Difference between revisions of "2018 USAJMO Problems/Problem 3"

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Now, <math>DP</math> is the radical axis of the circumcircles of <math>\triangle EDP</math> and <math>\triangle FDP.</math> Since <math>B</math> lies on <math>DP,</math> and <math>E, Q</math> lie on the circumcircle of <math>\triangle EPD</math> and <math>F, R</math> lie on the circumcircle of <math>\triangle FPD,</math> we have that <cmath>BF \cdot BR = BE \cdot BQ.</cmath> However, <math>BF=BE,</math> so <math>BR=BQ.</math> Since <math>E, B, Q</math> are collinear and so are <math>F, B, R</math> we can add these <math>2</math> equations to get <cmath>EQ=BE+BQ=BF+BR=FR,</cmath> which completes the proof.
 
Now, <math>DP</math> is the radical axis of the circumcircles of <math>\triangle EDP</math> and <math>\triangle FDP.</math> Since <math>B</math> lies on <math>DP,</math> and <math>E, Q</math> lie on the circumcircle of <math>\triangle EPD</math> and <math>F, R</math> lie on the circumcircle of <math>\triangle FPD,</math> we have that <cmath>BF \cdot BR = BE \cdot BQ.</cmath> However, <math>BF=BE,</math> so <math>BR=BQ.</math> Since <math>E, B, Q</math> are collinear and so are <math>F, B, R</math> we can add these <math>2</math> equations to get <cmath>EQ=BE+BQ=BF+BR=FR,</cmath> which completes the proof.
  
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~nukelauncher
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:52, 21 April 2018

Problem

($*$) Let $ABCD$ be a quadrilateral inscribed in circle $\omega$ with $\overline{AC} \perp \overline{BD}$. Let $E$ and $F$ be the reflections of $D$ over lines $BA$ and $BC$, respectively, and let $P$ be the intersection of lines $BD$ and $EF$. Suppose that the circumcircle of $\triangle EPD$ meets $\omega$ at $D$ and $Q$, and the circumcircle of $\triangle FPD$ meets $\omega$ at $D$ and $R$. Show that $EQ = FR$.

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Solution 1

First we have that $BE=BD=BF$ by the definition of a reflection. Let $\angle DEB = \alpha$ and $\angle DFB = \beta.$ Since $\triangle DBE$ is isosceles we have $\angle BDE = \alpha.$ Also, we see that $\angle BDE = \angle CAB = \angle CDB = \alpha,$ using similar triangles and the property of cyclic quadrilaterals. Similarly, \[\angle DFB = \angle FDB = \angle ACB = \angle ADB = \beta.\] Now, from $BE=BD=BF$ we know that $B$ is the circumcenter of $\triangle DEF.$ Using the properties of the circumcenter and some elementary angle chasing, we find that \[\angle DPE = 90^{\circ} + \beta - \alpha.\]

Now, we claim that $Q$ is the intersection of ray $\overrightarrow{EB}$ and the circumcircle of $ABCD.$ To prove this, we just need to show that $DEPQ$ is cyclic by this definition of $Q.$ We have that \[\angle DQE = \angle DCB = \angle DCA + \angle ACB = (90^{\circ}-\alpha)+\beta.\] We also have from before that \[\angle DPE = 90+\beta-\alpha,\] so $\angle DQE=\angle DPE$ and this proves the claim.

We can use a similar proof to show that $F, B, R$ are collinear.

Now, $DP$ is the radical axis of the circumcircles of $\triangle EDP$ and $\triangle FDP.$ Since $B$ lies on $DP,$ and $E, Q$ lie on the circumcircle of $\triangle EPD$ and $F, R$ lie on the circumcircle of $\triangle FPD,$ we have that \[BF \cdot BR = BE \cdot BQ.\] However, $BF=BE,$ so $BR=BQ.$ Since $E, B, Q$ are collinear and so are $F, B, R$ we can add these $2$ equations to get \[EQ=BE+BQ=BF+BR=FR,\] which completes the proof.

~nukelauncher

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

See also

2018 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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