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

Latest revision as of 19:30, 4 November 2020

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 $EQ = FR$ .

$[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

Then

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

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.

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

Latest revision as of 19:30, 4 November 2020

Contents

Problem

Solution 1

Solution 2

See also

@@ Line 4: / Line 4: @@
 <asy>
 unitsize(3cm);
-real labelscalefactor = 0.5; /* changes label-to-point distance */
+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 */
@@ Line 30: / Line 30: @@
 /* dots and labels */
 dot((0,0),dotstyle);
-label("$O$", (0.03388760411534265,0.08889671794036069), NE * labelscalefactor);
+label("$O$", (0.03388760411534265,0.08889671794036069), NE * 0.5);
 dot((0.8660254037844386,-0.5),dotstyle);
-label("$B$", (0.9043736119372844,-0.4159851665963708), NE * labelscalefactor);
+label("$B$", (0.9043736119372844,-0.4159851665963708), S * 3.5 * labelscalefactor);
 dot((-0.9987724554622847,0.04953364724950819),dotstyle);
-label("$D$", (-0.9671713048798903,0.13242101833145825), NE * labelscalefactor);
+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), NE * labelscalefactor);
+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), NE * labelscalefactor);
+label("$F$", (0.5561792088085077,-2.3310543838046627), S * 3.5 * labelscalefactor);
 dot((1.5189031419104242,-0.6923952683993904),linewidth(4pt) + dotstyle);
-label("$P$", (1.5572381178037407,-0.6249018084736391), NE * labelscalefactor);
+label("$P$", (1.5572381178037407,-0.6249018084736391), SE * 3);
 dot((0.3454211217688861,-0.9384477868458769),linewidth(4pt) + dotstyle);
-label("$Q$", (0.38208200724411934,-0.8686378906637853), NE * labelscalefactor);
+label("$Q$", (0.38208200724411934,-0.8686378906637853), S * 3);
 dot((0.9854301844182564,0.17008042696736536),linewidth(4pt) + dotstyle);
 label("$R$", (1.0175367929541368,0.23687933927009236), NE * labelscalefactor);
@@ Line 52: / Line 52: @@
 ==Solution 1==
+First we have that <math>BE=BD=BF</math> by the definition of a reflection. Let <math>\angle DEB = \alpha</math> and <math>\angle DFB = \beta.</math> Since <math>\triangle DBE</math> is isosceles we have <math>\angle BDE = \alpha.</math> Also, we see that <math>\angle BDE = \angle CAB = \angle CDB = \alpha,</math> using similar triangles and the property of cyclic quadrilaterals. Similarly, <cmath>\angle DFB = \angle FDB = \angle ACB = \angle ADB = \beta.</cmath> Now, from <math>BE=BD=BF</math> we know that <math>B</math> is the circumcenter of <math>\triangle DEF.</math> Using the properties of the circumcenter and some elementary angle chasing, we find that <cmath>\angle DPE = 90^{\circ} + \beta - \alpha.</cmath>
+Now, we claim that <math>Q</math> is the intersection of ray <math>\overrightarrow{EB}</math> and the circumcircle of <math>ABCD.</math> To prove this, we just need to show that <math>DEPQ</math> is cyclic by this definition of <math>Q.</math> We have that <cmath>\angle DQE = \angle DCB = \angle DCA + \angle ACB = (90^{\circ}-\alpha)+\beta.</cmath> We also have from before that <cmath>\angle DPE = 90+\beta-\alpha,</cmath> so <math>\angle DQE=\angle DPE</math> and this proves the claim.
+We can use a similar proof to show that <math>F, B, R</math> are collinear.
+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.
+~nukelauncher
+(Monday G. Fern)
+==Solution 2==
+We begin with the following claims.
+Claim. <math>B</math> is the circumcenter of <math>\triangle DEF</math>.
+Proof. By reflection <math>BD=BE=BF</math>.
+Claim. <math>A</math> is the circumcenter of <math>\triangle EPD</math>.
+Proof. First, we have
+    <cmath>\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</cmath>
+Then
+    <cmath> \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}</cmath>
+Then <math>2\angle ADE = 180^{\circ}-2(180^{\circ}-\angle DPE) \implies \angle DAE = 2(180^{\circ}-\angle DPE)</math>. This is enough to imply what we desire. \newline
+Claim. <math>C</math> is the circumcenter of <math>\triangle FPD</math>.
+Proof. Similar to above.
+Claim. <math>E, B, Q</math> are collinear.
+Proof. We have
+    <cmath>\angle ABE = \angle ABD = \angle AQD = \angle ADQ = 180^{\circ}-\angle ABQ</cmath>
+Claim.<math>F, B, R</math> are collinear.
+Proof. Similar to above.
+Since <math>DEPQ</math> is cyclic, <math>\triangle EBP \sim \triangle DBQ</math>. However, <math>BE=BD</math>, so <math>BP=BQ</math>. Similarly, <math>BP=BR</math>. Finishing, we have <cmath>EQ = EB + BQ = BD + BP = FB + BR = FR</cmath>, as desired. <math>\blacksquare</math> ~MSC
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