Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

$P$ is a point inside triangle $A B C$ , and lines $A P, B P, C P$ intersect the opposite sides $B C, C A, A B$ in points $D, E, F$ , respectively. It is given that $\angle A P B=90^{\circ}$ , and that $A C=B C$ and $A B=B D$ . We also know that $B F=1$ , and that $B C=999$ . Find $A F$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let $A C=B C=s, A B=B D=t$ . Since $B P$ is the altitude in isosceles triangle $A B D$ , it bisects angle $B$ . So, the Angle Bisector Theorem in triangle $A B C$ given $A E / E C=A B / B C=t / s$ . Meanwhile, $C D / D B=(s-t) / t$ . Now Ceva's theorem gives us $\begin{gathered} \frac{A F}{F B}=\left(\frac{A E}{E C}\right) \cdot\left(\frac{C D}{D B}\right)=\frac{s-t}{s} \\ \Rightarrow \frac{A B}{F B}=1+\frac{s-t}{s}=\frac{2 s-t}{s} \Rightarrow F B=\frac{s t}{2 s-t} \end{gathered}$ Now we know $s=999$ , but we need to find $t$ given that $s t /(2 s-t)=F B=1$ . So $s t=2 s-t \Rightarrow t=2 s /(s+1)$ , and then $A F=F B \cdot \frac{A F}{F B}=1 \cdot \frac{s-t}{s}=\frac{\left(s^{2}-s\right) /(s+1)}{s}=\frac{s-1}{s+1}=\frac{499}{500}$

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