Difference between revisions of "2019 USAMO Problems/Problem 2"
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We have | We have | ||
− | + | \begin{align*} | |
AP \cdot AB = AD^2 \iff AB^2 \cdot AP &= AD^2 \cdot AB \\ | 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 &= \frac{AB}{AP} \\ | ||
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\iff \frac{AB^2 - AD^2}{AD^2} &= \frac{BP}{AP} \\ | \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} | \iff \left(\frac{BC}{AD} \right)^2 &= \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} | ||
− | \end{align*} | + | \end{align*} |
as desired. <math>\square</math> | as desired. <math>\square</math> | ||
Revision as of 08:41, 5 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:
(1)
(2)
Claim:
Proof:
The conditions imply the similarities
and
whence
as desired.
Claim: is a symmedian in
Proof:
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 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |