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Berkeley Math Circle Monthly Contest 4

Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Let $ABCDEF$ be a hexagon circumscribing a circle $\omega$ . The sides $AB$ , $BC$ , $CD$ , $DE$ , $EF$ , $FA$ touch $\omega$ at $U$ , $V$ , $W$ , $X$ , $Y$ , and $Z$ respectively; moreover, $U$ , $W$ , and $Y$ are the midpoints of sides $AB$ , $CD$ , and $EF$ , respectively. Prove that $UX$ , $VY$ , and $WZ$ are concurrent.

Solution

Solution:

Since $U$ is the midpoint of $AB$ , we have $ZA = AU = UB = BV$ and so (letting $O$ be the center of $\omega$ ) $\triangle$ OZA $\cong \triangle$ OUA $\cong \triangle$ OUB $\cong \triangle$ OVB. Thus arcs $ZU$ and $UV$ are equal, and so $UX$ is the bisector of $VXZ$ . Similarly, $VY$ and $WZ$ are the bisectors of the other two angles of $\triangle VXZ$ , so these three lines are concurrent.

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