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Difference between revisions of "2019 USAMO Problems/Problem 2"
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==Solution==
 
==Solution==
  
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Realize that there is only one point <math>P</math> on <math>\overline{AB}</math> satisfying the conditions, because <math>\angle APD</math> decreases and <math>\angle BPC</math> increases as <math>P</math> moves from <math>A</math> to <math>B</math>. Therefore, if we prove that there is a single point <math>P</math> that lies on <math>\overline{AB}</math> such that <math>\angle APD \cong \angle BPC</math>, and that <math>PE</math> bisects <math>CD</math>, it must coincide with the point from the problem, so we will be done.
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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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Since <math>AD^2 + BC^2 = AB^2</math>, there is some <math>P</math> on <math>AB</math> such that <cmath>AD^2 = AP \times AB \text{ and } BC^2 = BP \times BA.</cmath> Thus, <math>\frac{AP}{AD} = \frac{AD}{AB}</math> and <math>\frac{BP}{BC} = \frac{BC}{BA}</math>. Thus we have that <math>\triangle APD \sim
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[list]
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\triangle ADB</math> and <math>\triangle BPC \sim \triangle BCA</math>, meaning that <math>\angle APD = \angle ADB = \angle ACB = \angle BPC</math>.
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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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We next must show that <math>PE</math> bisects <math>CD</math>. Define <math>K</math> as the intersection of <math>AC</math> and <math>PD</math> and <math>L</math> as the intersection of <math>BD</math> and <math>PC</math>. We know that <math>APLD</math> and <math>BPKC</math> are cyclic, because
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[/list]
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<cmath>\angle'ADL = \angle'ACB = \angle'BPC = \angle'AP L,</cmath> where <math>\angle'</math> represents an angle which is measured <math>\text{mod } \pi</math>. Furthermore, quadrilateral <math>AKLB</math> is also cyclic, because
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[b]Claim:[/b]<math>P = P'</math>
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<cmath>\angle'AKB = \angle'CKB = \angle'CP B</cmath> and <math>\angle'ALB = \angle'APD</math>, and these are equal.
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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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As the quadrilaterals are cyclic, we have that <math>\angle'KCD = \angle'ABD = \angle'ABL = \angle'AKL = \angle'CKL</math>, meaning that <math>CD \parallel KL</math>. Thus <math>CDKL</math> is a trapezoid whose legs intersect at <math>P</math> and whose diagonals intersect at <math>E</math>. Therefore, line <math>PE</math> bisects the bases <math>CD</math> and <math>KL</math>, as wished. ~ciceronii
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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 02:33, 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: [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. $\square$

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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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