Difference between revisions of "2019 USAJMO Problems/Problem 3"
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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: | 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: | ||
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− | + | <math>\dot </math>AP' \cdot AB = AD^2<math> | |
+ | </math>\dot BP' \cdot AB = CD^2<math> | ||
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− | + | 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> | |
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+ | Claim: </math>PE<math> is a symmedian in </math>AEB<math> | ||
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+ | Proof: | ||
We have | We have | ||
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− | Since <math>P< | + | <cmath>AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB</cmath> |
+ | <cmath>\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}</cmath> | ||
+ | <cmath>\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1</cmath> | ||
+ | <cmath>\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}</cmath> | ||
+ | <cmath>\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} </cmath> | ||
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+ | as desired. </math>\square<math> | ||
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+ | 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$ | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:34, 25 June 2019
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:
$\dot$ (Error compiling LaTeX. ! Extra }, or forgotten $.)AP' \cdot AB = AD^2$$ (Error compiling LaTeX. ! Missing $ inserted.)\dot BP' \cdot AB = CD^2P = P'$Proof:
The conditions imply the similarities$ (Error compiling LaTeX. ! Missing $ inserted.)ADP \sim ABDBCP \sim BAC\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB\squarePEAEB$Proof:
We have
<cmath>AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB</cmath> <cmath>\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}</cmath> <cmath>\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1</cmath> <cmath>\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}</cmath> <cmath>\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} </cmath>
as desired.$ (Error compiling LaTeX. ! Missing $ inserted.)\squarePN\measuredangle PEA = \measuredangle MEC = \measuredangle BEN\measuredangle MEC = \measuredangle BENMCD\square$
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |