Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 USAJMO Problems/Problem 3"

(→‎Solution)
(→‎Solution)
Line 7: Line 7:
 
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:
 
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:
  
−
1) <math>AP' \cdot AB = AD^2</math>
+
<math>AP' \cdot AB = AD^2 \quad \text{and}  BP' \cdot AB = CD^2</math>
−
2) <math>BP' \cdot AB = CD^2</math>
+
 
  
  

Revision as of 11:36, 25 June 2019

Problem

$(*)$ Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2+BC^2=AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD=\angle BPC$. Show that line $PE$ bisects $\overline{CD}$.

Solution

Let $PE \cap DC = M$. Also, let $N$ be the midpoint of $AB$.

Note that only one point $P$ satisfies the given angle condition. With this in mind, construct $P'$ with the following properties:

$AP' \cdot AB = AD^2 \quad \text{and} BP' \cdot AB = CD^2$


Claim:$P = P'$

Proof:

The conditions imply the similarities $ADP \sim ABD$ and $BCP \sim BAC$ whence $\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB$ as desired. $\square$

Claim: $PE$ is a symmedian in $AEB$

Proof:

We have

\[AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB\] \[\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}\] \[\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1\] \[\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}\] \[\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP}\]

as desired. $\square$

Since $P$ is the isogonal conjugate of $N$, $\measuredangle PEA = \measuredangle MEC = \measuredangle BEN$. However $\measuredangle MEC = \measuredangle BEN$ implies that $M$ is the midpoint of $CD$ from similar triangles, so we are done. $\square$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_USAJMO_Problems/Problem_3&oldid=106976"