HMMT-Nov 2023 team, Problem 7

Let $A B C D$ be a convex trapezoid such that $\angle B A D=\angle A D C=90^{\circ}, A B=20, A D=21$, and $C D=28$. Point $P \neq A$ is chosen on segment $A C$ such that $\angle B P D=90^{\circ}$. Compute $A P$.

Solution: Solution 1: Construct the rectangle $A B X D$. Note that \[ \angle B A D=\angle B P D=\angle B X D=90^{\circ} \] so $A B X P D$ is cyclic with diameter $B D$. By Power of a Point, we have $C X \cdot C D=C P \cdot C A$. Note that $C X=C D-X D=C D-A B=8$ and $C A=\sqrt{A D^{2}+D C^{2}}=35$. Therefore, \[ C P=\frac{C X \cdot C D}{C A}=\frac{8 \cdot 28}{35}=\frac{32}{5} \] and so \[ A P=A C-C P=35-\frac{32}{5}=\frac{143}{5} \] Solution 2: Since $A B P D$ is cyclic and $\angle B D P=\angle B A P=\angle A C D$, it follows that $\triangle B P D \sim \triangle A B C$. Thus, if $B D=5 x, B P=4 x$, and $D P=3 x$, then by Ptolemy's theorem, we have \[ \begin{aligned} A P \cdot(5 x) & =A D \cdot(3 x)+A B \cdot(4 x) \\ A P & =\frac{21 \cdot 3+20 \cdot 4}{5}=\frac{143}{5} \end{aligned} \] $\fbox{9}$.

Answer: 9