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 11:32, 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.
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 |