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
Since is the isogonal conjugate of
,
. However
implies that
is the midpoint of
from similar triangles, so we are done.
These problems are copyrighted © by the Mathematical Association of America, as part of the 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 |