2018 USAJMO Problems/Problem 3


($*$) 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 $EQ = FR$.

[asy] unitsize(3cm);  real labelscalefactor = 1.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882);   draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819)--cycle, linewidth(2) + rvwvcq);  /* draw figures */ draw(circle((0,0), 1), linewidth(2) + wrwrwr);  draw((0.8660254037844386,-0.5)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + wrwrwr);  draw((0.5922362871684147,0.8057643453026269)--(0.8660254037844386,-0.5), linewidth(2) + rvwvcq);  draw((0.8660254037844386,-0.5)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + rvwvcq);  draw((0.06064145095757084,-0.9981596137020172)--(-0.9987724554622847,0.04953364724950819), linewidth(2) + rvwvcq);  draw((-0.9987724554622847,0.04953364724950819)--(0.5922362871684147,0.8057643453026269), linewidth(2) + rvwvcq);  draw((0.5922362871684147,0.8057643453026269)--(0.06064145095757084,-0.9981596137020172), linewidth(2) + wrwrwr);  draw((0.5249726058304045,-2.4139334560841545)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr);  draw((-0.9987724554622847,0.04953364724950819)--(1.5189031419104242,-0.6923952683993904), linewidth(2) + wrwrwr);  draw((-0.9987724554622847,0.04953364724950819)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr);  draw((-0.9987724554622847,0.04953364724950819)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr);  draw((0.8660254037844386,-0.5)--(2.3530139989292476,0.7523271151020314), linewidth(2) + wrwrwr);  draw((0.8660254037844386,-0.5)--(0.5249726058304045,-2.4139334560841545), linewidth(2) + wrwrwr);  draw(circle((0.5922362871684147,0.8057643453026269), 1.7615883990890795), linewidth(2) + linetype("4 4") + wrwrwr);  draw(circle((0.06064145095757076,-0.9981596137020177), 1.4899728165839203), linewidth(2) + linetype("4 4") + wrwrwr);  draw((0.5249726058304045,-2.4139334560841545)--(0.9854301844182564,0.17008042696736536), linewidth(2) + wrwrwr);  draw((2.3530139989292476,0.7523271151020314)--(0.3454211217688861,-0.9384477868458769), linewidth(2) + wrwrwr);  /* dots and labels */ dot((0,0),dotstyle);  label("$O$", (0.03388760411534265,0.08889671794036069), NE * 0.5);  dot((0.8660254037844386,-0.5),dotstyle);  label("$B$", (0.9043736119372844,-0.4159851665963708), S * 3.5 * labelscalefactor);  dot((-0.9987724554622847,0.04953364724950819),dotstyle);  label("$D$", (-0.9671713048798903,0.13242101833145825), W * 3.5 * labelscalefactor);  dot((0.5922362871684147,0.8057643453026269),dotstyle);  label("$A$", (0.625818089434263,0.8897438451365556), NE * labelscalefactor);  dot((0.06064145095757084,-0.9981596137020172),linewidth(4pt) + dotstyle);  label("$C$", (0.09482162466287856,-0.9295719112113219), S * 3);  dot((2.3530139989292476,0.7523271151020314),dotstyle);  label("$E$", (2.3841998252345853,0.8375146846672384), NE * labelscalefactor);  dot((0.5249726058304045,-2.4139334560841545),dotstyle);  label("$F$", (0.5561792088085077,-2.3310543838046627), S * 3.5 * labelscalefactor);  dot((1.5189031419104242,-0.6923952683993904),linewidth(4pt) + dotstyle);  label("$P$", (1.5572381178037407,-0.6249018084736391), SE * 3);  dot((0.3454211217688861,-0.9384477868458769),linewidth(4pt) + dotstyle);  label("$Q$", (0.38208200724411934,-0.8686378906637853), S * 3);  dot((0.9854301844182564,0.17008042696736536),linewidth(4pt) + dotstyle);  label("$R$", (1.0175367929541368,0.23687933927009236), NE * labelscalefactor); [/asy]

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

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


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

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

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