# 2019 USAMO Problems/Problem 2

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## 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:

(1) $AP' \cdot AB = AD^2$

(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{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. $\square$

## Solution 2

Let $\omega$ be the circle centered at $A$ with radius $AD.$

Let $\Omega$ be the circle centered at $B$ with radius $BC.$

We denote $I_\omega$ and $I_\Omega$ inversion with respect to $\omega$ and $\Omega,$ respectively. $$B'= I_\omega (B), C'= I_\omega (C), D = I_\omega (D) \implies$$ $$AB' \cdot AB = AD^2, \angle ACB = \angle AB'C'.$$ $$A'= I_\Omega (A), D'= I_\Omega (D), C = I_\Omega (C) \implies$$ $$BA' \cdot AB = BC^2, \angle BDA = \angle BA'D'.$$ Let $\theta$ be the circle $ABCD.$ $I_\omega (\theta) = B'C'D,$ straight line, therefore $$\angle AB'C' = \angle AB'D' = \angle ACB.$$ $I_\Omega (\theta) = A'D'C,$ straight line, therefore $$\angle BA'D' = \angle BA'C = \angle BDA.$$ $ABCD$ is cyclic $\implies \angle BA'C = \angle AB'D.$ $$AB' + BA' = \frac {AD^2 + BC^2 }{AB} = AB \implies$$ points $A'$ and $B'$ are coincide.

Denote $A' = B' = Q \in AB.$

Suppose, we move point $Q$ from $A$ to $B.$ Then $\angle AQD$ decreases monotonically, $\angle BQC$ increases monotonically. So, there is only one point where $$\angle AQD = \angle BQC \implies P = Q.$$ $$B = I_\omega (P), D' = I_\omega (D'), C' = I_\omega (C), A = I_\omega (\infty) \implies$$ $\hspace{19mm} I_\omega (CD'P) = AC'D'B$ is cyclic. $$\angle ACD = \angle ABD = \angle CC'D \implies C' D' || CD \implies$$ $\hspace{19mm} C'D'CD$ is trapezoid.

It is known that the intersection of the diagonals, intersection point of the lines containing the lateral sides of the trapezoid and the midpoints of two parallel sides are collinear.

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