2019 USAJMO Problems/Problem 3

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

[b]Claim:[/b] $P = P'$

[i]Proof:[/i]

The conditions imply the similarities $ADP \sim ABD$ and $BCP \sim BAC$ whence $\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB$ as desired. $\square$

[b]Claim:[/b] $PE$ is a symmedian in $AEB$

[i]Proof:[/i]

We have \begin{align*} AP \cdot AB = AD^2 \iff AB^2 \cdot AP &= AD^2 \cdot AB \\ \iff \left( \frac{AB}{AD} \right)^2 &= \frac{AB}{AP} \\ \iff \left( \frac{AB}{AD} \right)^2 - 1 &= \frac{AB}{AP} - 1 \\ \iff \frac{AB^2 - AD^2}{AD^2} &= \frac{BP}{AP} \\ \iff \left(\frac{BC}{AD} \right)^2 &= \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} \end{align*} 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. $\square$

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

2019 USAJMO (Problems • Resources)
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2019 USAJMO Problems/Problem 3

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