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Difference between revisions of "2019 USAJMO Problems/Problem 3"

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==Solution==
 
==Solution==
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[THERE IS NO AVAILIBLE SOLUTION]
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Let <math>PE \cap DC = M</math>. Also, let <math>N</math> be the midpoint of <math>AB</math>.
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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]  
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[*] <math>AP' \cdot AB = AD^2</math>
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[*] <math>BP' \cdot AB = CD^2</math>
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[/list]
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[b]Claim:[/b]<math>P = P'</math>
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[i]Proof:[/i]
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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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[i]Proof:[/i]
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We have
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\begin{align*}
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AP \cdot AB = AD^2 \iff AB^2 \cdot AP &= AD^2 \cdot AB \\
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\iff \left( \frac{AB}{AD} \right)^2 &= \frac{AB}{AP} \\
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\iff \left( \frac{AB}{AD} \right)^2 - 1 &= \frac{AB}{AP} - 1 \\
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\iff \frac{AB^2 - AD^2}{AD^2} &= \frac{BP}{AP} \\
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\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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as desired. <math>\square</math>
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Since <math>P</math> is the isogonal conjugate of <math>N</math>, <math>\measuredangle PEA = \measuredangle MEC = \measuredangle BEN</math>. However <math>\measuredangle MEC = \measuredangle BEN</math> implies that <math>M</math> is the midpoint of <math>CD</math> from similar triangles, so we are done. <math>\square</math>
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:32, 25 June 2019

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

[b]Claim:[/b]$P = P'$

[i]Proof:[/i]

The conditions imply the similarities $ADP \sim ABD$ and $BCP \sim BAC$ whence $\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB$ as desired. $\square$

[b]Claim:[/b] $PE$ is a symmedian in $AEB$

[i]Proof:[/i]

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