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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>.
 +
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:
 +
 +
(1) <math>AP' \cdot AB = AD^2</math>
  
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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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(2) <math>BP' \cdot AB = CD^2</math>
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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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Claim: <math>P = P'</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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Proof:
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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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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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<cmath>\angle'AKB = \angle'CKB = \angle'CP B</cmath> and <math>\angle'ALB = \angle'APD</math>, and these are equal.
 
  
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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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Claim: <math>PE</math> is a symmedian in <math>AEB</math>
  
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{{MAA Notice}}
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Proof:
 +
We have
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<cmath>\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*}</cmath>
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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>
 +
 
 +
==Solution 2==
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[[File:2019 USAMO 2.png|450px|right]]
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[[File:2019 USAMO 2a.png|450px|right]]
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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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<cmath>B'= I_\omega (B), C'= I_\omega (C), D = I_\omega (D) \implies</cmath>
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<cmath>AB' \cdot AB = AD^2,  \angle ACB = \angle AB'C'.</cmath>
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<cmath>A'= I_\Omega (A), D'= I_\Omega (D), C = I_\Omega (C) \implies</cmath>
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<cmath>BA' \cdot AB = BC^2,  \angle BDA = \angle BA'D'.</cmath>
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Let <math>\theta</math> be the circle <math>ABCD.</math>
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 +
<math>I_\omega (\theta) = B'C'D,</math> straight line, therefore <cmath>\angle AB'C' = \angle AB'D' = \angle ACB.</cmath>
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<math>I_\Omega (\theta) = A'D'C,</math> straight line, therefore
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<cmath>\angle BA'D' = \angle BA'C = \angle BDA.</cmath>
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<math>ABCD</math> is cyclic <math>\implies \angle BA'C = \angle AB'D.</math>
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<cmath>AB' + BA' = \frac {AD^2 + BC^2 }{AB} = AB \implies</cmath> points <math>A'</math> and <math>B'</math> are coincide.
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Denote <math>A' = B' = Q \in AB.</math>
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Suppose, we move point <math>Q</math> from <math>A</math> to <math>B.</math> Then <math>\angle AQD</math> decreases monotonically, <math>\angle BQC</math> increases monotonically. So, there is only one point where <cmath>\angle AQD = \angle BQC \implies P = Q.</cmath>
 +
 
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<cmath>B = I_\omega (P), D' = I_\omega (D'), C' = I_\omega (C), A = I_\omega (\infty) \implies</cmath>
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<math>\hspace{19mm} I_\omega (CD'P) = AC'D'B</math> is cyclic.
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<cmath>\angle ACD = \angle ABD = \angle CC'D \implies C' D' || CD \implies</cmath>
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<math>\hspace{19mm} C'D'CD</math> is  trapezoid.
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 +
It is known that the intersection of the diagonals, intersection point of the lines containing the lateral sides of the trapezoid and the midpoints of two parallel sides are collinear.
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 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
  
 
==See also==
 
==See also==
 
{{USAMO newbox|year=2019|num-b=1|num-a=3}}
 
{{USAMO newbox|year=2019|num-b=1|num-a=3}}
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 +
{{MAA Notice}}

Latest revision as of 21:51, 18 October 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. $\square$

Solution 2

2019 USAMO 2.png
2019 USAMO 2a.png

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. \[B'= I_\omega (B), C'= I_\omega (C), D = I_\omega (D) \implies\] \[AB' \cdot AB = AD^2, \angle ACB = \angle AB'C'.\] \[A'= I_\Omega (A), D'= I_\Omega (D), C = I_\Omega (C) \implies\] \[BA' \cdot AB = BC^2, \angle BDA = \angle BA'D'.\] 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 $\implies \angle BA'C = \angle AB'D.$ \[AB' + BA' = \frac {AD^2 + BC^2 }{AB} = AB \implies\] points $A'$ and $B'$ are coincide.

Denote $A' = B' = Q \in AB.$

Suppose, we move point $Q$ from $A$ to $B.$ Then $\angle AQD$ decreases monotonically, $\angle BQC$ increases monotonically. So, there is only one point where \[\angle AQD = \angle BQC \implies P = Q.\]

\[B = I_\omega (P), D' = I_\omega (D'), C' = I_\omega (C), A = I_\omega (\infty) \implies\] $\hspace{19mm} I_\omega (CD'P) = AC'D'B$ is cyclic. \[\angle ACD = \angle ABD = \angle CC'D \implies C' D' || CD \implies\] $\hspace{19mm} C'D'CD$ is trapezoid.

It is known that the intersection of the diagonals, intersection point of the lines containing the lateral sides of the trapezoid and the midpoints of two parallel sides are collinear.

vladimir.shelomovskii@gmail.com, vvsss

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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