2019 USAJMO Problems/Problem 3
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. Unknown error_msg)AP' \cdot AB = AD^2$$ (Error compiling LaTeX. Unknown error_msg)\dot BP' \cdot AB = CD^2P = P'$Proof:
The conditions imply the similarities$ (Error compiling LaTeX. Unknown error_msg)ADP \sim ABDBCP \sim BAC
\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB
\square
PE
AEB$Proof:
We have
<cmath>AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB</cmath> <cmath>\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}</cmath> <cmath>\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1</cmath> <cmath>\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}</cmath> <cmath>\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} </cmath>
as desired.$ (Error compiling LaTeX. Unknown error_msg)\squareP
N
\measuredangle PEA = \measuredangle MEC = \measuredangle BEN
\measuredangle MEC = \measuredangle BEN
M
CD
\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 |