2019 USAJMO Problems/Problem 3

Revision as of 12:34, 25 June 2019 by Sriraamster (talk | contribs) (Solution)


$(*)$ 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 $\overline{CD}$.


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:

$\dot$ (Error compiling LaTeX. ! Extra }, or forgotten $.)AP' \cdot AB = AD^2$$ (Error compiling LaTeX. ! Missing $ inserted.)\dot BP' \cdot AB = CD^2$Claim:$P = P'$Proof:

The conditions imply the similarities$ (Error compiling LaTeX. ! Missing $ inserted.)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

<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. ! Missing $ inserted.)\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$

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See also

2019 USAJMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAJMO Problems and Solutions
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