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Difference between revisions of "2019 USAMO Problems/Problem 2"

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==Solution 2==
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Let <math>\omega</math> be the circle centered at <math>A</math> with radius <math>AD.</math>
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Let <math>\Omega</math> be the circle centered at <math>B</math> with radius <math>BC.</math>
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We denote <math>I_\omega</math> and <math>I\Omega</math> inversion with respect to <math>\omega</math> and <math>\Omega,</math> respectively.
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<math>B'= I_\omega (B), C'= I_\omega (C), D = I_\omega (D) \implies AB' \cdot AB = AD^2,  \angle ACB = \angle AB'C'</math>
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<math>A'= I_\Omega (A), D'= I_\Omega (D), C = I_\omega (C) \implies AA' \cdot AB = BC^2,  \angle BDA = \angle BA'D'</math>
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Let <math>\theta</math> be the circle <math>ABCD.  I_\omega (\theta) = B'C'D,</math> straight line, therefore <math>\angle AB'C' = \angle AB'D' = \angle ACB.</math>
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<math>I_\Omega (\theta) = A'D'C,</math> straight line, therefore <math>\angle BA'D' = \angle BA'C = \angle BDA.</math>
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<math>ABCD</math> is cyclic <math>\implies \angle BA'C = \angle AB'D.</math>
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==See also==
 
==See also==
 
{{USAMO newbox|year=2019|num-b=1|num-a=3}}
 
{{USAMO newbox|year=2019|num-b=1|num-a=3}}

Revision as of 11:52, 16 September 2022

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. $\square$

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

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. $B'= I_\omega (B), C'= I_\omega (C), D = I_\omega (D) \implies AB' \cdot AB = AD^2, \angle ACB = \angle AB'C'$ $A'= I_\Omega (A), D'= I_\Omega (D), C = I_\omega (C) \implies AA' \cdot AB = BC^2, \angle BDA = \angle BA'D'$ 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 $\implies \angle BA'C = \angle AB'D.$


See also

2019 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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