 1, then d(x1, x2) = 2 and |Γ1(x1) ∩ Γ1(x2)| = 3. In-deed, in this case we may without loss of generality suppose that x1 ={{1, 2}, {3, 4}, {5, 6}, {7, 8}} and that x2={{1, 2}, {3, 5}, {6, 7}, {4, 8}}.Then the three points x3 = {{1, 2}, {3, 4}, {6, 7}, {5, 8}}, x′

3 = {{1, 2},{5, 6}, {4, 8}, {3, 7}} and x′′

3 = {{1, 2}, {7, 8}, {3, 5}, {4, 6}} are all thecommon neighbours of x1 and x2.

(4) If x1 ∩ x2 = ∅ and Γ is the union of two cycles of length 4, thend(x1, x2) = 2 and |Γ1(x1) ∩ Γ1(x2)| = 2. Indeed, in this case, wemay without loss of generality suppose that x1 = {{1, 2}, {3, 4}, {5, 6},{7, 8}} and x2 = {{1, 3}, {2, 4}, {5, 7}, {6, 8}}. Then x3 = {{1, 2},{3, 4}, {5, 7}, {6, 8}} and x′

3 = {{1, 3}, {2, 4}, {5, 6}, {7, 8}} are the twocommon neighbours of x1 and x2.

(5) If x1 ∩ x2 = ∅ and Γ is a cycle of length 8, then d(x1, x2) = 3. Indeed,without loss of generality we may suppose that x1 = {{1, 2}, {3, 4},{5, 6}, {7, 8}} and x2 = {{2, 3}, {4, 5}, {6, 7}, {8, 1}}. Then one cancheck that x1 and x2 have no common neighbours. Since x3 = {{1, 4},{2, 3}, {5, 6}, {7, 8}} is a neighbour of x1 at distance 2 from x2, wemust have d(x1, x2) = 3.

Now, take an arbitrary point-line pair (p, L). Without loss of generality wemay suppose that one of the following five cases occurs:

(1) p = {{1, 2}, {3, 4}, {5, 6}, {7, 8}} and L = {{1, 2}, {3, 4}, {5, 6, 7, 8}}.In this case p ∈ L.

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(2) p = {{1, 2}, {3, 4}, {5, 6}, {7, 8}} and L = {{1, 2}, {3, 5}, {4, 6, 7, 8}}.In this case, p′ = {{1, 2}, {3, 5}, {4, 6}, {7, 8}} is the unique point of Lcollinear with p.

(3) p = {{1, 2}, {3, 4}, {5, 6}, {7, 8}} and L = {{1, 3}, {2, 4}, {5, 6, 7, 8}}.In this case p′ = {{1, 3}, {2, 4}, {5, 6}, {7, 8}} is the unique point of Lcollinear with p.

(4) p = {{1, 2}, {3, 4}, {5, 6}, {7, 8}} and L = {{1, 3}, {4, 6}, {2, 5, 7, 8}}.Then no point of L is collinear with p. Using the above observations re-garding the distances between points, we see that {{1, 3}, {4, 6}, {2, 5},{7, 8}} is the unique point of L at distance 2 from x.

(5) p = {{1, 2}, {3, 4}, {5, 6}, {7, 8}} and L = {{1, 3}, {6, 8}, {2, 4, 5, 7}}.Then no point of L is collinear with p. Using the above observations re-garding the distances between points, we see that {{1, 3}, {2, 4}, {6, 8},{5, 7}} is the unique point of L at distance 2 from p.

So, we see that S is a near hexagon and that every line of S is incident withprecisely three points. �

Solution of Problem 44. If S is a linear space distinct from a line, thenthe diameter of S is equal to 1 and there exists an anti-flag, showing that Smust be a near 3-gon. Conversely, suppose that S is a near 3-gon. Then thediameter of S is equal to 1 and so S is a linear space. Anti-flags should existand so S cannot be a line. �

Solution of Problem 45. Observe that Γ is not an empty nor a completegraph (as t ≥ 1 and α �= s+ 1).

We determine the total number v of points of S. Let L be a line of S.Every point of L is collinear with st points outside L and every point outsideL is collinear with α points of L. So, the total number of points of S is equalto s+ 1 + (s+1)st

α= (s+1)(st+α)

α.

Clearly, Γ has valency k = s(t + 1). Indeed, on each of the t + 1 linesthrough a fixed point x, there are s points distinct from x.

Let x and y be two distinct points incident with some line L. We deter-mine the number λ of points at distance 1 from x and y. There are s − 1such points on the line L. The number of such points not on L is equal tot(α − 1). Indeed, each such point lies on a line through x distinct from xy.Now, on each of the t lines through x distinct from L, there are α points

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collinear with y, but one of these α points coincides with x. So, we haveλ = (s− 1) + t(α− 1).

Let x and y be two distinct noncollinear points of S. Then x and y haveprecisely (t + 1)α common neighbours, namely α on each of the t + 1 linesthrough x.

We conclude that Γ is a strongly regular graph with parameters v =(s+1)(st+α)

α, k = s(t+ 1), λ = (s− 1) + t(α− 1) and μ = (t+ 1)α. �

Solution of Problem 46. If (x, L) is an anti-flag of S, then x is collinearwith α = 1 points of L. So, in order to show that S is a generalized quad-rangle, we still must show that there exist two disjoint lines. Let L1 be a lineof S, x a point incident with L1, K a line through x distinct from L1 and ya point of K \ {x}. The line K is the unique line through y meeting L1. So,if L2 is any of the t ≥ 1 lines through y distinct from K, then L1 and L2 aredisjoint. �

Solution of Problem 47. If S is a linear space of order (s, t), then forevery anti-flag (x, L), we know that the point x is collinear with s+1 pointsof L. So, S is then a partial geometry with parameters (s, t, s+ 1).

Conversely, suppose that S is a partial geometry with parameters (s, t, s+1) where s, t ≥ 1. If x and y are two distinct points of S and L is any linethrough y not containing x, then x is collinear with s + 1 points of L, i.e.with all points of L, in particular also with y. So, S is then a finite linearspace of order (s, t). �

Solution of Problem 48. Suppose S = (P,L, I) is a dual net of order(s, t). The following relation T on P is then an equivalence relation: ifx, y ∈ P, then (x, y) ∈ T if and only if x = y or (x and y noncollinear). If(x, L) is an anti-flag of S, then L contains a unique point equivalent with x,i.e. noncollinear with x and hence S is a partial geometry with parameters(s, t, α) = (s, t, s).

Conversely, suppose that S = (P,L, I) is a partial geometry with param-eters (s, t, α) = (s, t, s). Define a relation T on P as follows: if x, y ∈ P,then (x, y) ∈ T if and only if x = y or (x and y are noncollinear). We showthat T is an equivalence relation.

Obviously, T is reflexive and symmetric. We now show that T is alsotransitive. Suppose (x, y) ∈ T and (y, z) ∈ T . We need to show that (x, z) ∈T . We may suppose that x, y and z are mutually distinct. If (x, z) �∈ T ,

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then x, z are collinear and the line xz contains two points noncollinear with y(namely x and z), which is contrary to the fact that α = s. Hence, (x, z) ∈ Tand T is an equivalence relation.

We show that each line L contains a unique point of each equivalenceclass C. Let x be an arbitrary point of C. If x ∈ L, then x is the uniquepoint of L belonging to C. If x �∈ L, then since α = s, L contains a uniquepoint noncollinear with x and this is the unique point of L belonging to C.

As any line contains at least two points, the total number of equivalenceclasses is at least 2. We now show that each equivalence class C contains atleast two points. Let x be a point not belonging to C. Through x, there aretwo lines L1 and L2. As each of these two lines contains a unique point ofC, we have |C| ≥ 2. �

Solution of Problem 49. Suppose S is a net of order (s, t). Let (x, L) bean anti-flag of S. Through x, there is a unique line parallel with L and tlines nonparallel with L. Each of these t lines nonparallel with L meets L ina unique point, showing that there are precisely t points on L collinear withx. Hence, S is a partial geometry with parameters (s, t, t).

Conversely, suppose that S = (P,L, I) is a partial geometry with param-eters (s, t, t).

If (x, L) is an anti-flag, then through x, there are t lines meeting L andhence a unique line disjoint from L.

Define the following relation R on L: if K,L ∈ L, then (K,L) ∈ R if andonly if K = L or (K and L are disjoint). We show that R is an equivalencerelation. Obviously, R is reflexive and symmetric. We now show that R isalso transitive. We need to show that if K, L and M are three lines suchthat (K,L) ∈ R and (L,M) ∈ R, then also (K,M) ∈ R. This clearly holdsif two of K, L, M are equal. So, we may suppose that K, L and M aremutually distinct. If (K,M) �∈ R, then K and M meet in a unique point y.As K �= L and (K,L) ∈ R, we have y �∈ L. But that is impossible as thiswould imply that K and M are two distinct lines through y disjoint from L.

Clearly, two distinct lines meet if and only if they belong to distinctequivalence classes. Since for every anti-flag (x, L), there exists a unique linethrough x disjoint from L, we also see that every point is incident with pre-cisely one line of each equivalence class. So, the total number of equivalenceclasses is equal to t + 1 ≥ 2. �

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Solution of Problem 50. Clearly, Γ is not empty. Since t ≥ 1, there existsan anti-flag (x, L) and L contains a point noncollinear with x, showing thatΓ is also not complete.

We count the total number v of points of S. Let x be an arbitrary pointof S. The total number of points at distance 1 from x is equal to s(t + 1),since there are t + 1 lines through x, and each of these contains besides xprecisely s points. So, Γ is regular with valency k = s(t+ 1). Through eachpoint y ∈ Γ1(x), there are t lines distinct from xy and each of these containsa unique point collinear with x (namely y) and s points at distance 2 from x.Conversely, if z ∈ Γ2(x), then x and z have precisely μ common neighbours.

We conclude that |Γ2(x)| = |Γ1(x)|·stμ

= s2t(t+1)μ

and v = 1+ s(t+ 1) + s2t(t+1)μ

.Let x and y be two distinct points incident with some line L. No point

outside L is collinear with x and y. So, the vertices in Γ adjacent to x andy are precisely the points of L \ {x, y}. There are s− 1 such vertices.

We conclude that Γ is a strongly regular graph with parameters v =

1 + s(t+ 1) + s2t(t+1)μ

, k = s(t + 1), λ = s− 1 and μ. �

Solution of Problem 51. Suppose S is a partial quadrangle. Then S hasdiameter 2 and for every point-line pair (x, L) with d(x, L) ≤ 1, there existsa unique point on L nearest to x. So, if there exists a point-line pair (x, L)with d(x, L) = 2, then S is a near 5-gon. Suppose now that there does notexist a point-line pair (x, L) with d(x, L) = 2. Then we show that there aretwo disjoint lines, implying that S is a generalized quadrangle.

Let x be a point of S, L1 and L2 two lines through x, y a point of L2\{x}and L3 a line through y distinct from L2. Since S is a partial quadrangle,the lines L1 and L3 are disjoint. �

Solution of Problem 52. Suppose Γ is a strongly regular graph whoseparameters (v, k, λ, μ) satisfy λ = 0, k = t + 1 and μ > 0. Then Γ canbe regarded as a partial linear space of order (1, t). Since λ = 0, S has notriangles, i.e. for every anti-flag (x, L), there exists at most one point on Lcollinear with x. If x and y are two points at distance 2 from each other, then|Γ1(x)∩Γ1(y)| = μ. It follows that S is a partial quadrangle with parameters(s, t, μ) = (1, t, μ).

Conversely, suppose that S is a partial quadrangle with parameters(s, t, μ) = (1, t, μ). Then S can be regarded as a graph with valency t+1 ≥ 2.If x and y are two adjacent vertices of Γ and z ∈ Γ1(x)∩Γ1(y), then z wouldbe collinear with two points of the line {x, y}, which is impossible. So,

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every two adjacent vertices of Γ have λ = 0 common neighbours. Every twovertices at distance 2 from each other have precisely μ common neighbours.The graph Γ is neither complete nor empty. So, Γ is a strongly regular graphwhose parameters (v, k, λ, μ) satisfy λ = 0, k = t+ 1 and μ > 0. �

Solution of Problem 53. Observe that Γ is not an empty graph, nor acomplete graph since t ≥ 1 and α �= s+1. In fact, since μ > 0, the diameterof Γ is equal to 2.

We determine the total number v of points of S. Let x be a fixed point ofS. Then |Γ0(x)| = 1 and |Γ1(x)| = s(t + 1). Put M := |Γ2(x)|. We count intwo different ways the number N of ordered pairs (y, z) of vertices for whichz ∈ Γ2(x) and y ∈ Γ1(x)∩Γ1(z). On the one hand, we know that there areMchoices for z, and for fixed z there are μ choices for y, showing thatN = M ·μ.On the other hand, there are s(t+1) choices for y, and for given y, there aret(s+1−α) choices for z. Indeed, on each of the t lines through y distinct fromxy, there are precisely α points collinear with x and hence precisely s+1−αpoints at distance 2 from x. So, we also have N = s(t + 1)t(s + 1 − α).

Together with N = μ · M , this implies that |Γ2(x)| = M = s(t+1)t(s+1−α)μ

.

Hence, v = |Γ0(x)|+ |Γ1(x)|+ |Γ2(x)| = 1 + (t+1)s(μ+t(s−α+1))μ

.

Clearly, Γ has valency k = s(t + 1). Indeed, on each of the t + 1 linesthrough a fixed point x, there are s points distinct from x.

Let x and y be two distinct collinear points, incident with some line L.We determine the number λ of points at distance 1 from x and y. Thereare s− 1 such points on the line L. The number of such points not on L isequal to t(α − 1). Indeed, each such point lies on a line through x distinctfrom xy. On each of the t lines through x distinct from L, there are α pointscollinear with y, but one of these α points coincides with x. So, we haveλ = (s− 1) + t(α− 1).

From the above discussion, it is now clear that Γ is a strongly regulargraph. The μ-parameter of this strongly regular graph coincides with theμ-parameter of the semipartial geometry. �

Solution of Problem 54. Suppose S is a partial geometry with parameters(s, t, α). In order to show that S is a semi-partial geometry with parameters(s, t, α, (t + 1)α), we must show that any two noncollinear points x and yhave precisely (t+1)α common neighbours. This is indeed the case. On eachof the t + 1 lines through y, there are α points collinear with x.

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Suppose S is a semipartial geometry with parameters (s, t, α, μ) that isnot a linear space. Take then two noncollinear points x and y. If L is aline through y, then L contains at most s points collinear with x, showingthat α �= s + 1. Now, each of the t + 1 lines through y contains either 0or α points collinear with x, and so the number μ of points collinear with xand y is at most (t + 1)α, with equality if and only if each line through ycontains precisely α points collinear with x. Now, if μ = (t+1)α, then sinceα �= s + 1, there exists for every anti-flag (x, L) a point y on L noncollinearwith x. By the above discussion, we then know that L contains precisely αpoints collinear with x. So, if μ = (t+1)α, then S is a partial geometry withparameters (s, t, α). �

Solution of Problem 55. Suppose S is a finite linear space of order (s, t),where s, t ≥ 1. Then for every anti-flag (x, L), all s + 1 points of L arecollinear with x, showing that S is a semi-partial geometry with parameters(s, t, s+ 1, μ) for any μ ∈ N \ {0}.

Suppose S is a semipartial geometry with parameters (s, t, α, μ). If (x, L)is an anti-flag, then L contains at most |L| = s + 1 points collinear with x,implying that α ≤ s + 1. Suppose now that α = s+ 1 and that x and y aretwo noncollinear points of S. Let z be a common neighbour of x and y. Thenthe line yz contains a point collinear with x, namely the point z. It followsthat yz contains precisely α = s + 1 points collinear with x. In particular,the point y must be collinear with x, a contradiction. So, if α = s+ 1, thenS must be a linear space. �

Solution of Problem 56. Observe that partial linear spaces of order (1, t)can be regarded as graphs with valency t + 1. The condition that for eachanti-flag (x, L), either one or all points of L are noncollinear with x translatesto the nonexistence of triangles. �

Solution of Problem 57. Suppose S is a dual net and let (x, L) be an anti-flag. Let T denote the transversality relation in S. This is an equivalencerelation. The line L contains a unique point y such that (x, y) ∈ T . Thepoint y is the unique point of L noncollinear with x, showing that S is acopolar space. �

Solution of Problem 58. Suppose S is the disjoint union of the copolarspaces Si where i belongs to some index set I. Let (x, L) be an anti-flag ofS. If (x, L) is an anti-flag of some Si, i ∈ I, then either one or all points

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of L are noncollinear with x. If x is a point of Si and L is a line of Sj forsome i, j ∈ I with i �= j, then no point of L is collinear with x. Hence, S isa copolar space. �

Solution of Problem 59. Let x ∈ K. Then x is contained in precisely q+1lines and each of these lines contains besides x d − 1 extra points of K. So,|K| = 1 + (q + 1)(d− 1) = qd+ d− q.

Every line of S is incident with precisely q + 1 − d ≥ 2 points of S andevery point of S is incident with precisely |K|

d= q + 1 − q

d≥ q

2+ 1 ≥ 2 lines

of S. Since q + 1− qd∈ N, we have d | q.

Now, let (p, L) be an antiflag of S. The point p is incident with |K|d

linesof S. Each of these lines meets L (in the projective plane π). But |L∩K| = d

of these |K|d

lines meet L in a point that is not a point of S. So, there are|K|d− d = (q − d) + 1− q

dpoints on L collinear with p.

So, S is a partial geometry. S is a generalized quadrangle if and only if(q− d) + 1− q

d= 1. Since d | q, we have (q− d) + 1− q

d≥ q− q

2+1− q

2= 1.

So, S is a generalized quadrangle if and only if d = q2and d = 2, i.e. if and

only if q = 4 and K is a hyperoval of PG(2, 4). �

Solution of Problem 60. Each line of S is incident with precisely q pointsand each point of S is incident with precisely |K| = 1 + (d− 1)(q + 1) lines.Indeed, through each x ∈ K, there are q + 1 lines and each of these linescontains besides x d− 1 other points of K.

Now, let (x, L) be an anti-flag of S and denote by u the unique point ofK incident with L. The plane 〈x, L〉 meets PG(2, q) in a line through u andthis line contains besides u d − 1 other points of K which we will denote byy1, y2, . . . , yd−1. If z is a point of L collinear with x, then xz contains oneof the points y1, y2, . . . , yd−1. Now, each line xyi, i ∈ {1, 2, . . . , d − 1}, willintersect L in a unique point. So, L contains precisely d− 1 points collinearwith x.

It follows that S is a partial geometry with parameters s = q − 1, t =(d − 1)(q + 1) and α = d − 1. This partial quadrangle is a generalizedquadrangle if and only if d = 2, i.e. if and only if K is a hyperoval ofPG(2, q) (and so, q must be even). �

Solution of Problem 61. Let ζ denote the polarity of PG(3, q) associatedwith Q, i.e the polarity ζ such that for every x ∈ Q, xζ is the hyperplaneof PG(3, q) which is tangent to Q at the point x. Observe that ζ is anorthogonal polarity if q is odd, and a symplectic polarity if q is even.

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If x ∈ PG(3, q) \ Q, then xζ ∩ Q is a conic and so there are q + 1 linesthrough x meeting Q in a singleton, namely the q+1 lines through x meetingxζ ∩ Q. It follows that there are |Q|−(q+1)

2= q2−q

2lines through x meeting Q

in precisely two points.Clearly, every line of S is incident with precisely s + 1 := q points and

every point of S is incident with precisely t+ 1 := |Q| = q2 + 1 lines.Now, let (x, L) be an anti-flag of S, let p be the unique point of Q con-

tained in L and let M be the line of PG(3, q) obtained by intersecting theplane 〈x, L〉 with PG(3, q). For every point y of (M ∩ Q) \ {p}, xy ∩ L isa point of L collinear with x. Conversely, every point of L collinear withx is obtained in this way. As |M ∩ Q| ∈ {1, 2}, the number of points of Lcollinear with x is equal to |M ∩Q| − 1 ∈ {0, 1}.

Let x1 and x2 be two noncollinear points of S. Let y be the unique pointof PG(3, q) \ Q contained on the line x1x2. If z is a common neighbour ofx1 and x2, then 〈x1, x2, z〉 is a plane intersecting PG(3, q) in a line throughy containing the points u and v of Q, where {u} = x1z ∩ PG(3, q) and

{v} = x2z ∩ PG(3, q), showing that 〈x1, x2, z〉 ∩ PG(3, q) is one of the q2−q2

lines through y intersecting Q in precisely two points. Conversely, if u andv are two distinct points of Q such that y ∈ uv, then x1u∩ x2v is a commonneighbour of x1 and x2. It follows that x1 and x2 have precisely μ := q2 − qcommon neighbours. �

Solution of Problem 62. Let ζ denote the symplectic polarity ofPG(2n − 1, q) that defines W (2n − 1, q). Let S denote the geometry ofthe hyperbolic lines of W (2n− 1, q). Suppose (x, L) is an anti-flag of S. Wecan distinguish two cases.

• Suppose L ⊆ xζ . Then every line through x meeting L is totallyisotropic and so no point of L is collinear with x in S.

• Suppose xζ intersects L in a unique point y. Then the points of Lcollinear with x are precisely the points of L distinct from y. Indeed,the line xy is totally isotropic, and every line xz, where z ∈ L \ {y} isa hyperbolic line.

So, either one or all points of L are noncollinear with x (in S), proving thatS is a copolar space. We now show that S is also a semipartial geometryand determine its parameters (s, t, α, μ).

Since every hyperbolic line contains precisely q+1 points, the parameters is equal to q.

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Every point of PG(2n−1, q) is a point of S. If x is a point of PG(2n−1, q),

then x is contained in q2n−1−1q−1

lines, q2n−2−1q−1

of which are totally isotropic,

namely those in the hyperplane xζ of PG(2n − 1, q). So, x is contained in

precisely q2n−1−1q−1

− q2n−2−1q−1

= q2n−2 lines of S. So, the parameter t is equal to

q2n−2 − 1.Above, we already remarked that for every anti-flag (x, L) either 0 or q

points of L are collinear with x. So, the parameter α is equal to q.Now, suppose that x and y are two distinct noncollinear points of S. Then

the line xy is totally isotropic. A point z �∈ {x, y} is a common neighbour ofx and y (in S) if and only if xz and yz are hyperbolic lines. So, the commonneighbours of x and y are the points of PG(2n−1, q) not contained in xζ∪yζ.The number of such common neighbours is equal to

q2n − 1

q − 1− |xζ | − |yζ |+ |xζ ∩ yζ | = q2n − 1

q − 1− 2 · q

2n−1 − 1

q − 1+

q2n−2 − 1

q − 1= q2n−1 − q2n−2.

So, the parameter μ is equal to q2n−1 − q2n−2. �

Solution of Problem 63. Every line of S is incident with precisely q2+q+1points. The number of lines of S incident with a given point is equal to thenumber of points in a PG(n− 2, q), i.e. equal to qn−1−1

q−1.

If L1 and L2 are two lines of PG(n, q), then L1 and L2 are collinear aspoints of S if and only if they meet.

Take now an anti-flag (L, π) of S, where L is a line and π is a plane notcontaining L. If L ∩ π is a singleton, then precisely α := q + 1 points ofS incident with π are collinear with L, namely the q + 1 lines of π throughL ∩ π. If L ∩ π = ∅, then there are no points of S incident with π that arecollinear with L. Note that the latter possibility cannot occur if and only ifn = 3.

Let L1 and L2 be two noncollinear points of S. Then L1 and L2 aredisjoint lines of PG(n, q). Every common neighbour of L1 and L2 is a linemeeting L1 and L2. There are (q + 1)2 such lines.

We conclude that S is a semi-partial geometry with parameters(s, t, α, μ) = (q2 + q, q

n−1−qq−1

, q + 1, (q + 1)2). Moreover, S is a partialgeometry if and only if n = 3. �

Solution of Problem 64. First observe that every line of PG(2, q2) inter-sects U in either a singleton or a Baer subline (containing precisely q + 1points).

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Clearly, every line of S is incident with precisely s + 1 := q2 points andevery point of S is incident with precisely t+ 1 := |U| = q3 + 1 lines.

Now, let (x, L) be an antiflag of S, let p be the unique point of U containedin L and let M be the line of PG(2, q2) obtained by intersecting the plane〈x, L〉 with PG(2, q2). For every point y of (M∩U)\{p}, xy∩L is a point of Lcollinear with x. Conversely, every point of L collinear with x is obtained inthis way. So, the number of points of L collinear with x is equal to |M∩U|−1,i.e. equal to either 0 or α := q.

Let x1 and x2 be two distinct noncollinear points of S. Let y be the uniquepoint of PG(2, q2) \ U contained in the line x1x2. Suppose y is contained inα lines of PG(2, q2) meeting U in a unique point and β lines meeting U inprecisely q+1 points. From α+ β = q2 +1 and α+ β(q+ 1) = |U| = q3 +1,it follows that α = q + 1 and β = q2 − q. If z is a common neighbour ofx1 and x2, then 〈x1, x2, z〉 is a plane intersecting PG(2, q2) in a line throughy containing the points u, v of U , where {u} = x1z ∩ PG(2, q2) and {v} =x2z ∩PG(2, q2), showing that 〈x1, x2, z〉 ∩PG(2, q2) is one of the q2 − q linesthrough y intersecting PG(2, q2) in precisely q + 1 points. Conversely, if uand v are two distinct points of U on a line through y (intersecting U inprecisely q + 1 points), then x1u ∩ x2v is a common neighbour of x1 andx2. This shows that μ := |Γ1(x1) ∩ Γ1(x2)| is equal to the total number ofordered pairs (u, v) of distinct points of U such that uv is one of the q2 − qlines through y meeting U in precisely q + 1 points, i.e. μ = q2(q2 − 1).

We conclude that S is a semi-partial geometry with parameters(s, t, α, μ) = (q2 − 1, q3, q, q2(q2 − 1)). �

Solution of Problem 65. First observe that every line of PG(2, q2) inter-sects PG(2, q) in either a singleton or a Baer subline (containing preciselyq + 1 points).

Clearly, every line of S is incident with precisely s + 1 := q2 points andevery point of S is incident with precisely t+ 1 := q2 + q + 1 lines.

Now, let (x, L) be an antiflag of S, let p be the unique point of PG(2, q)contained in L and let M be the line of PG(2, q2) obtained by intersectingthe plane 〈x, L〉 with PG(2, q2). For every point y of (M ∩ PG(2, q)) \ {p},xy ∩L is a point of L collinear with x. Conversely, every point of L collinearwith x is obtained in this way. So, the number of points of L collinear withx is equal to |M ∩ PG(2, q)| − 1, i.e. equal to either 0 or α := q.

Let x1 and x2 be two distinct noncollinear points of S. Let y be theunique point of PG(2, q2) \ PG(2, q) contained in the line x1x2. Suppose y

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is contained in α lines of PG(2, q2) meeting PG(2, q) in a unique point andβ lines meeting PG(2, q) in precisely q + 1 points. From α+ β = q2 + 1 andα+ β(q+1) = |PG(2, q)| = q2 + q+1, it follows that α = q2 and β = 1. If zis a common neighbour of x1 and x2, then 〈x1, x2, z〉 is a plane intersectingPG(2, q2) in a line through y containing the points u, v of PG(2, q), where{u} = x1z ∩ PG(2, q2) and {v} = x2z ∩ PG(2, q2), showing that 〈x1, x2, z〉 ∩PG(2, q2) coincides with the unique line L∗ through y intersecting PG(2, q)in precisely q + 1 points. Conversely, if u and v are two distinct points ofL∗ ∩ PG(2, q), then x1u ∩ x2v is a common neighbour of x1 and x2. Thisshows that μ := |Γ1(x1) ∩ Γ1(x2)| is equal to the total number of orderedpairs (u, v) of distinct points of L∗ ∩ PG(2, q), i.e. μ = q(q + 1).

We conclude that S is a semipartial geometry with parameters (s, t, α, μ)=(q2 − 1, q2 + q, q, q(q + 1)). �

Solution of Problem 66. With Q, there is associated a symplectic polarityζ such that for every point x ∈ Q, xζ equals the hyperplane that is tangentto Q at the point x. The quadric Q has 27 points. If x ∈ PG(5, 2) \Q, thenxζ ∩ Q is a nonsingular parabolic quadric Q(4, 2) of xζ for which x is thenucleus. The lines through x intersecting Q in precisely 1 point are preciselythe 15 lines through x contained in xζ . So, the number of lines through xintersecting Q in precisely two points is equal to 27−15

2= 6, and the total

number of lines through x disjoint from Q is equal to 31 − 15 − 6 = 10. Itfollows that S is a partial linear space of order (s, t) = (2, 9).

Now, let (x, L) be an anti-flag. Then the plane 〈x, L〉 intersects Q(5, 2) ina quadric of 〈x, L〉 which is disjoint from L, i.e. in a singleton or a nonsingularconic of 〈x, L〉. In the former case, there are precisely α = 2 lines of S throughx meeting L. In the latter case, no line of S through x meets L.

Now, let x and y be two noncollinear points of S. Then the line xy meetsQ(5, 2) in a point u. If z is a common neighbour of x and y in S, then theplane 〈x, y, z〉 intersects Q(5, 2) in the singleton {u} since none of the linesxz, yz contains points of Q(5, 2). So, the number of common neighbours ofx and y (in S) equals four times the number of planes through xy meetingQ(5, 2) in {u}. By looking at the hyperplane which is tangent to Q at thepoint u, we see that the number of planes through xy meeting Q(5, 2) in {u}is equal to the number of lines through a fixed point u′ ∈ PG(3, 2) \Q−(3, 2)that are disjoint from Q−(3, 2). This number can be computed in a similarway as the number of lines through a given point of PG(5, 2) \Q−(5, 2) thatare disjoint from Q−(5, 2). The total number of lines through u′ meeting

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Q−(3, 2) in a singleton is equal to 3, the total number of lines through u′

meeting Q−(3, 2) in precisely two points is equal to |Q−(3,2)|−32

= 1 and thetotal number of lines of PG(3, 2) through u′ disjoint from Q−(3, 2) is equalto 7− 3− 1 = 3.

So, every two noncollinear points of S have precisely μ = 4 · 3 = 12common neighbours.

We conclude that S is a semipartial geometry with parameters (s, t, α, μ)=(2, 9, 2, 12). �

Solution of Problem 67. Suppose α is a plane intersecting PG(n−2, q) ina singleton {x}. Then α contains q2 lines disjoint from PG(n− 2, q), namelythe q2 lines of α not containing x. So, every line of S contains preciselyq2 > 2 points.

Suppose L is a line of PG(n, q) disjoint from PG(n− 2, q). Then each of

the qn−1−1q−1

planes through L intersects PG(n− 2, q) in a singleton. So, every

point of S is incident with precisely qn−1−1q−1

> 2 lines.

Two lines L1 and L2 of PG(n, q) disjoint from PG(n− 2, q) are collinearin the point-line geometry S if and only if they meet, in which case they areincident with the line 〈L1, L2〉 of S.

Now, let L be an arbitrary line of PG(n, q) disjoint from PG(n−2, q) andlet α be a plane not containing L and intersecting PG(n−2, q) in a singleton{x}. If L∩α = ∅, then L is not collinear with any point of α in the geometryS. If L ∩ α is a singleton {y}, then L is collinear with precisely q points ofα in the geometry S, namely the q lines of α through y distinct from xy.

Suppose L1 and L2 are two lines of PG(n, q) disjoint from PG(n − 2, q)which are not collinear regarded as points of S. Then L1 and L2 are disjointand hence 〈L1, L2〉 is a 3-dimensional subspace which intersects PG(n− 2, q)in a line L3. Through every point of L3, there exists a unique line meetingL1 and L2. So, among the (q + 1)2 lines meeting L1 and L2, there are(q + 1)2 − (q + 1) = q2 + q lines which are disjoint from L3. These are theμ = q2 + q points of S collinear with L1 and L2.

It follows that S is a semipartial geometry with parameters s = q2 − 1,t = qn−1−q

q−1, α = q and μ = q2 + q. �

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Solution of Problem 68. Suppose S is a generalized Moore geometry ofdiameter 2. Then:

(1) S has an order (s, t) for some s, t ≥ 1;

(2) for every point-line pair (x, L) with d(x, L) = 1, there is a unique pointon L collinear with x;

(3) there is a constant c ≥ 1 such that if x and y are two points at distance 2from each other, then there are precisely c lines through y containing a(necessarily unique) point at distance 1 from x, i.e. |Γ1(x)∩Γ1(y)| = c.

We conclude that S is a partial quadrangle whose μ-parameter is equal to c.Conversely, suppose that S is a partial quadranglewith parameters (s, t, μ).

Then S has order (s, t). The fact that for every anti-flag (x, L), there existsat most one point on L collinear with x implies the following:

(∗) if x and y are two points at distance 1 from each other, then there existsa unique line through y (namely xy) containing 1 point at distance 0from x and s points at distance 1 from x. Any other line through ycontains besides y precisely s points at distance 2 from x.

Property (∗) implies the following:

If x and y are two points at distance 2 from each other, then thereare precisely μ lines through y containing a (necessarily unique)point collinear with x.

So, S is a generalized Moore geometry of diameter 2. �

Solution of Problem 69. Suppose S is a generalized Moore geometry ofdiameter d. Then the following holds:

• if x and y are two points at distance i ∈ {1, 2, . . . , d − 1} from eachother, then there exists a unique line through y containing 1 point atdistance i− 1 from x and s points at distance i from x. Any other linethrough y contains besides y precisely s points at distance i+1 from x.

This implies that for every point-line pair (x, L) with d(x, L) ≤ d− 1, thereexists a unique point on L nearest to x. So, if d(x, L) ≤ d − 1 for everypoint-line pair (x, L), then S is a near 2d-gon. If there exists a point-linepair (x, L) with d(x, L) = d, then S is a near (2d+ 1)-gon. �

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Solution of Problem 70. Let x be a vertex of Γ. We show by induction oni ∈ {1, 2, . . . , d} that |Γi(x)| ≤ k(k−1)i−1. Observe that this is true for i = 1since |Γ1(x)| = k. Suppose now that |Γi−1(x)| ≤ k(k − 1)i−2 for a certaini ∈ {2, 3, . . . , d}. The number of edges between a vertex of Γi−1(x) and avertex of Γi(x) is at most |Γi−1(x)| · (k − 1) and at least |Γi(x)| · 1, implyingthat |Γi(x)| ≤ |Γi−1(x)| · (k − 1) ≤ k(k − 1)i−1. So, if v is the total numberof vertices of Γ, then

v ≤ 1 + k + k(k − 1) + · · ·+ k(k − 1)d−1.

Equality occurs if and only if the following holds for every vertex x:

(a) if i ∈ {1, 2, . . . , d} and y ∈ Γi(x), then precisely one neighbour of y liesat distance i− 1 from x;

(b) if i ∈ {1, 2, . . . , d− 1} and y ∈ Γi(x), then each of the k− 1 neighboursof y not contained in Γi−1(x) lies at distance i+ 1 from x.

Properties (a) and (b) are equivalent with saying that Γ is a GMd(1, k−1, 1),i.e. a Moore graph of valency k. �

Solution of Problem 71. Suppose Γ is a Moore graph of diameter 2 andvalency 3. Then Γ satisfies the following properties:

(I) Γ has 1 + 3 + 3 · 2 = 10 vertices.

(II) Γ has valency 3.

(III) Γ has no triangles.

(IV) Any two vertices of Γ at distance 2 have a unique common neighbour.

Let x1 and x3 be two vertices at distance 2 from each other and let x2 be theunique common neighbour of x1 and x3. Let x4 be a neighbour of x3 distinctfrom x2. By (IV), d(x1, x4) = 2 and so x1 and x4 have a unique commonneighbour x5. The subgraph of Γ induced on {x1, x2, x3, x4, x5} is a pentagon.Let yi, i ∈ {1, 2, 3, 4, 5}, denote the unique neighbour of xi not contained in{x1, x2, x3, x4, x5}. By (III) and (IV), the vertices x1, x2, x3, x4, x5, y1, y2, y3,y4, y5 are mutually distinct and hence constitute all the vertices of Γ. A vertexyi, i ∈ {1, 2, 3, 4, 5}, is adjacent to three other vertices, one of them is xi andthe other two lie in the set {y1, y2, . . . , y5} by (III) and (IV). It is not possible

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that yi and yi+1 are adjacent where i ∈ {1, 2, . . . , 5} and indices are takenmodulo 5. Indeed, if this were the case, then {xi, yi, yi+1, xi+1} would definea quadrangle, in contradiction with (IV). The remaining missing adjacenciesare now uniquely determined: y1 ∼ y3 ∼ y5 ∼ y2 ∼ y4 ∼ y1. The graph Γhas now completely been described and it is straightforward to verify thatthis graph satisfies the properties (I), (II), (III) and (IV) above. �

Solution of Problem 72. Let Γ be a Moore graph of diameter 2 andvalency k. Then Γ does not contain triangles and so Γ is a copolar space.Let S be the partial linear space associated with Γ. Then every line of S isincident with precisely 2 points and every point is incident with precisely klines. Since there are no triangles, it holds that for every anti-flag (x, L) ofS, x is collinear with either 0 or 1 points of L. For every two noncollinearpoints of S, there is a unique point collinear with both. So, S is a partialquadrangle. �

Solution of Problem 73. Let Γ be a Moore graph of diameter 2 andvalency k. Note that two distinct vertices x and y of Γ are collinear in S ifand only if they lie at distance 2 from each other, in which case the uniqueline containing them is equal to z⊥, where z is the unique common neighbourof x and y. So, S is a partial linear space.

Every point of S is incident with precisely k lines and every line is incidentwith precisely k points. Let (x, L) be an anti-flag of S. Then L = y⊥ whereeither x = y or d(x, y) = 2. In case x = y, all vertices of L are adjacentwith x and so no point of L is collinear with x (in S). In case d(x, y) = 2, Lcontains a unique vertex adjacent to x and so |L| − 1 = k − 1 vertices of Lare collinear with x in S.

Now, let x and y be two noncollinear points of S. Then x and y are twoadjacent vertices of Γ. The number of vertices of Γ that are not adjacent tox nor to y is equal to 1+k+k(k−1)−2−2(k−1) = (k−1)2. So, the pointsx and y have precisely (k − 1)2 common neighbours in S. We conclude thatS is a semipartial geometry with parameters (s, t, α, μ) = (k−1, k−1, k−1,(k − 1)2). �

Solution of Problem 74. A circular space on 4 vertices is isomorphic tothe affine plane of order 2, i.e. to the projective plane of order 2 in whichone line L and all three points on L have been removed. So, the point-linedual of the circular space on four vertices is isomorphic to the partial linear

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space obtained from the Fano plane by removing one points and all threelines incident with that point. �

Solution of Problem 75. Observe that each of these linear spaces only haslines of size 2 and 3. We now show that each of them also satisfies Property(F2). We need to show that for every anti-flag (x, L), σx(L) is again a lineof the geometry. Here, σx is the involutory permutation of the point set asdefined in Section 2.20.

Suppose S is the point-line dual of the circular space on 4 vertices. Thenwe can label the points by x, u1, v1, u2, v2, y such that {x, u1, v1}, {x, u2, v2},{x, y}, {u1, v2}, {v1, u2}, {u1, u2, y} and {v1, v2, y} are lines. There are fourlines L not containing x, and for each of these lines, we indeed have that σx(L)is a line: σx({u1, v2}) = {v1, u2}, σx({v1, u2}) = {u1, v2}, σx({u1, u2, y}) ={v1, v2, y} and σx({v1, v2, y}) = {u1, u2, y}.

Suppose (x, L) is an anti-flag in AG(2, 3). In AG(2, 3), there is a uniqueline L′ parallel with L and disjoint from L ∪ {x}. Every line through xmeeting L also meets L′ and so we have σx(L) = L′. �

Solution of Problem 76. Suppose S = (P,L, I) is a partial linear spacesatisfying the conditions (F1) and (F2) of Section 2.20, and denote by S ′ thepoint-line geometry derived from S as described in that section. Clearly, S ′

satisfies property (F1’). We now also show that S ′ satisfies Property (F2’).Suppose L1 = {u, x1, y1} and L2 = {u, x2, y2} are two distinct lines of size 3of S through the same point u.

Suppose one of the lines meeting L1\{u} and L2\{u} has size 2. Withoutloss of generality, we may suppose that this is the line {x1, x2}. By applyingthe automorphism σu, we see that also {y1, y2} is a line of S. Now, theautomorphism σy1 maps the line {u, x2, y2} to the line {x1, y2, σy1(x2)}. So,y1x2 and x1y2 are two lines of size three meeting in the same point. Call thispoint v. Since σu(x1y2) = y1x2, we see that σu(v) = v, i.e. {u, v} is a line. Itfollows that the subspace of S ′ generated by L1 and L2 is isomorphic to thepoint-line dual of the circular space on 4 points.

Suppose now that all lines meeting L1 \ {u} and L2 \ {u} have size three.Put x1x2 = {x1, x2, x3} and y1y2 = {y1, y2, y3}. Clearly, σu maps the linex1x2 to the line y1y2 and hence x3 to y3. If x3 = y3, then ux3 is a line of size2. But then the previous paragraph applied to the lines x2x1 and x2u of size3 implies that the geometry induced on the subspace generated by x1, x2 andu is isomorphic to the point-line dual of the circular space on 4 points. This

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would contradict the fact that every line meeting L1 \ {u} and L2 \ {u} hassize 3. So, x3 �= y3 and {u, x3, y3} is a line. The automorphism σy1 maps theline {x1, x2, x3} to the line {z1, z2, z3}, where z1 = u, z2 ∈ y1x2 and z3 ∈ y1x3.We now show that the lines of S contained in {x1, x2, x3, y1, y2, y3, z1, z2, z3}turn the latter set in an affine plane of order 3. We already know eightof these lines: {x1, x2, x3}, {y1, y2, y3}, {z1, z2, z3}, {x1, y1, z1}, {z1, x2, y2},{z1, x3, y3}, {y1, x2, z2}, {y1, x3, z3}. By applying σz1 to the lines {y1, x2, z2}and {y1, x3, z3}, we see that also {x1, y2, z3} and {x1, z2, y3} are lines. Byapplying σx2 and σx3 to the line {x1, y1, z1}, we also see that {z2, y2, x3} and{x2, y3, z3} are lines. So, we have found all 12 lines which turn {x1, x2, . . . , z3}into an affine plane of order 3.

Conversely, suppose S ′ is a partial linear space satisfying Properties (F1’) and(F2’), and denote by S the point-line geometry derived from S ′ as describedin Section 2.20. Then S satisfies Property (F1). We now also show that Ssatisfies Property (F2). In view of the fact that σ2

x is the identity, it sufficesto prove that if x is a point and {x1, x2, . . . , xk} with k ∈ {2, 3} is a line notcontaining x, then {σx(x1), σx(x2), . . . , σx(xk)} is contained in a line. Thisobviously holds if k = 2 or if both the lines xx1 and xx2 have size 2. So, wemay suppose that k = 3 and that at least one of the lines xx1 and xx2 has size3. Then the geometry induced on the subspace generated by {x, x1, x2, x3}is isomorphic to either AG(2, 3) or the point-line dual of a circular space onfour vertices. The property now follows from the fact that these two lattergeometries are examples of Fischer spaces. �

Solution of Problem 77. • We show that every line of SO is incident withat least three points. Let α be an arbitrary plane intersecting O in at leasttwo points. Let x be an arbitrary point of α ∩ O and put πx ∩ α = L. In α,there are |F| ≥ 2 lines through x distinct from L and on each of these linesthere is a point of O distinct from x, indeed showing that |α ∩ O| ≥ 3.

• We show that every two distinct points x1 and x2 of SO are incidentwith at least one line of SO. Clearly, every plane of PG(3,F) through x1x2

is such a line.• We prove that the internal structure of S at a point x ∈ O is isomorphic

to the affine plane AG(2,F). The lines and planes of PG(3,F) through xdefine a projective plane Px isomorphic to PG(2,F). We denote by Ax theaffine plane isomorphic to AG(2,F) that arises from Px, by taking πx as the“line at infinity”. The points of Ax are the lines of PG(3,F) through x notcontained in πx. The lines of Ax are the planes through x distinct from

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πx, i.e. the planes through x containing at least two points of O. Observethat the map L �→ (L \ {x}) ∩ O defines a bijection between the set of linesthrough x not contained in πx and the set O \ {x}. From this latter fact, itshould now be clear that Ax is isomorphic to the local structure of S at thepoint x. �

Solution of Problem 78. Suppose S = (P,L, I) is a finite inversive planecontaining N points. If x is a point of S, then the internal structure of S atthe point x is an affine plane defined on the set P \{x}. If n ≥ 2 is the orderof this affine plane, then N = n2+1. So, all affine planes that arise as internalstructures of S have the same order n. In any internal structure of S, alllines have size n ≥ 2 and every two distinct points are incident with a uniqueline. Translated to properties of S that means that every block of S containsprecisely n + 1 points and that every three distinct points are incident witha unique block, i.e. S is a Steiner system of type S(3, n + 1, n2 + 1) withn ∈ N \ {0, 1}.

Conversely, suppose that S is a Steiner system of type S(3, n+1, n2 +1)with n ∈ N \ {0, 1}. Then any two distinct points of S are incident with ablock, and every block contains n + 1 ≥ 3 points. Every internal structureof S is a Steiner system of type S(2, n, n2), i.e. an affine plane of order n. �

Solution of Problem 79. We show that two distinct points x1, x2 ∈ X\{x}are collinear in SX if and only if the line x1x2 through them does not containx. Indeed, if x1x2 contains x, then any plane through x1x2 contains x andso the points x1 and x2 cannot be collinear in SX . On the other hand,if x1x2 does not contain x, then there exists a unique plane through x1x2

containing x and |F| planes through x1x2 not containing x. Each of these |F|planes determines a line of SX containing x1 and x2. So, x1 and x2 are thencollinear in Sx.

So, if x1 and x2 are two distinct points of X \ {x}, then (x1, x2) ∈ T ifand only if x ∈ x1x2. This implies that T defines an equivalence relation onX \ {x} having |F| + 1 ≥ 3 equivalence classes. Each equivalence class ofT is of the form L \ {x}, where L is one of the |O| = |F| + 1 lines throughx containing a point of O. Note that if L is such a line, then every planeof PG(3,F) not containing x has a unique point in common with L \ {x},showing that every line of SX is incident with precisely one point of each ofthe equivalence classes of T .

Now, let y be an arbitrary point of X \ {x}. Our aim is to prove that theinternal structure S ′ of SX at the point y is a net and a dual net.

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The line xy meets π in a point o ∈ O. Every plane α through xy meets πin a line through o, and the map α �→ α∩π defines a bijection between the setof planes through xy and the set of lines of π through o. Put α∗ := 〈xy, Lo〉,where Lo is the unique line of π through o intersecting O in the singleton {o}.

The points of S ′ are the points of X \ xy = X \ α∗. For every point z ofX \ xy = X \α∗, let θ(z) be the line yz through y. We show that θ defines abijection between the set X \ xy = X \α∗ and the set of lines through y notcontained in α∗.

Suppose θ were not injective, i.e. θ(z1) = yz1 = θ(z2) = yz2 for twodistinct z1, z2 ∈ X \α∗. Then the plane 〈x, z1, z2〉 would intersect π in a linewhich contains at least three points of O, namely the unique points in thesingletons xy ∩ π = {o}, xz1 ∩ π and xz2 ∩ π.

We show that θ is surjective. Let L be an arbitrary line through y notcontained in α∗. Then the plane 〈x, L〉 intersects π in a line L′ through owhich is distinct from Lo. Let z′ denote the unique point of L′ ∩ O distinctfrom o. Then the line xz′ intersects L in a point z and we must have thatL = θ(z).

Now, the lines and planes of PG(3,F) through y determine a projectiveplane isomorphic to PG(2,F). If we regard α∗ as the line at infinity of thisprojective plane, then we obtain an affine plane AG(2,F). We denote by Uthe parallel class of lines of AG(2,F) defined by the point of infinity xy, andby S∗ the geometry obtained from AG(2,F) by removing the lines containedin U (but keeping all remaining lines and all points of AG(2,F)). The mapθ which sends each z ∈ X \ α∗ to θ(z) then defines an isomorphism betweenS ′ and S∗. So, it suffices to prove that S∗ is a net and a dual net.

Since the line set of S∗ is the union of at least two parallel classes of linesof AG(2,F), we see that S∗ is a net. In order to show that S∗ is also a dualnet, we consider the transversality relation T ∗ of S∗. If x1 and x2 are twodistinct points of AG(2,F), then (x1, x2) ∈ T ∗ if and only if the line x1x2

belongs to U . It follows that T ∗ is an equivalence relation. Each U ∈ Udefines an equivalence class, namely the set of points of AG(2,F) incidentwith U . The number of equivalence classes is |U| ≥ 2, and each equivalenceclass contains |F| ≥ 2 points. Since in an affine plane, two lines belonging todistinct parallel classes meet, we should have that every line of S∗ containsa unique point of each equivalence class of T ∗. So, S∗ must indeed be a dualnet. �

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Solution of Problem 80. There are two types of planes in PG(3,F), planeswhich are tangent to Q+(3,F) and intersect Q+(3,F) in the union of twodistinct intersecting lines and nontangent planes which intersect Q+(3,F)in an oval. Since each oval intersection contains |F| + 1 ≥ 4 points, everyline of SQ is incident with at least three points. Through every two pointsof Q+(3,F) not collinear on this quadric, there are two tangent planes and|F|−1 ≥ 1 nontangent planes, implying that these points are collinear in SQ.

Let R1 and R2 denote the two reguli defined by Q+(3,F), i.e. the twofamilies of generators of Q+(3,F). For every i ∈ {1, 2} and every two pointsx, y ∈ Q+(3,F), we say that (x, y) ∈ Ti if either x = y or (x �= y and the linexy of PG(3,F) belongs to Ri). Clearly, T1 and T2 are equivalence relations,T1 ∪T2 is the transversality relation of SQ and T1 ∩T2 is the identity relationon the point set of SQ.

Now, let x be an arbitrary point and let L1 and L2 denote the two linesof Q+(3,F) through x. Then 〈L1, L2〉 is the tangent plane of the quadricQ+(3,F) at the point x. The lines and planes through x define a projectivespace Px isomorphic to PG(2,F). By regarding 〈L1, L2〉 as the line at infinityof PG(2,F), we have an associated affine plane Ax

∼= AG(2,F). The planesthrough x that are not tangent to Q+(3,F) are precisely the planes notcontaining L1 or L2. These correspond to lines of Ax not belonging to thetwo parallel classes defined by the points at infinity L1 and L2. The pointsof Q+(3,F) \ {x} which are collinear with x in SQ are precisely the points ofQ+(3,F) not contained in 〈L1, L2〉. For each such point y, the line xy is apoint of Ax. In this way, we obtain a bijective correspondence between thepoints of SQ collinear with but distinct from x and the points of Ax.

So, we see that the internal structure S ′ of SQ at the point x is an affineplane in which two parallel classes of lines have been removed. Since the lineset of S ′ is the union of at least two parallel classes of lines of this affinespace, S ′ must be a net. For every antiflag (p, L) of S ′, the line L containstwo points noncollinear with p, namely the two points r1 and r2 of L suchthat pr1 and pr2 belong to the two missing parallel classes. �

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365

Index

(σ, ε)-Hermitian form, 190σ-Hermitian variety, 24σ-sesquilinear form, 189k-partite graph, 4

absolute bound, 48absolutely universal embedding,

76adjacency matrix, 2adjacent vertices, 1affine space, 17alternating bilinear form, 192anti-automorphism of skew field,

188anti-flag, 6anti-Hermitian form, 193automorphism of geometry, 6automorphism of graph, 1

Baer subline, 233big convex subspace, 136bilinear form, 189bipartite graph, 5biplane, 291block of design, 19, 275Bose-Mesner algebra, 49, 52Bruck-Ryser theorem, 15, 290

Bruck-Ryser-Chowla theorem, 286Buekenhout-Shult polar space, 23,

173

Cameron’s Theorem, 260cap of projective space, 63cartesian product of graphs, 138cartesian product of near polygons,

139Cayley-Dickson division algebra, 246centre of skew field, 188centre of triad, 103circular space, 11classical convex subspace, 136classical inversive plane, 20, 36classical Laguerre plane, 37classical Minkowski plane, 37classical point-convex subspace rela-

tion, 135classical point-quad relation, 148clique, 1collinear points of geometry, 6collinear points of polar space, 22,

167collinearity graph, 7collinearity matrix, 7commutative quasigroup, 296


367

Index

complement of graph, 1complementary design, 279complementary graph, 1complete graph, 1complete multipartite graph, 4conference graph, 45conic, 221connected component of geometry, 7connected component of graph, 2connected geometry, 7connected graph, 2convex subspace, 10copolar space, 32cotriangular space, 33cross, 12cycle of graph, 2

de Bruijn-Erdos Theorem, 61decomposable copolar space, 33degenerate generalized quadrangle,

28, 129degenerate projective plane, 13degree of maximal arc, 310degree of vertex, 1dense near polygon, 150derived design, 20Desarguesian projective space, 14design, 19, 275diameter of geometry, 8diameter of graph, 2dimension of projective space, 14dimension of singular subspace, 22,

166direct product of near polygons,

139direct sum of polar spaces, 212direct sum of projective spaces, 15disjoint union of graphs, 2

distance in graph, 2distance-regular graph, 5, 50double of point-line geometry, 7dual grid, 12dual net, 18dual polar space, 26, 251dual twisted triality hexagon, 123dualized projective space, 215

edge of graph, 1egglike inversive plane, 36eigenvalues of geometry, 7eigenvalues of graph, 3elliptic polar space, 227, 257elliptic quadric, 24, 226empty graph, 1extendable design, 20extension of design, 20exterior line, 63external structure, 278

faithful pseudo-embedding, 80families of generators, 27, 245fan of ovoids, 123Fano plane, 14Feit-Higman Theorem, 29, 114finite graph, 2Fischer space, 34Fisher’s inequality, 278flag, 6Frobenius automorphism, 218full projective embedding, 75full subgeometry, 9full subquadrangle, 105Fundamental theorem of linear

spaces, 61

generalized n-gon, 89generalized 2-gon, 12, 89

368

Index

generalized digon, 12generalized Moore geometry, 33generalized polygon, 29, 89generalized quadrangle, 21generated subspace, 10generating index, 10generating set, 10generator of Hermitian variety,

24, 234generator of quadric, 23, 221geodesic, 2geodesic path, 2geodetically closed subspace, 10geometry, 6geometry of hyperbolic lines, 25girth of graph, 2GQ, 21Gram matrix, 46graph, 1Grassmannian, 16, 27grid, 12

Hadamard design, 282Hadamard matrix, 282Hadamard product of matrices, 52Haemers-Roos inequality, 30, 120half-idempotent latin square, 296half-idempotent quasigroup, 296half-spin geometry, 27, 270Hamming near polygon, 139Hermitian curve, 233Hermitian form, 193Hermitian polar space, 236, 237,

258Hermitian variety, 233hex, 151, 256Higman’s inequality, 21, 30, 60,

103, 120

Hoffman-Singleton graph, 34Husain graphs, 292hyperbolic basis, 220hyperbolic dual polar space, 257hyperbolic line, 24, 25hyperbolic polar space, 224hyperbolic quadric, 24, 224hyperoval, 65hyperplane, 9, 76hyperplane arising from

embedding, 76

idempotent latin square, 296idempotent quasigroup, 296imprimitive strongly regular graph,

42incidence graph, 7incidence matrix, 7incidence relation, 6indecomposable copolar space, 33induced subgraph, 2internal structure, 7intersection numbers of distance-

regular graph, 6, 50inversive plane, 20, 35irreducible polar spaces, 212irreducible projective plane, 13irreducible projective space, 13isomorphic full projective

embeddings, 76isomorphic geometries, 6isomorphic graphs, 1isomorphic polar spaces, 22isomorphic pseudo-embeddings, 80isomorphism between geometries,

6, 9isomorphism between graphs, 1

369

Index

isomorphism between polar spaces,167

Klein quadric, 245Krein parameters, 56

Laguerre plane, 36large Mathieu design, 20large Witt design, 20latin square, 294lemma of tangents, 67length of cycle, 2length of path, 2line as geometry, 11line of geometry, 6line system, 131line-quad relation, 147linear space, 9lines of polar space, 22, 167local space, 135

Mobius plane, 35Mathieu design, 20Mathon bound, 163max, 151, 256maximal arc, 310maximal clique, 2maximal singular subspace, 22, 167minimal idempotents, 55Minkowski plane, 37Miquelian inversive plane, 20, 36MOLS, 295Moore geometry, 33Moore graph, 33multipartite graph, 4mutually orthogonal latin squares,

295

near (2d+ 1)-gon, 28

near 2d-gon, 28, 129near polygon, 28, 129near-linear space, 8near-pencil, 12neighbours, 1net, 18next-to-maximal singular subspace,

22, 167nondegenerate projective plane, 13nondegenerate quadratic form, 206nondegenerate sesquilinear form,

190nonsingular Hermitian variety, 235nonsingular quadratic form, 206nonsingular quadric, 222nonsymmetrical dual grid, 12nonsymmetrical grid, 12nontrivial strongly regular graph, 42normalized Hadamard matrix, 283nucleus of Q(2n,F), 225nucleus of oval, 66

one or all axiom, 23, 174opposite points, 8opposite skew field, 16, 188order of geometry, 9order of projective plane, 13ordinary polygon, 12orthogonal latin squares, 295oval, 37, 64ovoid, 36, 64, 123ovoidal point-quad relation, 148

parabolic dual polar space, 257parabolic polar space, 225parabolic quadric, 24, 225parallel convex subspaces, 137parallel lines, 17, 18, 134

370

Index

parameters of distance-regular graph,6, 50

parameters of partial geometry, 30

parameters of partial quadrangle, 31

parameters of regular near polygon,28, 158

parameters of semipartial geometry,32

parameters of strongly regular graph,5, 39

partial geometry, 30

partial linear space, 8

partial plane, 8

partial quadrangle, 31

path in graph, 2permutation matrix, 2

Petersen graph, 34, 42

planes of polar space, 22, 167

point as geometry, 11

point graph, 7point of geometry, 6

point-line dual, 6

point-line geometry, 6

point-quad relation, 147points of polar space, 22, 166

polar Grassmannian, 27

polar space, 22, 166

primitive strongly regular graph, 42

product near polygon, 138, 139projection, 135, 138

projective Grassmannian, 16

projective plane, 13

projective space, 13

pseudo-elliptic polar space, 228, 257pseudo-elliptic quadric, 24, 228

pseudo-embedding, 79

pseudo-hyperplane, 78

pseudo-hyperplane arising frompseudo-embedding, 81

pseudo-quadratic form, 204pseudo-quadric, 204

quad, 135quadratic form, 204quadric, 205quasigroup, 296quotient of projective embedding,

76quotient of pseudo-embedding, 81quotient polar space, 182

rank of dual polar space, 26, 251,252

rank of polar space, 22, 166reducible polar spaces, 212reducible projective plane, 13reducible projective space, 13Ree-Tits octagon, 123reflection, 138reflexive sesquilinear form, 190regular graph, 1regular near polygon, 28, 158rosette of ovoids, 123

Schur product of matrices, 52secant line, 63self-complementary graph, 1self-dual geometry, 7semi-linear space, 8semipartial geometry, 32sesquilinear form, 189shortest path, 2singular Hermitian variety, 235singular point, 222, 235singular quadric, 222

371

Index

singular subspace, 206singular subspace of geometry, 10singular subspace of polar space,

22, 166small Mathieu design, 20small Witt design, 20split-Cayley hexagon, 122spread, 124square design, 280square lattice graph, 42standard sequence, 53star, 12Steiner quadruple system, 19Steiner system, 19, 275Steiner triple system, 19, 299strongly regular graph, 5, 39subgeometry, 9subgraph, 2subquadrangle, 105subspace, 9symmetric bilinear form, 192symmetric design, 280symmetric latin square, 296symmetrical dual grid, 12symmetrical grid, 12symplectic dual polar space, 257symplectic polar space, 25, 220system of vectors, 131

tangent hyperplane, 209tangent line, 63

tangent plane, 74tetrahedrally closed line system, 131tetrahedrally closed system of vec-

tors, 131thick line, 9thin line, 9thin-ovoidal point-quad relation,

148totally isotropic subspaces, 199,

200trace-valued form, 196transversality relation, 18, 36triad, 103triality, 272triangular graph, 42trivial sesquilinear form, 189trivial strongly regular graph, 42twisted triality hexagon, 123

unital, 19universal pseudo-embedding, 82

valency of regular graph, 1valency of vertex, 1Veblen-Young axiom, 13Veldkamp-Tits polar space, 22, 166vertex of graph, 1

Wedderburn’s theorem, 14, 188Witt design, 20Witt index, 23, 24, 221, 234

372

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