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

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

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

@@ Line 31: / Line 31: @@
 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}}
+==Solution 2==
+[[File:2019 USAMO 2.png|450px|right]]
+[[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>B</math> with radius <math>BC.</math>
+We denote <math>I_\omega</math> and <math>I_\Omega</math> inversion with respect to <math>\omega</math> and <math>\Omega,</math> respectively.
+<cmath>B'= I_\omega (B), C'= I_\omega (C), D = I_\omega (D) \implies</cmath>
+<cmath>AB' \cdot AB = AD^2,  \angle ACB = \angle AB'C'.</cmath>
+<cmath>A'= I_\Omega (A), D'= I_\Omega (D), C = I_\Omega (C) \implies</cmath>
+<cmath>BA' \cdot AB = BC^2,  \angle BDA = \angle BA'D'.</cmath>
+Let <math>\theta</math> be the circle <math>ABCD.</math>
+<math>I_\omega (\theta) = B'C'D,</math> straight line, therefore <cmath>\angle AB'C' = \angle AB'D' = \angle ACB.</cmath>
+<math>I_\Omega (\theta) = A'D'C,</math> straight line, therefore
+<cmath>\angle BA'D' = \angle BA'C = \angle BDA.</cmath>
+<math>ABCD</math> is cyclic <math>\implies \angle BA'C = \angle AB'D.</math>
+<cmath>AB' + BA' = \frac {AD^2 + BC^2 }{AB} = AB \implies</cmath> points <math>A'</math> and <math>B'</math> are coincide.
+Denote <math>A' = B' = Q \in AB.</math>
+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>
+<cmath>B = I_\omega (P), D' = I_\omega (D'), C' = I_\omega (C), A = I_\omega (\infty) \implies</cmath>
+<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>
+<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.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
 ==See also==
 {{USAMO newbox|year=2019|num-b=1|num-a=3}}
+{{MAA Notice}}

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

Latest revision as of 21:51, 18 October 2022

Contents

Problem

Solution

Solution 2

See also