Difference between revisions of "2019 USAJMO Problems/Problem 3"
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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: | ||
+ | [list] | ||
+ | [*] <math>AP' \cdot AB = AD^2</math> | ||
+ | [*] <math>BP' \cdot AB = CD^2</math> | ||
+ | [/list] | ||
+ | |||
+ | [b]Claim:[/b]<math>P = P'</math> | ||
+ | |||
+ | [i]Proof:[/i] | ||
+ | |||
+ | 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> | ||
+ | |||
+ | [b]Claim:[/b] <math>PE</math> is a symmedian in <math>AEB</math> | ||
+ | |||
+ | [i]Proof:[/i] | ||
+ | |||
+ | 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. <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> | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:32, 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: [list] [*] [*] [/list]
[b]Claim:[/b]
[i]Proof:[/i]
The conditions imply the similarities and whence as desired.
[b]Claim:[/b] is a symmedian in
[i]Proof:[/i]
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.
Since is the isogonal conjugate of , . However implies that is the midpoint of from similar triangles, so we are done.
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 |