on the harmonic hexagon of a triangle

On the Harmonic Hexagon of a TriangleAuthor(s): John CaseySource: Proceedings of the Royal Irish Academy. Science, Vol. 4 (1884 - 1888), pp. 545-556Published by: Royal Irish AcademyStable URL: http://www.jstor.org/stable/20635933 .

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Casey?On the Harmonic Hexagon of a Triangle. 545

XXX.?On the Harmonic Hexagon op a Triangle. By John

Caset, LLD., F.K.S.

[Read, January 26, 1886.]

Definition I.?If ABC be any triangle; AA', ?Bf, CC its

Bymmedian lines, produced to meet its circumcircle in the points A'y B\ C, the hexagon, whose vertices are the six points A, B'9 C, A1) B, C\ possessing several geometrical properties, it is convenient ta have a definite name for it. I propose to call it the harmonic hexagon of the triangle.

c'

Definition II.?Two triangles, having the same symmedian lines, are called cosymmedians.

Definition III.?The lines AAf, BB\ CC are called the sym median lines of the hexagon.

Proposition I.?The triangles ABC, A'B'C are cosymmedians.

(Fig. 1).



540 Proceedings of the Royal Irish Academy.

Rem.?Since the three lines AA', RR', CC are concurrent, the six points in which they intersect the circle are in involution. I Fence the anharmonic ratio of the four points B, A, C, A* is equal to the anharmonic ratio of their four conjugates B', A', C, A ; hut since A A' is a symmedian of the triangle AR C, the four points B, A, C, A' form a harmonic system. Hence the four points R', A', C, A form a harmonic system. Therefore AA' is a symmedian of the triangle

A'B'C. Similarly, BB', CC are symmedians, and therefore the

triangles are cosymmedians.

Proposition II.?If two triangles be cosymmedians, the sides of one

are proportional to the medians of the other.

Rem.-?The angle B'A'C is equal to the sum of the angles RAA, AA'C: that is, equal to the sum of RR A, ACC; and these angles are, respectively, equal to the angles which the medians from R and C of the triangle ARC make with RC. Hence, if G he the centroid oiABC, the angle RA C is the supplement oiBGC; and therefore the angles of the triangle A'B'C are equal to the angles of a tri angle whose sides are the medians of ARC. Hence the proposition is proved.

Proposition III.?Lemoine's First Circles of two cosymmedian tri

angles are identical.

Bern.?Let 0 be the common circumcentre of the triangles; R its

radius; p, pi, the radii of their first Lemoine Circles; then, by a well-known property of these circles,

Sp2 = R

Hence p = pi; and since the middle point of OR is the centre of each, the circles are identical.

Cor. 1.?Lemoine's Second Circles of two cosymmedian trianglos are

identical. For if p', pi denote the radii of their Lemoine's second circles,

we have

= pi'2. Hence p'

= pi'.

Cor. 2.?The Brocard angles of two cosymmedian triangles are equal. For if the Eroeard angles be <o, cj', we have

OR? 1-3 tan2cu = =1-3 tan2 o/. Hence w = a/.

Rr

Cor. 3.?The Brocard points of a triangle are also the Brocard

points of its cosymmedian.




Let Q, O' be the Brocard points of ABC; then the angle QOK= a> = to'. Hence ? is a Brocard point of the triangle A'B'C.

Cor. 4.?If x, y, %; x', f, %' be the perpendiculars let fall from

their common symmedian point on the sides of the two cosymmedian triangles ABC, A'B'C';

x xf y y' z %' then

- = ? = t = 7- = - =

-7* a a o o c c

Tor - = 2 tan w,

? == 2 tan a/. Hence - = a a a a

Proposition IV.?If the circumeirele of two cosymmedian triangles be inverted from any arbitrary point, the inverses of their six vertices will be the vertices of two other cosymmedian triangles.

Bern.?Let the points inverse to A, B, C; A', B', C, respectively, be Au Bi, Ci; A/, Bx', Cf Now since the anharmonic ratio of any four coneyclic points is equal to the anharmonic ratio of the four

points inverse to them (Sequel, Book YL, Sect, iv., Prop. 8); the four points Bx, Ax, Clf A{ form a harmonic system. Hence

AiAi is a symmedian of the triangle A1B1Ci, and the proposition is proved.

Cor. 1.?The circumeirele of two cosymmedian triangles can be

inverted, so that the points inverse to their vertices will be the vertices of two equilateral triangles.

For if the inverses of A, B, C be the vertices of an equilateral triangle, the inverses of A', B', C will evidently be the vertices of another; but if the centre of inversion be either of the two points common to two of the Apollonian circles of the triangle ABC, it will invert into an equilateral triangle. Hence the proposition is proved.

Cor. 2. ? When two cosymmedian triangles invert into two

equilateral triangles, the harmonic hexagon inverts into a regular hexagon.

Cor. 3.?The inverse of the angular points of a harmonic hexagon from any arbitrary point are the angular points of another harmonic hexagon.

Peopositton Y.?The anharmonic ratio of any four consecutive ver

tices of a harmonic hexagon is constant.

Bern.?The anharmonic ratio is equal to that of four consecutive vertices of a regular hexagon, which is constant.

Cor. 1.?Any side of a triangle is divided in a given anharmonic ratio by the two non-corresponding sides of its cosymmedian.

Eor consider the line AB; the anharmonic ratio of the four points ALMB is equal to the pencil (C AB'A'B), which is given, being equal to that of a corresponding pencil for a regular hexagon.




Cor. 2.?The rectangle contained by the six sides of two cosym median triangles is equal to 27 times the continued product of the sides of the harmonic hexagon.

Rem.?The anharmonic ratio of the four points A, B'f C, A' is

equal to that of corresponding points of a regular hexagon. Hence A C. RA' = 3 . AR. CA; and, multiplying these and two other

corresponding equations, the proposition is proved.

Cor. 3.?The continued product of the three symmedian lines of a harmonic hexagon is equal to eight times the continued product of three alternate sides of the hexagon.

Proposition VI.?If the points of intersection of the sides of two

cosymmedian triangles ARC, A'B'C be denoted by L, M, N, P, Q, R ; then the Brocard points of the triangles are isogonal conjugates, with

respect to the angles of intersection L, M, N, P, Q, R.

Bern.?Join OA, QBf, Q'B, Q'C. JSow the angle QAB is equal to the angle QB'C, being, respectively, the Brocard angles of the triangles ABC, A'B'C. Hence the four points A, L, O, B' are

concyclic. Hence the angle QLB' is equal to ClAB'. In like man ner, the angle WLB - ti'CB. But since the angles BCB', BAB' are equal and the angles Q' CB' and LAO, are Brocard angles, the

angles Of C'B, UAB' are equal; hence the angles Q'LB and B'LQ are equal. Hence the proposition is proved.

Cor. 1.?The six angles of intersection of the sides of the triangles ABC, A'B'C, at the points L, M, N, P, Q, R, are equal, respec tively, to those subtended at either Brocard point by the six sides of the harmonic hexagon.

For, since the points A, Z, O, B are concyclic, the angle ALB' = AQB'.

Cor. 2.?The angle subtended at O by the three alternate sides BC, AB', CA' of the harmonic hexagon are, respectively, equal to those subtended at W by the same side, taken in the order CA', BC, AB'; and, similarly, for the other three sides.

Cor. 3.?The feet of the perpendiculars, let fall from the points ti, Q' on the sides of the two cosymmedian triangles ABC, A'B'C'. are concyclic.

Cor. 4.?If R' be the radius of the circle of Cor. 3, R' = R sin ox

Proposition VII.?If in any triangle ABC a triangle similar to iU cosymmedian be inscribed, the centre of similitude of the inscribed tri angles is the symmedian point of the original triangle ABC.

Bern.?Let K be the symmedian point; then the angle BKC is

equal to the sum of the angles BA C, ABK, &CA; that is, equal to the sum of the angles BAC, B'A'A, AA'C; or the sum of BAC, BA'C. Hence (Sequel, Book III. Prop. 17) the propositon is proved.



Casey?On the Harmonic Hexagon of a Triangle.

Pbopositiow VIII.?If K be the centre of similitude of the triangle ABC (Fig. 2), and a homothetie triangle a?y, the circumeirele of ihn triangle a?y will meet the symmedian lines AA', BB', CC in three new

points a' ?' y', which will be the vertices of a triangle homothetie with the

cosymmedian of the triangle ABC.

Bern.?Since K is the homothetie centre of the triangles ABC, a?y, it is the centre of similitude of their circumcircles. Hence the lines KA', KB', KC1 are divided proportionally in the points a', ?', y'; and therefore the triangles A'B'C', a'?'y' are homothetie. Hence the

proposition is proved. Cor. 1.?The triangles a?y, a'?'f are cosymmedians. Cor. 2.?If the sides of the triangle a?y produced, if necessary,

meet those of ABC in six points; and the sides of a'?'y' meet the sides of A'B'C in six other points, the twelve points are concyclic.

Bern.?The first six points lie on a Tucker's circle of the triangle ABC, and the other six on a corresponding Tucker's circle of A'B'C'; and the proposition will be proved by showing that these circles are identical.



Proceedings of the Bog a I Irish Academy.

Let a? meet AC in E', and ay meet AB in i^, and a" be the middle point of FE'; and 0, ?' the circumcentres of ABC, a?y. Through a" draw a"0" parallel to AO; then 0" will be the middle point of 00', and will be the centre of the Tucker's circle of ABC. N"ow leti?, B! be the circumradii of the triangles ABC, a?y; then we have

2

Also, if pi denote the radius of Lemoine's second circle of the tri angle ABC, we have

Pl : Fa" : : KA : a"A ::?:%(?-?'); hence

Fa =P\~^\

Hence the square of the radius of the Tucker's circle of the triangle ABC is equal to

/B + B'V ^/B-B'Y 2R

and it can be shown that the square of the radius of the correpond ing Tucker's circle of the triangle A'B' C is the same, and they have the same centre. Hence they coincide. q.e.d.

Proposition IX.?If the sides of the triangle a?y meet the sides of ABC in the points B, B'; E, E; F, F'f the three triangles

AFE', BF'B, B'CE are directly similar. Their invariable points are the centroids of these triangles, and their double points are the other points of intersection of the circle through the invariable points with the symmedians of the triangle ABC.

For if a, b, c be the centroids of the triangles AFE', BF'B, B'CE; join aF, bF', and produce them to meet c'; and since a is the centroid of AFE', the angle KaF is the angle between two medians of AFE'; but AFE' is similar to ABC. Hence KaF is equal to the angle between two medians of ABC, and therefore equal to an angle of the

cosymmedian triangle A'B'C, which is easily seen to be A'B'C; therefore KaF is equal to A!AC. Hence aF is parallel to AC. Similarly bF' is parallel to BC. Therefore the figures KA C'B, Kae'b are homothetic. Hence the point is on the line EC; and it is evident that the figures c'FAE', c'F'BB are directly similar. Hence c' is a double point. Similarly a', b' are double points, and it is easy to see that a, b, c are the invariable points.

Cor. 1.?If we make a corresponding construction for the triangles

a'?'y', A'B' C, we shall find that for the new system of three figures directly similar, a, b, c are the double points; and a', V, c' the inva

riable points. Hence the two systems are so related, that the double

points of either system are the invariable points of the other. Cor. 2.?The triangles abc, a'b'c' are cosymmedians.




Proposition X.?If through the symmedian point of a harmonic hexagon parallels be drawn to the tangents at its vertices; then taking the points in which each parallel meets the sides of the hexagon passing through the corresponding vertex, the twehe points of intersection are

concyclic.

Bern.?Let JTbe the symmedian point of the triangle ABC\ then BA\ A'C are two sides of the harmonic hexagon. Through K draw KU parallel to AT, then U will he one of the twelve points of inter section. Draw KV parallel to AT. Now the angle A UK = TA' U = A A V. Similarly the angle A VK = KA' U. Hence the triangles AKV, UK A' are equiangular. Hence KU. KV = AK. KA'. Now KV is the radius of the second Lemoine circle, both for the triangle ABC and its cosymmedian. Hence KU has the same value for each of the twelve points of intersection, and therefore the twelve inter sections are concyclic.

Dep.? We shall, from its analogy to the case of the triangle, call the circle through these twelve points Lemoine1 s second hexagon circle.

Cor. 1.?The intercepts which Lemoine's second hexagon circle make on the sides of the hexagon are proportional to the cosines of the angles which these sides subtend in the circumcircle. Hence it may be called the cosine circle of the hexagon.

Cor. 2.?If the sides of the harmonic hexagon a?'ya'?y' (see flg., Prop, viii.), be produced to meet the sides of AB'CA'BC, viz., each side of the former intersecting the two sides adjacent to the side parallel to it in the latter, the twelve points of intersection are

concyclic. If R, R' be the circumradii of the hexagon AB' CA'BC,

a?'ya'?y', and R" the radius of the cosine circle of AB'CA'BC ', then it may be proved, as Prop. vrn. Cor. 2, that the twelve points of intersection lie on a circle, the square of whose radius is

(??*)\




Cor. 3.?When the hexagon a?'ya'?y' reduces to a point, the circle of Cor. 2 will he called, from analogy, Lemoine's first circle of the hexagon; its radius squared is equal to

R* + R'n

4

Proposition XI.?The perpendiculars from the symmedian point of a harmonic hexagon, on the sides of the hexagon, are proportional to the sides.

j)eMt?The perpendiculars from K on the lines CA, AB' (fig., Prop, viii.), are in the ratio of sin CAR: sin KAB'; that is, as sin CAA': smAAB' or : : CA': AB'; but CA': AB': : CA : AB', since the points C, A', B', A' form a harmonic system. Hence the

perpendiculars from K on the lines CA, AB' are proportional to these lines, &c. Hence the proposition is proved.

Definition I.?If 0 be the circumcentre of the hexagon, the

circle on OR as diameter is called the Brocard circle of the hexagon.

Definition II.?If the sides of the harmonic hexagon be denoted by a, b, c, d, e, f, and the perpendiculars from K on three sides by x, y, %, u, v, w; then the auxiliary angle ? determined by any of the equations x = % a tan %, y = \b tan &c, is called the Brocard angle of the hexagon. ^

Cor.?The radius of the cosine circle of the hexagon is *equal to R tan %.

For, let fall the perpendicular KX on A'C (see fig., Prop, x.); then

KX -r KU= sin U= sin A'A C=A'C+2R. Hence

KU? 2R = KX ? A'C = i tan ? ; therefore

KZT= Rtzn?.

Proposition XII.?The perpendiculars from the centre of the circum circle on the sides of a harmonic hexagon meet its Brocard circle in six

points, which connect concurrently in two waysjaith the vertices of the hexagon.

Bern.?Let the points (Fig. 4) of intersection of the perpendicular with the Brocard circle be L, M, N, P, Q, R. Join KL, KM; then because the angle KLO is right, KL is parallel to BA'. Hence LL' is equal to the perpendicular from iT on BA, Similarly, M'M is equal to the perpendicular from K on A'C. Hence (Prop, xi.) LL': MM': : BL' : A'M'. Hence the triangles BL'L, A'MM are

equiangular; there fore the angle BLL' is equal to A'MM'. Hence, if the lines BL and A'M intersect in O, the four points 0, M, L, Q are concyclic ; there




fore ? is a point on the Brocard circle. Similarly the lines CN, B'P, A Q, C'R each intersect BL on the Brocard circle, and therefore each

passes through ?. In the same manner it may be shown that the six lines A'L, CM, B'N, AP, C'Q, BR are concurrent, and meet in another point ?' on the Brocard circle.

Def.??, ?' are caUed the Brocard points; and L, M, N, P, Q, R the invariable points of the hexagon.

Cor. 1.??, ?' are isogonal conjugates with respect to each angle of the harmonic hexagon AB'CA'BC.

For the angles A'BQ, CBQ,' are each equal to the Brocard angle of the hexagon. Hence, &c.

Cor. 2.?The feet of the perpendiculars from ?, ?' on the sides of the hexagon are coneyclic.

Cor. 3.?If m be the Brocard angle of the triangle ABC, and <s> the Brocard angle of the hexagon, tan 0 = 3 tan o>.




Bern.?2 tan ? = LL! -f A'B = perpendicular from K on A'R -f ^'i?;

and 2 tan o> = perpendicular from Eon BC ? BC. Hence

tan ? perp. from Kon ^l'i? BC tan to perp. from ?Ton BC A'B

sin ̂̂ JT BC 7^7 " ^TCBK

* ?!B~ ITC

* A7!? '

that is, equal to the anharmonic ratio of the four points B, A', Ct B', and therefore equal to the anharmonic ratio of the corresponding

points in a regular hexagon, and therefore equal to 3. Hence

tan 0 = 3 tan <*>.

Proposition XIII.?If figures directly similar be described on the sides of the harmonic hexagon, the middle points of its symmedian lines AA'f BB', CC are each a double point for three pairs of figures.

Bern.?Let A" he the middle point of AA'; then it may he proved, as in Conies, page 247, that A" is the double point of the figures on CA, AB'-, and also of the figures described on BA', A'C; and it remains to be proved that it is a double point of the figures described on BC, B'C. Join BA," A"C; CA", A"B; then, since A" is a double point of the figures CA, AB', we have CA" : A!'A : : A"A : A"B'. Hence CA!'. A"B' = A"A2. Similarly BA" .A"C^ A"A'\

Hence BA" .A"C = CA" . A"Bf; and the angles BA"C, B'A"C are

equal. Hence the triangles BA" C, B'A"C are equiangular, and

they are directly similar. Hence the proposition is proved. Cor. 1.?The four points B, C, K, A" are concyclic. Cor. 2.?If figures directly similar be described on the six sides of

the harmonic hexagon AB'CA'BC, the symmedian lines of the harmonic hexagon, formed by any six corresponding lines, pass

respectively through the middle points A", B", C" of the symmedian lines of AB'CA'BC.

Cor. 3.?In the same case, the locus of the symmedian point of

the hexagon, formed by six corresponding lines of these figures, is the Brocard circle of the original hexagon.

Cor. 4.?The centre of similitude of any two hexagons, each formed by six corresponding lines of figures directly similar described on the sides of AB'CA'BC, is a point on its Brocard circle.

Cors. 2, 3, 4 may be proved exactly as in the corresponding cases

for triangles (see Conies, page 248). Cor. 5.?The six lines joining, respectively, the invariable points

L, M, N, P, Q, R to six corresponding points are concurrent. The locus of their point of concurrence is the Brocard circle of AB'CA'BC, and they form a pencil in involution.

Proposition XIV.?The triangle formed by three alternate sides of the harmonic hexagon is in perspective with the triangle formed by ihe three invariable points corresponding to the three remaining sides.




Bern.?Let us consider the triangle formed by the three sides

BC, AB', CA'. Let them meet in the points A'", C", B'"; it is required to prove that the triangle A"fBf"C" is in perspective with the triaugle NLQ, formed by the invariable points corresponding to the sides B'C, A'B, CA. Join the points JSf, L, Q, respectively, to

K, and produce to meet the sides BC, AB', CA'. They will meet them in points where the same sides are intersected by Lemoine's first hexagon circle; and since Lemoine's first circle and the Brocard circle are concentric, the parts intercepted by them on the line KN will be equal. Hence the lines A"N, A"'K are isotomic conjugates with

respect to the angle A". Similarly, B'"Q, B"'K are isotomic conju gates with respect to the angle B"'; and C"L, C"K with respect to the angle C". Hence the lines A"'N% B"'Q, C"L are concur

rent. q.e.d.

NOTE ADDED IN THE PBESS.

Since writing the preceding Paper, I have succeeded in showing that the propositions contained in it are capable of remarkable exten sions. I do this by proving that we can construct a harmonic polygon of any number of sides that is a cyclic polygon, having a point in its plane called its symmedian point, such that perpendiculars from it on the sides of the polygon are proportional to the sides. The solution of this problem is contained in the following theorem :?

The inverses of the angular points of a regular polygon of any number of sides form the angular points of a harmonic polygon of the same number of sides.

Bern.?Let A, B, C, &c, be the angular points of the original polygon; A, B', C the points diametrically opposite to them. Now invert from any arbitrary point. The circumcircle of the original polygon will invert into a circle, and the lines AA', BB', CC, &c, into a coaxal system, and the radical axes of each circle of this system and the inverse of the circumcircle will be a concurrent system of lines (Sequel, Book YL, Sect, v., Prop. 4). Now if the inverses of the points A, B, C, &c.; A', B', C, &c., be a, ?, y, &c.; a', ?', y', &c.; the concurrent lines will be aaf, ??', yy', &c, respectively; let K be their point of intersection. Now, since evidently the pionts A, B, C, B' form a harmonic system, the points a, ?, y, ?' form a harmonic system. Hence the perpendiculars from the point K in

??' on the lines a?, ?y are proportional to these lines. Hence the proposition is proved.

It is evident now that the whole theory of Brocard circles, Brocard points, Lemoine circles, cosine circles, similar figures, invariable

points, double points, &c, can be extended to harmonic polygons of any number of sides. The following arc a fcAv of the numerous

H.r.A. PROC, SER. II., VOL. IV.?SCIENCE. 3 B




additional propositions that can be given in connexion with these polygons:?

1. Ji the alternate vertices 1, 3, 5 ... 2n - 1 of a harmonic poly gon of 2n sides be joined, the lines of connexion form a harmonic polygon of n sides, and so also do the lines joining the remaining vertices 2, 4, 6 ... 2n.

2. If n be an odd number, the three polygons of Ex. 1 are

cosymmedians. 3. If the symmedian lines AK, BK7 CK, &c, of a harmonic

polygon of an odd number of sides be produced to meet the circum circle again in the points A!, B', C, &c, the points A', B*, C, &c, form the vertices of another harmonic polygon; and these two poly gons are cosymmedian, and have the same Brocard angles, Brocard

points, Lemoine circles, and cosine circles, &c.

4. The four symmedian chords of a harmonic octagon form a harmonic pencil.

5. The circles described through the extremities of the symmedian chords of a harmonic polygon, and intersecting the circumcircle ortho gonaBy, are coaxal, and intersect each other at equal angles.

6. A harmonic polygon of any number of sides can be projected into a regular polygon of the same number of sides, and the proj ec tion of the symmedian point of the former will be the circumcentre of the latter.

7. The symmedian point of any harmonic polygon is the mean centre of the feet of the perpendiculars let fall from it on the sides of the polygon.

8. If ??, ?3n he the Brocard angles of two harmonic polygons of n sides, and 2? sides, respectively ; then

tan ?2n = 4 cos2

^- . tan ?n. A n



on the harmonic hexagon of a triangle

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