2019 USAMO Problems/Problem 2

Problem

Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$ . The diagonals of $ABCD$ intersect at $E$ . Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD = \angle BPC$ . Show that line $PE$ bisects $\overline{CD}$ .

Solution

Let $PE \cap DC = M$ . Also, let $N$ be the midpoint of $AB$ . Note that only one point $P$ satisfies the given angle condition. With this in mind, construct $P'$ with the following properties:

[*] $AP' \cdot AB = AD^2$ [*] $BP' \cdot AB = CD^2$

Claim: $P = P'$ Proof: The conditions imply the similarities $ADP \sim ABD$ and $BCP \sim BAC$ whence $\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB$ as desired. $\square$ Claim: $PE$ is a symmedian in $AEB$ Proof: We have $\begin{aligned} A P \cdot A B = A D^{2} ⟺ A B^{2} \cdot A P & = A D^{2} \cdot A B \\ ⟺ {(\frac{A B}{A D})}^{2} & = \frac{A B}{A P} \\ ⟺ {(\frac{A B}{A D})}^{2} - 1 & = \frac{A B}{A P} - 1 \\ ⟺ \frac{A B^{2} - A D^{2}}{A D^{2}} & = \frac{B P}{A P} \\ ⟺ {(\frac{B C}{A D})}^{2} & = {(\frac{B E}{A E})}^{2} = \frac{B P}{A P} \end{aligned}$ as desired. $\square$ Since $P$ is the isogonal conjugate of $N$ , $\measuredangle PEA = \measuredangle MEC = \measuredangle BEN$ . However $\measuredangle MEC = \measuredangle BEN$ implies that $M$ is the midpoint of $CD$ from similar triangles, so we are done. $\square$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

2019 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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2019 USAMO Problems/Problem 2

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