Difference between revisions of "2019 USAMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | + | 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> | ||
− | + | (2) <math>BP' \cdot AB = CD^2</math> | |
− | + | ||
+ | Claim: <math>P = P'</math> | ||
− | + | Proof: | |
− | + | 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> | |
− | |||
− | + | Claim: <math>PE</math> is a symmedian in <math>AEB</math> | |
− | {{ | + | Proof: |
+ | We have | ||
+ | <cmath>\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*}</cmath> | ||
+ | as desired. <math>\square</math> | ||
+ | |||
+ | 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== | ||
+ | [[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== | ==See also== | ||
{{USAMO newbox|year=2019|num-b=1|num-a=3}} | {{USAMO newbox|year=2019|num-b=1|num-a=3}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 22:51, 18 October 2022
Contents
Problem
Let be a cyclic quadrilateral satisfying
. The diagonals of
intersect at
. Let
be a point on side
satisfying
. Show that line
bisects
.
Solution
Let . Also, let
be the midpoint of
.
Note that only one point
satisfies the given angle condition. With this in mind, construct
with the following properties:
(1)
(2)
Claim:
Proof:
The conditions imply the similarities and
whence
as desired.
Claim: is a symmedian in
Proof:
We have
as desired.
Since is the isogonal conjugate of
,
. However
implies that
is the midpoint of
from similar triangles, so we are done.
Solution 2
Let be the circle centered at
with radius
Let be the circle centered at
with radius
We denote and
inversion with respect to
and
respectively.
Let
be the circle
straight line, therefore
straight line, therefore
is cyclic
points
and
are coincide.
Denote
Suppose, we move point from
to
Then
decreases monotonically,
increases monotonically. So, there is only one point where
is cyclic.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.