tritangent circles - uoc.gr
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Tritangent circlesA file of the Geometrikon gallery by Paris Pamfilos
As the whole of nature is akin, and the soul haslearned everything, nothing prevents a man, afterrecalling one thing only β a process men calllearning β discovering everything else for himself, ifhe is brave and does not tire of the search, forsearching and learning are, as a whole, recollection.
Plato, Meno 81 d
Contents (Last update: 28β11β2021)
1 Inscribed, Escribed, Incenter and Excenters 2
2 Tangents from the vertices 3
3 Relations between the radii 6
4 Radii of the tritangent circles related to sideβlengths 7
5 Relations between angles and sideβparts 7
6 Heronβs triangle area formula 8
7 A symmetric equilateral hexagon 9
8 Eulerβs theorem for the incenter/excenter 12
9 Some properties of the bisectors 16
10 Eulerβs line and circle of the triangle 18
11 Some properties of Eulerβs circle 19
12 Feuerbachβs theorem 22
1 Inscribed, Escribed, Incenter and Excenters 2
1 Inscribed, Escribed, Incenter and Excenters
βTritangentβ circles are circles simultaneously tangent to three lines βin general positionβ,i.e. lines not all passing through a point. Such a triple of lines defines a triangle π΄π΅πΆ, sixβouter domainsβ and four tritangent circles (see figure 1). The βinscribedβ in the triangle andthe three βescribedβ lying in respective outer domains. Often in the literature by tritangentis meant only one of the escribed circles. The center of the inscribed is the βincenterβ of thetriangle and the centers of the escribed are the βexcentersβ of the triangle.
The configuration of the four tritangent circles has strong ties with the βbisectorsβ ofthe triangleπ΄π΅πΆ, since these carry the centers of the circles and are, each, a symmetry axisof two of the circles. In this file we study the tritangent circles configuration and some ofthe related to it topics. Complementary material can be found in [Cou80, p.72].
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Figure 1: The tritangent circles of three lines forming a triangle
Theorem 1. The internal bisector of one triangle angle and the external bisectors of the other twoangles pass through a common point. The same is true for the three internal bisectors.
Proof. We will show that the internal bisector at π΅ and the external bisectors at π΄ andπΆ intersect at the same point, which we denote by πΌπ΅. Similar things will hold also forthe other angles, which will define the points πΌπ΄ and πΌπΆ (See Figure 1). The proof for thethree internal bisectors is the same, defining the incenter πΌ of the triangle. Let πΌπ΅ be theintersection point of two out from the three bisectors of the triangle and specifically ofthe external bisectors of the angles π΄ and πΆ. We will show that the internal bisector ofangle οΏ½ΜοΏ½ also passes through πΌπ΅. Indeed, the distances of πΌπ΅ from the sides of angle π΄ areequal |πΌπ΅πΈ| = |πΌπ΅π·|. Similarly, also the distances of πΌπ΅ from the sides of angle πΆ are equal|πΌπ΅πΈ| = |πΌπ΅π·|. Consequently the three distances will all be equal |πΌπ΅π| = |πΌπ΅πΈ| = |πΌπ΅π·|,therefore they are radii of a circle with center πΌπ΅ and radius ππ΅ = |πΌπ΅π·|. The equality ofthe distances |πΌπ΅π| = |πΌπ΅π·| from the sides of angle οΏ½ΜοΏ½ show that πΌπ΅ is to be found also onthe bisector of angle οΏ½ΜοΏ½. The fact that the sides are orthogonal to these radii of this circle,shows that the circle is tangent to all three sides of the triangle.
Exercise 1. Let {πΌ, πΌπ΄, πΌπ΅, πΌπΆ} be respectively the incenter and the excenters of the triangle π΄π΅πΆ(see figure 1). Show that:
2 Tangents from the vertices 3
1. Each of the triples {(π΄, πΌ, πΌπ΄), (π΅, πΌ, πΌπ΅), (πΆ, πΌ, πΌπΆ)} consists of collinear points.2. The line defined by each of these triples is an altitude of the triangle πΌπ΄πΌπ΅πΌπΆ.3. Point πΌ is the βorthocenterβ of the triangle πΌπ΄πΌπ΅πΌπΆ.4. Triangle π΄π΅πΆ is the βorthicβ of triangle πΌπ΄πΌπ΅πΌπΆ.
2 Tangents from the vertices
Theorem 2. The length of the tangent π΅π·, from the vertex π΅ to the corresponding escribed circlewith center πΌπ΅, is equal to half the perimeter π of the triangle π΄π΅πΆ (see figure 1).
|π΅π·| = 12(π + π + π) = π.
Proof. Here, as usual, with {π, π, π} we denote the lengths of the sides of the triangle. Theproof follows directly from the equality of the tangents from π΅ βΆ |π΅π| = |π΅π·|, as well asfrom π΄ and πΆ βΆ |π΄π| = |π΄πΈ|, |πΆπΈ| = |πΆπ·|. The perimeter therefore is written as
π + π + π = (|π΅πΆ| + |πΆπΈ|) + (|π΄πΈ| + |π΅π΄|)= (|π΅πΆ| + |πΆπ·|) + (|π΅π΄| + |π΄π|)= 2(|π΅πΆ| + |πΆπ·|) = 2|π΅π·|.
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Figure 2: Tangents from the vertices Tangent parallel to base
Theorem 3. The tangents {π΄π΅β², π΄πΆβ²} from the vertex π΄ of triangle π΄π΅πΆ to its inscribed circlehave length (see figure 2β(I)):
|π΄πΆβ²| = |π΄π΅β²| = π β π,where π = 1
2(π + π + π) is the half perimeter of the triangle.
Proof. As the proof of the previous proposition, so this one is also relying on the equalityof the tangents from one point to a circle: |π΄π΅β²| = |π΄πΆβ²|, |π΅πΆβ²| = |π΅π΄β²|, |πΆπ΄β²| = |πΆπ΅β²| . Itsuffices therefore to write the perimeter as
π + π + π = 2π = 2(|π΄πΆβ²| + |π΅π΄β²| + |π΄β²πΆ|) = 2(|π΄πΆβ²| + π),
from which the wanted equality follows directly.
Exercise 2. In the triangle π΄π΅πΆ, with |π΄πΆ| β₯ |π΄π΅|, the circle π is tangent to the sides {π΄π΅, π΄πΆ}and passes through the point π΄β² of the base π΅πΆ (see figure 2β(I)). Show that π coincides with theinscribed circle of the triangle π΄π΅πΆ, if and only if, it holds
|π΄β²πΆ| β |π΄β²π΅| = |π΄πΆ| β |π΄π΅|.
Then point π΄β² coincides with the contact point of the circle with the base π΅πΆ.
2 Tangents from the vertices 4
Hint: For {π₯ = |π΄β²π΅|, π¦ = |π΄β²πΆ|}, the above relation is equivalent with the two relations{π¦ β π₯ = π β π, π¦ + π₯ = π}.Exercise 3. Let πΈ be the contact point of the tangent π»πΊ of the inscribed circle, which is parallelto the base π΅πΆ of the triangle π΄π΅πΆ. Show that the line π΄πΈ intersects π΅πΆ at a point π·, such that|π΄π΅| + |π΅π·| = |π·πΆ| + |πΆπ΄| = π. Conclude that |π΅π·| = |πΉπΆ|, where πΉ is the contact point of theincircle with π΅πΆ (see figure 2β(II)).
Hint: {π΄π»π½, π΄π΅πΆ} are similar and π΄π» + π»πΈ = π΄π½ + π½πΈ.Theorem 4. Next table gives the centers of the inscribed and escribed circles of triangle π΄π΅πΆ, aswell as their respective projections on the sides π΄π΅, π΅πΆ and πΆπ΄ (see figure 3).
π΄π΅ π΅πΆ πΆπ΄πΌ π· π» ππΌπ΄ π π½ ππΌπ΅ π πΏ ππΌπΆ πΎ π π
The following relations are valid:
π β π = |π΄π·| = |π΄π| = |πΆπΏ| = |πΆπ| = |π΅π| = |π΅πΎ|π β π = |π΅π·| = |π΅π»| = |πΆπ½| = |πΆπ| = |π΄πΎ| = |π΄π|π β π = |πΆπ| = |πΆπ»| = |π΄π | = |π΄π| = |π΅π| = |π΅π½|
|π»π½| = |π β π|, |ππ| = |π β π|, |πΎπ·| = |π β π|.
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Figure 3: Segments on the sides
Proof. That it holds π β π = |π΄π·| = |π΄π|, we saw in the theorem 3. For the other equalitieson the same line write
|πΆπ½| = |π΅πΆ| β |π΅π½| = π β π.Similarly follow also the equalities in the second and third line. Equality |π»π½| = |π β π|follows from the previous
|π»π½| = |π΅πΆ| β |π΅π»| β |πΆπ½| = π β (π β π) β (π β π) = π β π.
Similarly follow also the two last equalities.
2 Tangents from the vertices 5
Exercise 4. Using figure 3 and the previous relations show that1. |ππ½| = |πΎπ| = π, |π½πΏ| = |ππ| = π, |ππ| = |π πΎ| = π.2. |π·π| = |ππ| = π, |π»πΏ| = |π·π | = π, |ππ| = |π»π| = π.3. |ππ| = π + π, |ππΏ| = π + π, |ππ | = π + π.
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Figure 4: Common tangents of two circles
Exercise 5. Given are two not congruent and external to each other circles π (π) and πΏ(π). Showthat the relations suggested by figure (see figure 4).
1. |πΎπΏ| = |πΏπ| = |π·πΈ| = |πΈπ| etc.2. |πΏπΈ| = |ππΆ| etc.3. The circles π = (πΆπ·ππ»), π = (π΅πΈπΏπ½), π½ = (π΄ππΎπΌ) are concentric.4. The circle π passes through the centers π and π.
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Figure 5: The common tangent π of two circles
Exercise 6. Show that point πΆ is contained in the internal common tangent of two external to eachother circles {π 1(π1), π 2(π2)} and external to the segment π΄β²π΅β² of the contact points, if and onlyif, for the other (external) tangents {πΆπ΄, πΆπ΅} the following relation is valid: ||πΆπ΄| β |πΆπ΅|| = |π΄β²π΅β²|(see figure 5).
Hint:Obviously the relation is valid if πΆ is on π and outside π΄β²π΅β². For the converse conβsider themovement of πΆ on a line π parallel to the radical axis π of the two circles. Showthat the function π (π₯) = ||πΆπ΄| β |πΆπ΅|| = π/(|πΆπ΄| + |πΆπ΅|), with π constant, is a decreasingfunction of the distance π₯ of πΆ from the line of centers π1π2.
3 Relations between the radii 6
3 Relations between the radii
Most of the following formulas are simple consequences of the relations discussed in secβtion 2. All of them have simple proofs and are formulated as exercises. In some of themit is of help to use the formula for the area Ξ of the triangle in dependence from thecircumradius π βΆ
Ξ = π β π β π4π , (1)
which results from the area formula Ξ = ππ sin(πΎ)/2 = πππ/(4π ), using the well knownsine rule π/ sin(πΎ) = 2π .
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Figure 6: π bisects the perimeter (π΄π΅πΆ) = 12(π + π β π) β ππΆ
Exercise 7. Find a point π on the side π΄π΅ of triangle π΄π΅πΆ (See Figure 6βI), such that
|ππ΄| + |π΄πΆ| = |ππ΅| + |π΅πΆ|.
Exercise 8. Show that for the radii {π, ππ΄, ππ΅, ππΆ} of the inscribed and escribed circles, the altitudes{βπ΄, βπ΅, βπΆ} and the area Ξ = (π΄π΅πΆ) of the triangle π΄π΅πΆ holds
Ξ = π β π (2)= (π β π)ππ΄ = (π β π)ππ΅ = (π β π)ππΆ, (3)
1π = 1
ππ΄+ 1
ππ΅+ 1
ππΆ= (4)
= 1βπ΄
+ 1βπ΅
+ 1βπΆ
. (5)
Hint: For the first three equalities see figure 6βII. For the rest see that 1βπ΄
= π2Ξ , which
implies that 1βπ΄
+ 1βπ΅
+ 1βπΆ
= πΞ = 1
π .
Exercise 9. Show the following formulas:
4π + π = ππ΄ + ππ΅ + ππΆ, (6)
π 2 = π2π2π2
2(π2π2 + π2π2 + π2π2) β (π4 + π4 + π4). (7)
Hint: The last equality follows from πππ = 4Ξπ and the identity
2(π2π2 + π2π2 + π2π2) β π4 β π4 β π4 = (π + π + π)(π + π β π)(π + π β π)(π + π β π). (8)
Exercise 10. Show that, if the incenter of the triangle π΄π΅πΆ coincides with its centroid or itsorthocenter, then the triangle is equilateral.
4 Radii of the tritangent circles related to sideβlengths 7
4 Radii of the tritangent circles related to sideβlengths
Theorem 5. The radius π of the inscribed and ππ΄ of the escribed circle of the triangle π΄π΅πΆ aregiven respectively by the formulas
π2 = (π β π)(π β π)(π β π)π , π2
π΄ = π(π β π)(π β π)π β π . (9)
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Figure 7: π, ππ΄ as functions of the sides
Proof. In figure 7 are seen two circles: the inscribed πΌ(π) and the escribed πΌπ΄(ππ). The proofuses the relations of section 3 and the similarity between two pairs of triangles. The firstpair of triangles is (π΄π·πΌ, π΄πΈπΌπ΄). The second pair is (π·πΌπ΅, πΈπ΅πΌπ΄). Both pairs consist of righttriangles and we have:
|π·πΌ||πΈπΌπ΄| = |π΄π·|
|π΄πΈ| β πππ΄
= π β ππ , (10)
|π·πΌ||π·π΅| = |πΈπ΅|
|πΈπΌπ΄| β ππ β π = π β π
ππ΄. (11)
Solving the second relative to ππ΄ and substituting into the first expression, we find theformula for π2. Squaring the first formula and substituting with the found expression forπ2, we prove the second formula as well.
5 Relations between angles and sideβparts
Next formulas relating the side parts {π β π, π β π, π β π} and π, defined by the tritanβgent circles, with the angles follow by inspecting figure 7 and are formulated as exercises.In these {πΌ, π½, πΎ} denote the angles of the triangle and π its circumradius.
Exercise 11. Show that for every triangle hold the formulas
cot(πΌ) = π2 + π2 β π2
4Ξ β cot(πΌ) + cot(π½) + cot(πΎ) = π2 + π2 + π2
4Ξ . (12)
Remark 1. For every triangle π΄π΅πΆ there is a special angle π, called βBrocard angleβ ofthe triangle and satisfying cot(π) = cot(πΌ) + cot(π½) + cot(πΎ). It follows that
π2 + π2 + π2 = 4Ξ cot(π).
6 Heronβs triangle area formula 8
Exercise 12. Show that in every triangle the following formulas are valid
sin(πΌ2) = β(π β π)(π β π)
ππ , sin(π½2 ) = β(π β π)(π β π)
ππ , sin(πΎ2 ) = β(π β π)(π β π)
ππ .(13)
Hint: Start with the formula for the cosine and show first that
2 sin(πΌ2)
2= (π β π + π)(π + π β π)
2ππ . (14)
Exercise 13. Show that for every triangle the following formulas are valid
cos(πΌ2) = βπ(π β π)
ππ , cos(π½2 ) = βπ(π β π)
ππ , cos(πΎ2 ) = βπ(π β π)
ππ . (15)
tan(πΌ2) = β(π β π)(π β π)
π(π β π) , tan(π½2 ) = β(π β π)(π β π)
π(π β π) , tan(πΎ2 ) = β(π β π)(π β π)
π(π β π) .
(16)
Exercise 14. Show that for every triangle holds next formula and the similars resulting by cyclicpermutations of the letters
cos(πΌ/2) cos(π½/2) sin(πΎ/2) = ππ΄4π . (17)
Exercise 15. Show that for every triangle the following formulas are valid
π β ππ =
sin (12(πΌ β π½))
cos (12πΎ)
, π + ππ =
cos (12(πΌ β π½))
sin (12πΎ)
. (18)
Exercise 16. Given is a triangle π΄π΅πΆ with centroid π and an arbitrary point π. Show that forthe sum of squares holds
|ππ΄|2 + |ππ΅|2 + |ππΆ|2 = |ππ΄|2 + |ππ΅|2 + |ππΆ|2 + 3|ππ|2. (19)
Also show the converse, that is, if the point π satisfies the above equation, for every point π of theplane, then it coincides with the centroid of the triangle.
Exercise 17. Show that, for a triangle π΄π΅πΆ with circumcircle π(π, π ) and centroid π, the folowβing formula is valid
|ππ|2 = π 2 β 19(π2 + π2 + π2). (20)
6 Heronβs triangle area formula
This formula relates the area Ξ of the triangle π΄π΅πΆ to its sideβlengths {π, π, π} and thehalfβperimeter π = (π + π + π)/2 βΆ
Ξ2 = π β (π β π) β (π β π) β (π β π). (21)
The proof of Heronβs formula follows by substituting in the formula for the area Ξ = π β πthe radius π through the formula of theorem 5 (see figure 8β(I)).
7 A symmetric equilateral hexagon 9
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Figure 8: Heronβs proof Symmetric inscribed hexagon
Exercise 18. Let {π΅β²πΆβ², πΆβ³π΄β², π΄β³π΅β³} be tangents of the inscribed circle π (π) of the triangle π΄π΅πΆ,respectively parallel to the sides {π΅πΆ, πΆπ΄, π΄π΅}. Show that the hexagon π΄β²π΄β³π΅β³π΅β²πΆβ²πΆβ³ is symβmetric and has its opposite sides equal and parallel (see figure 8β(II)). Show also that the sum ofthe inradii of the small circles ππ΄ + ππ΅ + ππΆ = π. Finally show that the circumcircles of the smalltriangles are tangent to the circumcircle of triangle π΄π΅πΆ.
Exercise 19. Show that the area Ξ of the triangle π΄π΅πΆ is expressed with the help of its altitudes{βπ΄, βπ΅, βπΆ} through the formula:
1Ξ2 = ( 1
βπ΄+ 1
βπ΅+ 1
βπΆ) β (β 1
βπ΄+ 1
βπ΅+ 1
βπΆ) β ( 1
βπ΄β 1
βπ΅+ 1
βπΆ) β ( 1
βπ΄+ 1
βπ΅β 1
βπΆ) .(22)
Remark 2. In the articles of Baker [Bak85a], [Bak85b] are contained 110 formulas for thearea Ξ of the triangle among which the following interesting, in that they relate the areaΞ to the areas of other triangles intimately related to π΄π΅πΆ βΆ
Ξ = πΏ β 12 cos(πΌ) cos(π½) cos(πΎ) , (23)
= π β (π + π)(π + π)(π + π)2πππ , (24)
= π β 2π π . (25)
In these πΏ is the area of the βorthicβ triangle, with vertices the feet of the altitudes, π isthe area of the triangle having for vertices the traces of the internal bisectors, and π isthe area of the triangle having for vertices the points of tangency of the incircle.
7 A symmetric equilateral hexagon
Intimately related to the triangle π΄π΅πΆ is the hexagon created by projecting on the sidesof the triangle its excenters and extending them to their intersections {π΄3, π΅3, πΆ3}. Thiscreates the hexagon β = πΌπ΄π΅3πΌπΆπ΄3πΌπ΅πΆ3 (see figure 9) of which next theorem sums up thebasic properties.
Theorem 6. The sides of β are equal to the circumdiameter 2π of π΄π΅πΆ. The hexagon is symβmetric w.r. to the circumcenter π of π΄π΅πΆ and its exterior angles equal those of the triangle.
Proof. The claim about the outer angles is obvious. The claim on the equality of sidesfollows from the triangles {π΄3πΌπ΅πΌπΆ, π΅3πΌπΆπΌπ΄, πΆ3πΌπ΄πΌπ΅} easily seen to be isosceli. The symβmetry follows from the equality of sides and the parallelity of opposite sides, since these
7 A symmetric equilateral hexagon 10
are orthogonal, each to a corresponding side of π΄π΅πΆ. That the center is π follows byconsidering a diagonal, πΌπΆπΆ3 say. The symmetry center is on the parallel to πΌπΆπ΅3, whichpasses through the middle of π΄1π΄2, which is identical with the middle of π΅πΆ. Finally,the claim about 2π follows by calculating the side π΄3πΌπ΅ in terms of πΌπ΅πΌπΆ, which in turnis expressible in terms of {πΌπ΅π΅β³, πΌπΆπΆβ³}.
Corollary 1. The lengths of the sides of the triangle πΌπ΄πΌπ΅πΌπΆ are
πΌπ΄πΌπ΅ = 4π cos(πΎ/2), πΌπ΅πΌπΆ = 4π cos(πΌ/2), πΌπΆπΌπ΄ = 4π cos(π½/2). (26)
Proof. Follows by applying the cosine formula to triangle π΄3πΌπ΅πΌπΆ and his alikes.
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Figure 9: Equilateral symmetric hexagon
Corollary 2. The lengths of the diagonals of the hexagon β are
(πΌπ΄π΄3)2 = 4π (π + 2ππ΄), (πΌπ΅π΅3)2 = 4π (π + 2ππ΅), (πΌπΆπΆ3)2 = 4π (π + 2ππΆ). (27)
Proof. Follows by applying the parallelogram rule π₯2 + π¦2 = 2(π2 + π2) relating the diβagonals {π₯, π¦} to the sides {π, π}. Indeed, setting
π₯ = (πΌπ΄π΄3)2, π¦ = (πΌπ΅π΅3)3, π§ = (πΌπΆπΆ3)2, and π’ = (πΌπ΅πΌπΆ)2, π£ = (πΌπΆπΌπ΄)2, π€ = (πΌπ΄πΌπ΅)2
and applying this rule we obtain the simple linear system w.r. to {π₯, π¦, π§}:
π§ + π¦ = 2(4π 2 + π’),π₯ + π§ = 2(4π 2 + π£),π¦ + π₯ = 2(4π 2 + π€).
7 A symmetric equilateral hexagon 11
This has the solution
π₯ = π£ + π€ β π’ + 4π 2, π¦ = π€ + π’ β π£ + 4π 2, π§ = π’ + π£ β π€ + 4π 2,
which, in view of corollary 1, leads to the claimed formulas. To see this, after applyingcorollary 1, use the trigonometric identity
cos2(πΌ/2) + cos2(π½/2) β cos2(πΎ/2) = 2 cos(πΌ/2) cos(π½/2) sin(πΎ/2), (28)
and the formulas of exercise 14.
Remark 3. The formulas of corllary 2 are equivalent to βEulerβs theoremβ connecting theradii {π , ππ΄} with the distance |ππΌπ΄| of the centers of the corresponding circles. In fact,since it holds |ππΌπ΄| = |πΌπ΄π΄3|/2, and the likes for the other centers, these formulas becomeequivalent to
|ππΌπ΄|2 = π (π + 2ππ΄), |ππΌπ΅|2 = π (π + 2ππ΅), |ππΌπΆ|2 = π (π + 2ππΆ), (29)
which are Eulerβs formulas for the circumcenter and the excenters of the triangle. The corβresponding formula for the incenter is handled below.
Remark 4. The correspondence of the hexagon to the triangle π΄π΅πΆ β β is reversible. Thetriangle can be constructed if we know the symmetric equilateral hexagon β. For this itsuffices to construct the triangle of the diagonals πΌπ΄πΌπ΅πΌπΆ and take the orthic π΄π΅πΆ of thelatter. The selection of the other diagonals and the corresponding triangle π΄3π΅3πΆ3 leads,because of the symmetry w.r. to π to a congruent to π΄π΅πΆ triangle.
Exercise 20. Construct a triangle π΄π΅πΆ knowing its circumradius π and1. an angle πΌ and one of the sides {πΌπ΄πΌπ΅, πΌπ΄πΌπΆ.}2. a sum of two sides π + π and the corresponding angle πΌ.3. two sums of sides {π + π, π + π}.4. an exradius ππ΄ and the sum π + π .5. two exradii {ππ΄, ππ΅}.
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Figure 10: Double area (π΄π΅πΆπ·πΈπΉ) = 2(π΅π·πΉ)
Exercise 21. In the convex symmetric hexagon π΄π΅πΆπ·πΈπΉ show that the triangle of the diagonalsπ΅π·πΉ has half the area of the hexagon (see figure 10).
Hint: Show that the hexagon has area equal to the parallelogram π΅π·π΅β²πΉ.
8 Eulerβs theorem for the incenter/excenter 12
Exercise 22. Show that the triangles {π΄β²π΅β²πΆβ², πΌπ΄πΌπ΅πΌπΆ} are homothetic (see figure 11). Show alsothat their homothety center πΏ is described in trilinear coordinates through
πΏ = ( 1π β π βΆ 1
π β π βΆ 1π β π ) .
Hint: The homothety results from the parallelity of corresponding sides of the triangles.The length ratio of two parallel sides, {π΄πΆ/πΌπ΄πΌπΆ} say, is easily computed to be:
π΄β²πΆβ² = 2π΅πΆβ² sin(π½/2), πΌπ΄πΌπΆ = 4π cos(π½/2) β π΄β²πΆβ²
πΌπ΄πΌπΆ= π β π
2π tan(π½/2).
Using this and the formulas established so far, we find that
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Figure 11: Similar triangles
πΏπ΄β = Ξππ΄2ππ β Ξ = πππ΄
2π β π = ( π2π2π β π) 1
π β π,
in which the factor before 1/(π β π) is independent of the particular side of the triangle.
Remark 5. Point πΏ coincides with the βtriangle centerβ π57 in Kimberlingβs list ([Kim18]).It is the βisogonal conjugateβ of the soβcalled βMittenpunktβ π9 = (π β π βΆ π β π βΆ π β π) ofthe triangle π΄π΅πΆ.
8 Eulerβs theorem for the incenter/excenter
Every triangle has a circumscribed circle πΏ and an inscribed π . If we hide the trianglewe see two circles (π in the interior of πΏ) and the question rises, whether there are othertriangles which have these two circles respectively as inscribed and circumscribed. Moregenerally, for two circles {π , πΏ}, the first of which is contained in the second, one may ask,whether there is a triangle circumscribed to the first and inscribed to the second.
The next Eulerβs (1707β1783) theorem (one of many [Ric08], [Nah06]) leads to the exβpression of the power of the incenter πΌ relative to the circumcircle of the triangle as a
8 Eulerβs theorem for the incenter/excenter 13
function of the radii of the circumcircle and the incircle (see figure 12β(I)). The formulawhich results gives also a quantitative criterion which answers the previous question, ofthe existence of a triangle circumscribed/inscribed to two given circles.
Theorem 7. (Eulerβs theorem) In every triangle the radius π of its circumcircle, the radius π ofits incircle and the distance ππΌ of the centers of these circles are related through the formula
|ππΌ|2 = π (π β 2π). (30)
Proof. The formula reminds of the power π(πΌ) = |ππΌ|2 β π 2 of the incenter πΌ (center of theinscribed circle) relative to the circumcircle, especially if we write it as |ππΌ|2 β π 2 = β2ππ .It suffices therefore to show that the power of the incenter πΌ relative to the circumcircle isβ2ππ . The basic observations which lead to the proof are two.
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Figure 12: Eulerβs theorem Triangles with the same incircle and circumcircle
First, that the extension of the bisector π΄πΌ passes through the middle π· of the arc π΅πΆand the line segments {π·π΅, π·πΌ, π·πΆ} are equal (See Figure 12βI). Segments π·π΅ and π·πΆ areof course equal, since π· is the middle of the arc π΅πΆ . Triangle π΅π·πΌ is however isosceles,because its angle at π΅ is the sum πΌ+π½
2 and the same happens with its angle at πΌ, as externalof the triangle π΅πΌπ΄. The power of πΌ then is |π΄πΌ||πΌπ·| = |π΄πΌ||π·πΆ|.
The second observation is that the right triangles π΄πΌπ and πΈπ·πΆ are similar. Here, πis the projection of πΌ on π΄πΆ, therefore its length is |πΌπ| = π. Point πΈ is the diametricallyopposite of π·. Obviously the triangles are similar because they have their acute angles atπ΄ and πΈ equal. Then their sides are proportional:
|π΄πΌ||πΌπ| = |πΈπ·|
|π·πΆ| β |π΄πΌ||π·πΆ| = |πΌπ||πΈπ·|,
which, with what we said, translates to the claimed:
β π(πΌ) = |π΄πΌ||πΌπ·| = |π΄πΌ||π·πΆ| = |πΌπ||πΈπ·| = π(2π ).
Theorem 8. If for two circles π (πΌ, π) and πΏ(π, π ) holds the relation |ππΌ|2 = π (π β 2π), then forevery point π΄ of πΏ there exists a triangle π΄π΅πΆ inscribed in πΏ and cirumscribed to π .
Proof. To begin with, π is inside circle πΏ. This is seen by considering a point π of π andcalculating the difference
|ππ|2 < (|ππΌ| + |πΌπ|)2 < |ππΌ|2 + |πΌπ|2 = π2 + π 2 β 2π π = (π β π)2,
8 Eulerβs theorem for the incenter/excenter 14
where in the previous to last equality we used the assumed relation. We draw thereforefrom an arbitrary point π΄ of πΏ the tangents to π , which intersect again the circle πΏ at π΅and πΆ (See Figure 12βII). It suffices to show that line π΅πΆ is tangent to π . To see this, wedraw the bisector of the angle π΅π΄πΆ, which intersects the circle πΏ at the middle π· of arcπ΅πΆ . Suppose again that πΈ is the diametrically opposite ofπ· andπ is the projection of πΌ onπ΄πΆ. The right triangles π΄πΌπ and πΈπ·πΆ have their acute angles at π΄ and πΈ equal, thereforethey are similar. Consequently
|π΄πΌ||πΌπ| = |πΈπ·|
|π·πΆ| β |π΄πΌ||π·πΆ| = |πΌπ||πΈπ·| = π(2π ).
However, by hypothesis, the power of πΌ relative to πΏ is alsoβπ(πΌ) = |π΄πΌ||πΌπ·| = π 2 β|ππΌ|2 =2π π. Therefore
|π΄πΌ||πΌπ·| = |π΄πΌ||π·πΆ| β |πΌπ·| = |π·πΆ|.
Thus, there are defined two isosceli triangles, π΅π·πΌ and π·πΌπΆ. Let π = πΌπ΅π΄. Taking intoaccount that angles π·π΅πΆ and π·π΄π΅ are equal to πΌ/2 , where πΌ = π΅π΄πΆ, we have
π΅πΌπ· = π + πΌ2 = πΌπ΅π· = πΌπ΅πΆ + πΌ
2 β πΌπ΅πΆ = π.
This means that π΅πΌ is a bisector of the angle π΄π΅πΆ hence π΅πΆ is also tangent to π .
Next exercise settles the analogous relation between the circumcenter π and the exβcenter πΌπ΄ of a triangle.
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Figure 13: Eulerβs theorem for escribed Circle πΏ(πΌ, β2 β π)
Exercise 23. Show that the Eulerβs theorem 7 as well as 8 holds similarly also for the escribedcircle of triagle π΄π΅πΆ. Specifically, if πΌπ΄(ππ΄) is the escribed circle contained in the angle π΄, then
|ππΌπ΄|2 = π (π + 2ππ΄). (31)
Conversely, if for two circles π (πΌ, π) and πΏ(π, π ) holds |ππΌ|2 = π (π + 2π), then for every point π΄of πΏ there is a triangle π΄π΅πΆ inscribed in πΏ and escribed to π .
Hint:With the notation of figure 13β(I), the proof of the theorem 7 is transferred verbatimto this case. An alternative proof was given in section 7. The proof of the converse claimis similar. The key in the transfer of these proofs is the fact that the circle π with diameterπΌπΌπ΄ passes through points {π΅, πΆ} and has for center the point π·.
8 Eulerβs theorem for the incenter/excenter 15
Exercise 24. Assume that the tangent ππ· of the inscribed circle π (πΌ, π), which is parallel to thebaseπ΅πΆ of the triangleπ΄π΅πΆ, intersects sideπ΄π΅ at the pointπ. Show that the circleπ, with diameterππΆ, intersects the altitude π΄π΄β² at a point πΎ, whose distance from the incenter πΌ is equal to β2 β π.
Hint: Show that π΄ lies on the radical axis of circles πΏ(πΌ, β2 β π) and π (see figure 13β(II)), bycalculating its two powers {πΏ1, πΏ2} relative to the circles {πΏ, π}. If {π, π, π = |π΄π΄β²|} denoterespectively the half perimeter, the radius of the inscribed circle and the altitude from π΄,these powers are
πΏ1 = (π β π)2 β π2 = πΏ2 = (π β 2π)ππ cos(πΌ)π = 1
2(π2 + π2 β π2).
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Figure 14: Rectangle of tritangent centers
Exercise 25. For the triangles {π΄π΅πΆ, π΄β²π΅πΆ} with common base π΅πΆ we consider the four centersthe tritangent of the inscribed and escribed cirlces in angle π΄ (see figure 14). Show that the corβresponding quadrilateral πΌπΎπΏπ is a rectangle centerred at the middle π of the arc π΅πΆ of thecircumcircle π .
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Figure 15: Two similar rectangles
9 Some properties of the bisectors 16
Exercise 26. Continuing the preceding exercise, consider all four tritangent circles of the twotriangles {π΄π΅πΆ , π΄β²π΅πΆ}. Show that their centers are the vertices of two similar rectangles (seefigure 15). For this consider the diameter ππβ² of π orthogonal to π΅πΆ and show that the similaritywith center π΅, rotation angle ππ΅πβ² and ratio π‘ = π΅πβ²/π΅π transforms the rectangle with centerπ to that with center πβ².
Exercise 27. Using the notation of this section and figure 15βI, show the relations:
|πΆπ·| = π2 cos (πΌ
2 )= 2π sin(πΌ
2) , (32)
|πΌπ΅| = |πΌπΌπ΄| sin(πΎ2 ) = 4π sin(πΌ
2) sin(πΎ2 ) , (33)
π = 4π sin(πΌ2) sin(π½
2 ) sin(πΎ2 ) , (34)
π β π = |π΄πΌ| cos(πΌ2) = 4π sin(π½
2 ) sin(πΎ2 ) cos(πΌ
2) , (35)
cos(πΌ2) = π
|π΄πΌπ΄| , π΄π·πΈ = |π½ β πΎ|2 (36)
|π΄π·| = 2π cos(|π½ β πΎ|2 ) , |π΄πΈ| = 2π sin(|π½ β πΎ|
2 ) , (37)
π2 β π2 β 4π π = 12(π2 + π2 + π2), π2 + π2 + 4ππ = ππ + ππ + ππ. (38)
Hint: The relations in the first five lines result immediately from the figure and the ruleof sine for triangles. For the relations of the sixth line start from tan (πΌ
2 ) = ππβπ and the
relations of theorem 5. The last equation follows from the previous one and the identity2(ππ + ππ + ππ) = (π + π + π)2 β (π2 + π2 + π2).
9 Some properties of the bisectors
Exercise 28. Show that the line π΄π·, passing through a vertex of triangle π΄π΅πΆ and intersectingthe opposite side at π· and the circumcircle at πΈ is a bisector of the angle π΅π΄πΆ, if and only if it holds(see figure 16β(I)):
|πΈπ·||πΈπ΄| = |πΈπ΅|2 = |πΈπΆ|2. (39)
Hint: Point πΈ is the middle of the arc π΅πΆ and the triangles π΄π΅πΈ and π΅π·πΈ are similar.
Exercise 29. Show that for the isosceles triangle π΄π΅πΆ and the point π· of its base π΅πΆ and with thenotation {π = |π΄π΅|, π = |π΄π·|, π₯ = |π΅π·|, π¦ = |π·πΆ|}, the relation π2 = π2 + π₯ β π¦ holds true.
Hint: Almost identical to exercise 28 (see figure 16β(I)).
Exercise 30. Construct a triangle from its elements πΌ = |π΅π΄πΆ|, π = |π΅πΆ| and πΏπ΄ = |π΄π·|, whereπ΄π· is the bisector of angle π΄.
Hint: From the first two given elements the circumcircle of the wanted triangle π΄π΅πΆ canbe constructed and the position of π΅, πΆ on it can be determined. Using figure 16β(I), theconclusion of the previous exercise, and setting π₯ = |πΈπ΄|, we have
π₯(π₯ β πΏπ΄) = |π΅πΈ|2,
from which |πΈπ΄| is determined. With center πΈ and radius |πΈπ΄| we draw a circle whichintersects the previously constructed circumcircle at π΄.
9 Some properties of the bisectors 17
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Figure 16: Relation with bisector Relations of segments on the bisector
Exercise 31. Given is a triangle π΄π΅πΆ with incenter πΌ, and in the bisector from π΄ the excenterπ, the intersection πΈ with the circumcircle and the intersection point π· with π΅πΆ. Show that thecircles {π , π} with centers, respectively, {πΌ, π} and radii {|πΌπ·|, |ππ·|} define points {π·β², πβ²} on π΄πΆ(see figure 16β(II)), such that the triangles {π΄π΅πΈ, π΄π·β²πΌ, π΄π·πΆ, π΄πβ²π} are similar. Also similar arethe triangles {π΄πΌπ΅, π΄πΆπ}, as well as the triangles {π΄πΌπΆ, π΄π΅π} and the following relations hold:
1. |π΄πΈ| β |πΌπ·| = |ππΌ| β |π΄πΌ|.2. |π΄πΈ| β |π΄π·| = |π΄πΌ| β |π΄π| = |π΄π΅| β |π΄πΆ|.3. |πΌπ΄| β |ππ·| = |ππ΄| β |πΌπ·|.
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Figure 17: {π΄, π·, πΈ} collinear Traces of altitudes, bisectors and medians
Exercise 32. In the triangle π΄π΅πΆ, the middles of the sides are respectively π΄β², π΅β², πΆβ², the tracesof the altitudes are π΄β³, π΅β³, πΆβ³, the orthocenter is π» and π΄1, π΅1, πΆ1 are the middles of π»π΄, π»π΅,π»πΆ. Show that the points of intersection of the lines π· = (π΄β²πΆβ³, π΄1π΅β²) and πΈ = (π΄1πΆβ², π΄β²π΅β³)are contained in the parallel to π΅πΆ from π΄. Moreover line πΈπ΄1 is a bisector of the angle π΄πΈπ΄β² andline π·π΅β² is a bisector of the angle π΄π·π΄β². Show that the circle with diameter π΄π» is the inscribedcircle of triangle π·πΈπ΄β² (see figure 17β(I)).
Exercise 33. Let {π·, πΈ, π»} be respectively the traces of the altitude, bisector and median on theside π΅πΆ of the triangle π΄π΅πΆ. Let also π be the projection of the incenter πΌ on the side π΅πΆ. Showthat |π»π|2 = |π»π·| β |π»πΈ| (see figure 17β(II)). Using |ππ»| = |πβπ|
2 , construct the triangle, whosegiven are the lengths {ππ΄ = |π΄π·|, ππ΄ = |π΄π»|, |π β π|}.
10 Eulerβs line and circle of the triangle 18
Hint: The relation follows from the equality of the first and last term of the equations
|π·π||ππ»| = |π΄πΌ|
|πΌπ½| = |π΄πΌ||π΅π½| = |πΌπΎ|
|π»π½| = |πΌπ||π»π½| = |ππΈ|
|πΈπ»| β |πΈπ»| β |π»π· β π»π| = |ππ»| β |ππΈ|.
10 Eulerβs line and circle of the triangle
Theorem 9. The centroid πΊ of the triangle π΄π΅πΆ is contained in the line segment with end pointsthe circumcenter π and the orthocenter π» and divides it into ratio 1:2 (|πΊπ»| = 2|πΊπ|). The lineπ = ππ» is called βEuler lineβ of the triangle (see figure 18).
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Figure 18: The Euler line π»π
Proof. Consider the diametrically opposite πΌ of vertex π΅ relative to the circumcircle. π΄ andπΆ are also on the circumcircle, therefore they see the diameter π΅πΌ under a right angle. Itfollows easily thatπ΄π»πΆπΌ is a parallelogram, hence |π΄π»| = |πΌπΆ|. However, ifπ· is themiddleof π΅πΆ, ππ· joins side middles of the triangle π΅πΆπΌ, therefore |πΌπΆ| = 2|ππ·| and consequentlyπ΄π» has twice the length and is parallel to ππ·. Suppose πΊ is the intersection point of themedian π΄π· with ππ». Triangles π΄π»πΊ and π·ππΊ are similar, having corresponding anglesequal. Consequently, their sideswill be proportional and, because |π΄π»| = 2|ππ·|, the samewill happen also with the other corresponding sides, in other words:
|π΄πΊ| = 2|πΊπ·| and |π»πΊ| = 2|πΊπ|,
the first characterizing the centroid and the second proving the claim.
βEulerβs circleβ of the triangle, is the one which passes through the three middles ofthe sides. Its particularity lies in the fact that it also passes through six more noteworthypoints of the triangle. Thatβs why it is often called βcircle of nine pointsβ of the triangle (seefigure 19).
Theorem 10. The circle π , which passes through the middles π, π, π½ of the sides of triangle π΄π΅πΆ,has the following properties:
1. It passes also through the traces {π·, πΈ, π} of the altitudes of the triangle.2. It passes also through the middles {π, π, πΌ} of the line segments which join the vertices with
the orthocenter π» of the triangle.3. Its center π is the middle of the segment which joins the orthocenterπ» with the circumcenter
π of the triangle.4. Its radius |ππΌ| is half that of the radius π = |ππ΄| of the circumscribed circle πΏ of the triangle.5. Point π» is a similarity center of π and the circumcircle of the triangle, with similarity ratio
1 βΆ 2.
11 Some properties of Eulerβs circle 19
Proof. The proof relies on the existence of three rectangles which have, by two, a commondiagonal. The rectangles are ππππ½ , ππππΌ and πΌπ½ππ . First, let us see that these rectβangles exist. I show that ππππ½ is such a rectangle. The proof for rectangles ππππΌ andπΌπ½ππ is similar.
In ππππ½ then, ππ½ joins themiddles of sides of the triangleπ΅π»π΄. Therefore it is paralleland the half of π»π΄. Similarly ππ joins the middles of sides of triangle π΄π»πΆ. Therefore itis parallel and the half of π»π΄. Consequently ππ½ and ππ, being parallel and equal, theydefine a parallelogram ππππ½ . That this is actually a rectangle, follows from the fact that
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Figure 19: Eulerβs circle π of the triangle π΄π΅πΆ
ππ joins middles of sides of triangle π»π΅πΆ, therefore it is parallel and the half of π΅πΆ. Sinceπ΄π· and π΅πΆ are mutually orthogonal, the same will happen also with their parallels ππ½and ππ .
The three rectangles have by two a common diagonal, which is the diameter of theircircumscribed circle. This implies that the three circumscribed circles of these rectanglescoincide. This completes the proof of the first two claims of the proposition.
For the proof of the remaining two claims, it suffices to observe that in the triangleπ»ππ΄ the segment ππΌ joins the middles of the sides, therefore it is parallel and the halfof ππ΄. However ππ΄ is a radius of the circumscribed circle πΏ and ππΌ is a radius of circleπ . Latter follows from the fact |π΄π»| = 2|ππ| , therefore πΌπ»ππ is a parallelogram and itsdiagonals are bisected at π . However πΌπ, as seen previously, is a diameter of the circleπ . The last claim is a consequence of the two previous ones.
11 Some properties of Eulerβs circle
Here we discuss some properties of Eulerβs circle formulated as exercises.
Exercise 34. Show that the triangle π΄π΅πΆ coincides with the orthic triangle of triangle πΌπ΄πΌπ΅πΌπΆ,with vertices the excenters of π΄π΅πΆ. Conclude that the circumcircle of the triangle π΄π΅πΆ coincideswith the Euler circle of πΌπ΄πΌπ΅πΌπΆ (see figure 20).
11 Some properties of Eulerβs circle 20
Exercise 35. The orthocenter π» of the triangle π΄π΅πΆ is projected on the bisectors of angle π΄,internal and external, at the points π and π. Show that the line ππ passes through the middle ofπ΅πΆ and the center of its Euler circle.
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Figure 20: Circumcircle of π΄π΅πΆ as Euler circle of πΌπ΄πΌπ΅πΌπΆ
Exercise 36. Show that for every triangle π΄π΅πΆ with orthocenter π», the Euler circles of the triβangles π΄π΅πΆ, π΄π΅π», π΅πΆπ» and πΆπ΄π» coincide (see figure 21β(I)). Conclude that these four triangleshave circumscribed circles of equal radii.
Exercise 37. The diagonals of the quadrilateral π΄π΅πΆπ· define four triangles π΄π΅πΆ, πΆπ·π΄, π΅πΆπ·,π·π΄π΅ (see figure 21β(II)). Show that the Euler circles of these triangles pass through a commonpoint π.
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Figure 21: Common Euler circle Intersection of four Euler circles
Hint:Consider the parallelogram πΈππ»π½ of themiddles of the sides of the quadrilateral andthe middles {πΌ, πΎ} of the diagonals. Let also π be the intersection of two of these Euler
11 Some properties of Eulerβs circle 21
circles e.g. of {π΄π΅π·, π΄πΆπ·}. Show that the other Euler circles pass through the same point,proving f.e. that π½ππΎ and π½ππΎ are supplementary angles.
Exercise 38. The intersection point π of the diagonals of the quadrilateral π΄π΅πΆπ· defines fourtriangles π΄π΅π, π΅πΆπ, πΆπ·π, π·π΄π. Show that the circumcenters {πΈ, π, π», π½} of these triangles arevertices of a parallelogram. Show the same also for the centers {πΌ, πΎ, πΏ, π} of the Euler circles ofthese triangles (see figure 22β(I)).
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(Ξ) (ΞΞ)
Ξ
L
Ξ
ΞΞ Ξ
Figure 22: Four circumscribed circles Four Euler circles
Exercise 39. For the quadrilateral and the four triangles, defined in the previous exercise, showthat the radical axes of the Euler circles of these triangles define the sides and the diagonals of aparallelogram, which is similar to πΌπΎπΏπ (see figure 22β(II)).
Ξ
Ξ
CΞ''
Ξ''
C''
C'
Ξ'
Ξ'
Ξ
C1Ξ
1
Ξ1
Ξ
Ξ
Ξ2
Ξ2
C2
Figure 23: Property of the center of the Euler circle
Exercise 40. For a triangle π΄π΅πΆ the middles of its sides are respectively {π΄β², π΅β², πΆβ²}, the tracesof its altitudes are {π΄β³, π΅β³, πΆβ³}, the orthocenter, the center of mass and the center of its Eulercircle are respectively {π», π, πΈ} (see figure 23). Also {π΄1, π΅1, πΆ1} are respectively the middles of
12 Feuerbachβs theorem 22
{π»π΄, π»π΅, π»πΆ}. Show that the triangle with vertices the centroids {π΄2, π΅2, πΆ2} of the respectivetriangles {π»π΅πΆ, π»πΆπ΄, π»π΄π΅} is homothetic toπ΄π΅πΆ with center of homothecy the point πΈ and ratioof homothecy 1:3. Also the orthocenter of π΄2π΅2πΆ2 coincides with π.
Exercise 41. Given a triangle π΄π΅πΆ and a point π· not contained in the sideβlines of the triangle,show that the Euler circles {π π΄, π π΅, π πΆ} respectively of triangles {π·π΅πΆ, π·πΆπ΄, π·π΄π΅} intersect ata point πΏ contained in the Euler circle of the triangle π΄π΅πΆ.
Hint: Consider the second intersection point πΏ of circles π π΅, π πΆ (the first is the middle πΈof the segment π·π΄) (see figure 24β(I)). From the inscriptible quadrilateral πΏπ»π½πΈ, the angle
Ξ
ΞC
D
Ξ
Ξ Ξ
JΞ
Ξ
LΞΊΞ
ΞΊΞ
ΞΊC
Ξ΅
Ξ
Ξ
Ξ
Ξ
L
C
D
Ξ
(Ξ) (ΞΞ)
Figure 24: Concurring circles of Euler Symmetric circumscribed hexagon
πΈπΏπ» is supplementary to πΈπ½π», which is equal to πΈπ·π», because πΈπ½π»π· is a parallelogram.Similarly, ππΏπΈ is supplementary to ππ·πΈ and ππΏπ» is equal to ππ·π». The circle π π΄ will alsopass through πΏ. Next show that point πΏ lies on the Euler circle of π΄π΅πΆ by proving thatpoint πΏ sees πΎπΌ under an angle of measure 180β β π½.
Exercise 42. Show that for every acuteβangled triangle π΄π΅πΆ there exists a convex symmetrichexagon π΄π·π΅ππΆπΈ with equal sides, whose center coincides with the center πΏ of its Euler circle(see figure 24β(II)). Show that, conversely in each convex symmetric hexagon with equal sides, thetriangle which results by taking non successive vertices of it has as center of its Euler circle thecenter of symmetry of the hexagon.
Exercise 43. Construct a triangleπ΄π΅πΆ for which are given the position of the vertexπ΄, the positionof the projection π· of π΄ on the opposite side π΅πΆ and the position of the center π of its Euler circle.
12 Feuerbachβs theorem
In 1822 Feuerbach (1800β1834), who was a high school teacher, published a small book,which, among other noteworthy theorems on the triangle ([Joh60, p.190]), contained alsothe theorem we prove below. In this proof ([Aud02, p. 110], [CG67, p.117]) the key roleis played by the circle π with diameter πΎπΏ , where {πΎ, πΏ} are the projections on π΅πΆ of theincenter πΌ and of the excenterπ contained in the angleπ΄ of triangleπ΄π΅πΆ (see figure 25β(I)).From theorem 4we know that this circle has its center at themiddleπ΄β² of π΅πΆ. The proof of
12 Feuerbachβs theorem 23
Feuerbachβs theorem results from properties of the inversion relative to that circle. Nexttheorem formulates these properties beginning with the acute triangle with π > π, usingarguments which hold in all cases.
Theorem 11. With the previous definitions and notations, hold the properties:1. The circle π is orthogonal to the inscribed π as well as to the escribed π.2. The intersection point π½ of the bisector π΄πΌ with π΅πΆ is the inverse relative to π of the trace π΄β³
of the altitude.3. The angles with π΅πΆ, of the tangent π to π from π½ and the tangent to the Euler circle at the
middle π΄β² of π΅πΆ are equal to |π½ β πΎ|.4. The inversion relative to π maps the Euler circle π to the line π.
A'
O
L
I
B
A
C
K J
ΞΊ
Ξ»
ΞΌ
Ο-b Ο-b
A''
Ξ² Ξ³
Ξ²-Ξ³
(Ξ) (ΞΞ)
Ρν
F
K'
Figure 25: The circle π and its inversion
Proof. Nrβ1 is obvious, since the radii of the circles atπΎ and πΏ, respectively, are orthogonalto the corresponding ones of π .
Nrβ2 follows fromcalculationsweperformed inprevious sections. According to propoβsition 4, the radius of π is equal to π = |π β π|/2. The length |π΄β²π΄β³| is calculated by considβering the power of π΅ relative to the circle with diameter π΄πΆ, and is found to be equal to|π2βπ2|
2π . The length |π΄β²π½| is calculated through the ratio in which the bisector divides π΅πΆ,and is found to be equal to π|πβπ|
π+π . The claim follows from the fact π2 = |π΄β²π½||π΄β²π΄β³|.Nrβ3 follows, on one hand from the figure 25β(II), which shows that the measure of the
angle at π΄ is equal to |π½ β πΎ|, and on the other from the fact, that the angle at π½ will havemeasure |180β β 2π΅π½π΄|, which is also easily seen to be equal to |π½ β πΎ|.
Nrβ4 follows from the properties of inversion and the previous claims. Indeed, sincethe Euler circle π passes through the center of inversion and through π΄β³, π½, its image willbe a line passing through π½ and forming at π½ an angle equal to that formed by the circle πwith π΅πΆ, therefore coincident with π. This conclusion follows from nrβ3.
BIBLIOGRAPHY 24
Theorem 12. (Feuerbach) In every triangle the Euler circle is tangent to the inscribed, as well as,to its three escribed circles.
Proof. With the notation of the previous theorem, we consider the inversion relative tocircle π (see figure 25β(I)). In it the inverse of the inscribed π is itself and, according to theprevious proposition, the inverse of the Euler circle π is the line π, i.e. the second tangentto π from π½. Since π is tangent to ππππππ, its inverse, which is the circle π, will be alsotangent to π at a point πΉ, which is the inverse of the contact point πΎβ² of π with π. Notethat, because of the symmetry of π relative to the bisector π΄πΌ, point πΎβ² is the symmetric ofπΎ relative to π΄πΌ.
The same argument is applied also to the escribed circle π with center π. This circle,as well, is orthogonal to π and has π as a tangent. Therefore the inverse of this circle isitself and the inverse of the line π, which is the Euler circle, will be tangent to it.
What we said proves then that the Euler circle is tangent simultaneously to the inβscribed and escribed contained in the angle π΄. Similarly we prove that the Euler circle istangent to the escribed circles contained in the other angles.
Bibliography
[Aud02] Michele Audin. Geometry. Springer Verlag, 2002.
[Bak85a] Marcus Baker. A collection of formulae for the area of a plane triangle. Annalsof Mathematics, 1:134β138, 1885.
[Bak85b] Marcus Baker. A collection of formulae for the area of a plane triangle. Annalsof Mathematics, 2:11β18, 1885.
[CG67] H. Coxeter and L. Greitzer. Geometry Revisited. Math. Assoc. Amer.WashingtonDC, 1967.
[Cou80] Nathan Altshiller Court. College Geometry. Dover Publications Inc., New York,1980.
[Joh60] Roger Johnson. Advanced Euclidean Geometry. Dover Publications, New York,1960.
[Kim18] KlarkKimberling. Encyclopedia of Triangle Centers. http://faculty.evansville. edu/ck6/encyclopedia/ETC.html, 2018.
[Nah06] Paul J. Nahin. Dr. Eulerβs Fabulous Formula. Princeton University Press, Princeβton, 2006.
[Ric08] David Richeson. Eulerβs gem. Princeton University Press, Princeton, 2008.
Related material1. Circle Pencils2. Inversion
BIBLIOGRAPHY 25
3. Isodynamic points of the triangle4. Pedals5. Tritangent circles of the triangle
Any correction, suggestion or proposal from the reader, to improve/extend the exposition, is welcomeand could be send by eβmail to: [email protected]