2019 USAJMO Problems/Problem 3

Revision as of 11:32, 25 June 2019 by Sriraamster (talk | contribs) (Solution)

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: [list] [*] $AP' \cdot AB = AD^2$ [*] $BP' \cdot AB = CD^2$ [/list]

[b]Claim:[/b]$P = P'$

[i]Proof:[/i]

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

[b]Claim:[/b] $PE$ is a symmedian in $AEB$

[i]Proof:[/i]

We have APAB=AD2AB2AP=AD2AB(ABAD)2=ABAP(ABAD)21=ABAP1AB2AD2AD2=BPAP(BCAD)2=(BEAE)2=BPAP 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 (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAJMO Problems and Solutions