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

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(→‎Solution 2)
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<cmath>B = I_\omega (P), D' = I_\omega (D'), C' = I_\omega (C), A = I_\omega (\infty) \implies</cmath>
 
<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{10mm} I_\omega (CD'P) = AC'D'B</math> is cyclic.
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<math>\hspace{19mm} I_\omega (CD'P) = AC'D'B</math> is cyclic.
 
<cmath>\angle ACD = \angle ABD = \angle CC'D \implies C' D' || CD \implies</cmath>
 
<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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<math>\hspace{19mm} 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.
 
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.

Revision as of 11:12, 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 \[\angle AQD = \angle BQC \implies P = Q.\]

\[B = I_\omega (P), D' = I_\omega (D'), C' = I_\omega (C), A = I_\omega (\infty) \implies\] $\hspace{19mm} I_\omega (CD'P) = AC'D'B$ 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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