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:

$\dot$ (Error compiling LaTeX. Unknown error_msg)AP' \cdot AB = AD^2$$ (Error compiling LaTeX. Unknown error_msg)\dot BP' \cdot AB = CD^2 $Claim:$ P = P'$Proof:

The conditions imply the similarities$ (Error compiling LaTeX. Unknown error_msg)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

<cmath>AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB</cmath> <cmath>\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}</cmath> <cmath>\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1</cmath> <cmath>\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}</cmath> <cmath>\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} </cmath>

as desired.$ (Error compiling LaTeX. Unknown error_msg)\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 USAJMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
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2019 USAJMO Problems/Problem 3

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