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

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Let <math>PE \cap DC = M</math>. Also, let <math>N</math> be the midpoint of <math>AB</math>.
 
Let <math>PE \cap DC = M</math>. Also, let <math>N</math> be the midpoint of <math>AB</math>.
 
Note that only one point <math>P</math> satisfies the given angle condition. With this in mind, construct <math>P'</math> with the following properties:
 
Note that only one point <math>P</math> satisfies the given angle condition. With this in mind, construct <math>P'</math> with the following properties:
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[list]
+
 
[*] <math>AP' \cdot AB = AD^2</math>
 
[*] <math>AP' \cdot AB = AD^2</math>
 
[*] <math>BP' \cdot AB = CD^2</math>
 
[*] <math>BP' \cdot AB = CD^2</math>
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[/list]
+
−
[b]Claim:[/b]<math>P = P'</math>
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Claim: <math>P = P'</math>
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[i]Proof:[/i]
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Proof:
 
The conditions imply the similarities <math>ADP \sim ABD</math> and <math>BCP \sim BAC</math> whence <math>\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB</math> as desired. <math>\square</math>
 
The conditions imply the similarities <math>ADP \sim ABD</math> and <math>BCP \sim BAC</math> whence <math>\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB</math> as desired. <math>\square</math>
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[b]Claim:[/b] <math>PE</math> is a symmedian in <math>AEB</math>
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Claim: <math>PE</math> is a symmedian in <math>AEB</math>
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[i]Proof:[/i]
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Proof:
 
We have  
 
We have  
 
\begin{align*}  
 
\begin{align*}  

Revision as of 01:34, 2 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:

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

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

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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