HMMT, 2014

Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Let $ABC$ be an acute triangle with circumcenter $O$ such that $AB = 4$ , $AC = 5$ , and $BC = 6$ . Let $D$ be the foot of the altitude from $A$ to $BC$ , and $E$ be the intersection of $AO$ with $BC$ . Suppose that $X$ is on $BC$ between $D$ and $E$ such that there is a point $Y$ on $AD$ satisfying $XY \parallel AO$ and $YO \perp AX$ . Determine the length of $BX$ .

Solution

Solution:

Answer: $\dfrac{96}{41}$

Let $AX$ intersect the circumcircle of $\triangle ABC$ again at $K$ . Let $OY$ intersect $AK$ and $BC$ at $T$ and $L$ , respectively. We have $\angle LOA = \angle OYX = \angle TDX = \angle LAK$ , so $AL$ is tangent to the circumcircle. Furthermore, $OL \perp AK$ , so $\triangle ALK$ is isosceles with $AL = AK$ , so $AK$ is also tangent to the circumcircle. Since $BC$ and the tangents to the circumcircle at $A$ and $K$ all intersect at the same point $L$ , $CL$ is a symmedian of $\triangle ACK$ . Then $AK$ is a symmedian of $\triangle ABC$ . Then we can use $\dfrac{BX}{XC} = \dfrac{(AB)^2}{(AC)^2}$ to compute $BX = \dfrac{96}{41}$ .

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