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2018 USAJMO Problems/Problem 3

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 (Monday G. Fern)

Solution 2

We begin with the following claims.

Claim. $B$ is the circumcenter of $\triangle DEF$. Proof. By reflection $BD=BE=BF$.

Claim. $A$ is the circumcenter of $\triangle EPD$. Proof. First, we have 

   \[\angle DPE = 180^{\circ}-\angle PDE - \angle PED = 180^{\circ}-\angle BED - \frac{\angle DBF}{2} = \angle 180^{\circ}-\angle BED - \frac{180^{\circ}-2\angle BFD}{2} = 90^{\circ}-\angle BDE + \angle DFB\]

Then 

   \[\angle ADE = \angle ADP - \angle EDP = \angle ADB - \angle EDB = \angle ACB - \angle BDE = \angle FDB - \angle BDE = \angle DFB - \angle DEB = \angle DPE - 90^{\circ} = \frac{180^{\circ}-2(180^{\circ}-\angle DPE)}{2}\]

Then $2\angle ADE = 180^{\circ}-2(180^{\circ}-\angle DPE) \implies \angle DAE = 2(180^{\circ}-\angle DPE)$. This is enough to imply what we desire. \newline

Claim. $C$ is the circumcenter of $\triangle FPD$. Proof. Similar to above.

Claim. $E, B, Q$ are collinear. Proof. We have 

   \[\angle ABE = \angle ABD = \angle AQD = \angle ADQ = 180^{\circ}-\angle ABQ\]

Claim.$F, B, R$ are collinear. Proof. Similar to above.

Since $DEPQ$ is cyclic, $\triangle EBP \sim \triangle DBQ$. However, $BE=BD$, so $BP=BQ$. Similarly, $BP=BR$. Finishing, we have \[EQ = EB + BQ = BD + BP = FB + BR = FR\], as desired. $\blacksquare$ ~MSC

Solution 3

Let $J, K, L$ be the feet of the altitude from $D$ to $BA, BC, CA$, respectively. Note that $L=AC \cap BD$. $J, K, L$ are collinear by the Simson line. Let $h$ be the homothety with center $D$ and scale factor of 2. We see $h(J)=E$ and $h(K)=F$. Since $E, F, P$ are collinear and $P$ lies on $DL$, we have that $h(L)=P$. Therefore, $P$ is the reflection of $D$ over AC. Let $O_1$ and $O_2$ be the centers of $(EPD)$ and $(FPD)$, respectively. From earlier, the perpendicular bisector of $DP$ is $AC$. The perpendicular bisectors of $DE$ and $DF$ are $BA$ and $BC$, respectively. Then, $O_1=AC \cap BA$ and $O_2=AC \cap BC$, so $O_1=A$ and $O_2=C$. Let $O$ be the center of $(ABCD)$. $DQ$ is the radical axis of $(EPD)$ and $(ABCD)$, so $\angle OAD=\angle OAQ$. \[\angle AOD=2\angle ABD=2(90-\angle BAC) \Rightarrow \angle DAQ=\angle OAD+\angle OAQ=2\angle OAD=180^{\circ}-\angle AOD=2\angle BAC.\] Then, $\angle BAC=\angle BAQ+\angle CAD \Rightarrow \angle BAP+\angle CAP=\angle BAQ+\angle CAD \Rightarrow \angle BAP=\angle BAQ$. Therefore, $Q$ is the refection of $P$ across $BA$ because $AP=AQ$. Reflections preserve length, so $DP=EQ$. Similarly, $DP=FR$ ,so $EQ=FR$. $\blacksquare$ ~IS

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

See also

2018 USAJMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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