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

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==Solution 2==
 
==Solution 2==
 
[[File:2019 USAMO 2.png|450px|right]]
 
[[File:2019 USAMO 2.png|450px|right]]
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[[File:2019 USAMO 2a.png|450px|right]]
 
Let <math>\omega</math> be the circle centered at <math>A</math> with radius <math>AD.</math>
 
Let <math>\omega</math> be the circle centered at <math>A</math> with radius <math>AD.</math>
 
Let <math>\Omega</math> be the circle centered at <math>B</math> with radius <math>BC.</math>
 
Let <math>\Omega</math> be the circle centered at <math>B</math> with radius <math>BC.</math>
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Denote <math>A' = B' = Q \in AB.</math>
 
Denote <math>A' = B' = Q \in AB.</math>
  
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Suppose, we move <math>P</math> from <math>A</math> to <math>B.</math> Then <math>\angle AQD</math> decreases monotonically, <math>\angle AQC</math> increases monotonically. So, there is only one point where <math>\angle AQD = \angle BQC \implies P = Q.</math>
+
Suppose, we move <math>P</math> from <math>A</math> to <math>B.</math> Then <math>\angle AQD</math> decreases monotonically, <math>\angle AQC</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>
 +
<math>\hspace{10mm} I_\omega (CD'P) = AC'D'B</math> is cyclic.
 +
<cmath>\angle ACD = \angle ABD = \angle CC'D \implies C' D' || CD \implies</cmath>
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<math>\hspace{10mm} 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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<math>B = I_\omega (P), D' = I_\omega (D'), C' = I_\omega (C), A = I_\omega (\infty) \implies I_\omega (CD'P) = AC'D'B </math> is cyclic.
 
−
<math>\angle ACD = \angle ABD = \angle CC'D \implies C' D' || CD \implies C'D'CD</math> 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'''
 
'''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}}

Revision as of 11:10, 16 September 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. $\square$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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. \[B'= I_\omega (B), C'= I_\omega (C), D = I_\omega (D) \implies\] \[AB' \cdot AB = AD^2, \angle ACB = \angle AB'C'.\] \[A'= I_\Omega (A), D'= I_\Omega (D), C = I_\omega (C) \implies\] \[AA' \cdot AB = BC^2, \angle BDA = \angle BA'D'.\] 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 $P$ from $A$ to $B.$ Then $\angle AQD$ decreases monotonically, $\angle AQC$ increases monotonically. So, there is only one point where \[\angle AQD = \angle BQC \implies P = Q.\]

\[B = I_\omega (P), D' = I_\omega (D'), C' = I_\omega (C), A = I_\omega (\infty) \implies\] $\hspace{10mm} I_\omega (CD'P) = AC'D'B$ is cyclic. \[\angle ACD = \angle ABD = \angle CC'D \implies C' D' || CD \implies\] $\hspace{10mm} C'D'CD$ 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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