Library / Irish Mathematical Olympiad /138 of 236

Irish Mathematical Olympiad

Geometry Difficulty 5.9 AIME, harder Prove it Ireland

Triangle $ABC$ is inscribed in a circle $\Gamma$ . The points $X, Y$ and $Z$ are the midpoints of those arcs $BC, CA$ and $AB$ , respectively, on $\Gamma$ which do not contain the third point of the triangle. The intersection of the triangles $ABC$ and $XYZ$ form a hexagon $DEFGHK$ . Prove that the diagonals $DG, EH$ and $FK$ are concurrent.

Solution

Join $AX$ , $BY$ and $CZ$ . These lines are the bisectors of the angles of $\triangle ABC$ and intersect at its incentre $I$ . Label the vertices of the hexagon as shown below and join $KI$ and $ZA$ .

We have $\angle KZI = \angle XZC = \angle XAC = \angle KAI$ , hence $AZKI$ is a cyclic quadrilateral. This implies $\angle ZIK = \angle ZAK = \angle ZAB = \angle ZCB$ from which we infer that $KI$ and $BC$ are parallel. Similarly it follows that $IF$ and $BC$ are parallel. This shows that $K, I$ and $F$ are collinear, i.e. the diagonal $KF$ passes through $I$ .

The same argument shows that the other two diagonals, $DG$ and $EH$ , also pass through $I$ , thus the three diagonals of the hexagon are concurrent.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.