Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer United States

Problem:

In right triangle $A B C$ , a point $D$ is on hypotenuse $A C$ such that $B D \perp A C$ . Let $\omega$ be a circle with center $O$ , passing through $C$ and $D$ and tangent to line $A B$ at a point other than $B$ . Point $X$ is chosen on $B C$ such that $A X \perp B O$ . If $A B=2$ and $B C=5$ , then $B X$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$ . Compute $100 a+b$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

Note that since $A D \cdot A C = A B^{2}$ , we have the tangency point of $\omega$ and $A B$ is $B'$ , the reflection of $B$ across $A$ . Let $Y$ be the second intersection of $\omega$ and $B C$ . Note that by power of point, we have $B Y \cdot B C = B B'^{2} = 4 A B^{2} \Longrightarrow B Y = \frac{4 A B^{2}}{B C}$ . Note that $A X$ is the radical axis of $\omega$ and the degenerate circle at $B$ , so we have $X B^{2} = X Y \cdot X C$ , so
$B X^{2} = (B C - B X)(B Y - B X) = B X^{2} - B X(B C + B Y) + B C \cdot B Y$
This gives us
$B X = \frac{B C \cdot B Y}{B C + B Y} = \frac{4 A B^{2} \cdot B C}{4 A B^{2} + B C^{2}} = \frac{80}{41}$

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