Difference between revisions of "2019 USAMO Problems/Problem 2"
Sriraamster (talk | contribs) (→Solution) |
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Proof: | 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> | 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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Claim: <math>PE</math> is a symmedian in <math>AEB</math> | Claim: <math>PE</math> is a symmedian in <math>AEB</math> | ||
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Proof: | Proof: | ||
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We have | We have | ||
\begin{align*} | \begin{align*} |
Revision as of 01:35, 2 March 2020
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:
[*] [*]
Claim: Proof: The conditions imply the similarities and whence as desired.
Claim: is a symmedian in
Proof:
We have
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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 |