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

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We have  
 
We have  
  
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\begin{align*}  
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<math>\begin{align*}  
 
AP \cdot AB = AD^2 \iff AB^2 \cdot AP &= AD^2 \cdot AB \\
 
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 &= \frac{AB}{AP} \\
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\iff \frac{AB^2 - AD^2}{AD^2} &= \frac{BP}{AP} \\
 
\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}  
 
\iff \left(\frac{BC}{AD} \right)^2 &= \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP}  
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\end{align*}
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\end{align*}</math>
 
as desired. <math>\square</math>
 
as desired. <math>\square</math>
  

Revision as of 07:40, 5 March 2020

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

$\begin{align*} AP \cdot AB = AD^2 \iff AB^2 \cdot AP &= AD^2 \cdot AB \\ \iff \left( \frac{AB}{AD} \right)^2 &= \frac{AB}{AP} \\ \iff \left( \frac{AB}{AD} \right)^2 - 1 &= \frac{AB}{AP} - 1 \\ \iff \frac{AB^2 - AD^2}{AD^2} &= \frac{BP}{AP} \\ \iff \left(\frac{BC}{AD} \right)^2 &= \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} \end{align*}$ (Error compiling LaTeX. Unknown error_msg) 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

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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