Track / Stage 5 / 388 of 400 #988 of 2000

Problem 988

AIME late

Geometry Difficulty 6.0 Prove it

[Angle between two chords and two secants]

Triangles $T_{1}$ and $T_{2}$ are inscribed in a circle, and the vertices of triangle $T_{2}$ are the midpoints of the arcs into which the circle is divided by the vertices of triangle $T_{1}$ . Prove that in the hexagon formed by the intersection of triangles $T_{1}$ and $T_{2}$ , the diagonals connecting opposite vertices are parallel to the sides of triangle $T_{1}$ and intersect at one point.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let the vertices of triangle $T_{1}$ be denoted as $A, B$ , and $C$ ; the midpoints of the arcs $BC, CA, AB$ as $A_{1}, B_{1}, C_{1}$ . Then $T_{2} = A_{1} B_{1} C_{1}$ . The lines $A A_{1}, B B_{1}, C C_{1}$ are the angle bisectors of triangle $T_{1}$ , so they intersect at a single point $O$ . Let the lines $AB$ and $C_{1} B_{1}$ intersect at point $K$ . It is sufficient to verify that $K O \| A C$ . In triangle $A B_{1} O$ , the line $B_{1} C_{1}$ is both a bisector and an altitude, so this triangle is isosceles. Consequently, triangle $A K O$ is also isosceles. The lines $K O$ and $A C$ are parallel, since $\angle K O A = \angle K A O = \angle O A C$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.