Iran TST-3 2013 Problem 5

by liberator, Aug 19, 2014, 6:41 PM

Problem: In triangle $ABC$ , $AD$ and $AH$ are respectively the angle bisector and the altitude of vertex $A$ . The perpendicular bisector of $AD$ intersects the semicircles with diameters $AB, AC$ , drawn outside triangle $ABC$ , at $X, Y$ , respectively. Prove that the quadrilateral $XYDH$ is cyclic.

Proposed by Mahan Malihi

[b]My solution[/b]

$[asy] /* Iran TST-3 Problem 5, free script by liberator */ unitsize(4cm); pointpen=black; /* Initialize objects */ pair A = dir(120); pair B = dir(200); pair C = dir(340); pair H = foot(A,B,C); pair D1 = bisectorpoint(B,A,C); pair D2 = rotate(180,D1)*A; pair D = intersectionpoint(A--D2, B--C); pair P1 = bisectorpoint(A,D); pair Q1 = reflect(A,D)*P1; pair P = intersectionpoint(P1--Q1, A--B); pair Q = intersectionpoint(P1--Q1, C--A); pair X1 = rotate(180, P1)*Q; pair X = intersectionpoint(P--X1, circumcircle(A,B,H)); pair Y1 = rotate(180, Q1)*P; pair Y = intersectionpoint(Q--Y1, circumcircle(A,C,H)); /* Draw objects */ draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1)); draw(A--H, rgb(0.4,0.6,0.8)); draw(A--D, rgb(0.4,0.6,0.8)); draw(A--X, rgb(0.4,0.6,0.8)); draw(A--Y, rgb(0.4,0.6,0.8)); draw(B--X, rgb(0.4,0.6,0.8)); draw(C--Y, rgb(0.4,0.6,0.8)); draw(D--X, rgb(0.4,0.6,0.8)); draw(H--X, rgb(0.4,0.6,0.8)); draw(D--Y, rgb(0.4,0.6,0.8)); draw(H--Y, rgb(0.4,0.6,0.8)); draw(X--Y, rgb(0.4,0.6,0.8)); draw(A..Y..C, rgb(0.9,0,0)); draw(circumcircle(A,X,B), rgb(0.9,0,0)); draw(circumcircle(Y,D,H), rgb(0,0.6,0)); /* Place dots on and label each point */ dot(A); label("$A$", A, dir(A)); dot(B); label("$B$", B, dir(B)); dot(C); label("$C$", C, dir(C)); dot(H); label("$H$", H, dir(-90)); dot(D); label("$D$", D, dir(-90)); dot(P); label("$P$", P, dir(140)); dot(Q); label("$Q$", Q, dir(Q)); dot(X); label("$X(X')$", X, 2*dir(-118)); dot(Y); label("$Y$", Y, dir(Y)); /* Clip Image */ clip((-1.6, -0.5)--(1.6, -0.5)--(1.6, 1.6)--(-1.6,1.6)--cycle); [/asy]$

Let $P \equiv XY \cap AB, Q \equiv XY \cap AC, X' \equiv XY \cap (YDH)$ , with $X'$ distinct from $Y$ .

$AHCY$ is cyclic, with diameter $AB$ . Now $\angle AX'Y = \angle DX'Y = \angle DHY = \angle CAY$ . Thus

\[\begin{align*} \angle X'AB = \angle APQ - \angle AX'P &= \angle AQP - \angle YAQ \\ &= \angle AYX' \\ &= \angle DYX' \\ &= \angle X'HB, \end{align*}\]

hence $AX'BH$ is cyclic, so $X \equiv X'$ , and the result follows.

Could anybody find a solution using the fact that $A$ is the center of the spiral similarity $\overline{XH} \to \overline{DY}$ ?

This post has been edited 2 times. Last edited by liberator, Aug 31, 2014, 3:44 PM

Angle Chase

geometry

Iran TST

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Iran TST-3 2013 Problem 5

by liberator, Aug 19, 2014, 6:41 PM

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