2018 USAMO Problems/Problem 5

Problem 5

In convex cyclic quadrilateral $ABCD,$ we know that lines $AC$ and $BD$ intersect at $E,$ lines $AB$ and $CD$ intersect at $F,$ and lines $BC$ and $DA$ intersect at $G.$ Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$ , and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$ , where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$ , then $\angle MAC = 90^{\circ}.$

Solution

[asy]

/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */

import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.573333333333343, xmax = 3.56, ymin = -4.74, ymax = 8.473333333333338; /* image dimensions */ pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen cczzff = rgb(0.8,0.6,1); pen ccwwff = rgb(0.8,0.4,1); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);

/* draw figures */

draw(circle((-4.80390015600624,-0.5952574102964114), 2.6896620042551294), linewidth(1) + sexdts); draw((-5.58,1.98)--(-4.06,-3.18), linewidth(1) + cczzff); draw((-3.8846455730896308,1.9324397054436309)--(-7.42,-1.22), linewidth(1) + ccwwff); draw(circle((-7.256640463424001,0.8150682664688008), 2.0416143581437347), linewidth(1) + sexdts); draw(circle((-4.74356842508035,1.555353052957629), 0.9380526686465809), linewidth(1) + sexdts); draw((-13.275816730447621,2.195893092761112)--(-3.912501589632712,1.120300499610314), linewidth(1) + wvvxds); draw((-3.775365275873233,5.118495172394378)--(-8.947688716232484,-0.3288482488643849), linewidth(1) + wvvxds); draw((-5.58,1.98)--(-3.775365275873233,5.118495172394378), linewidth(1) + wvvxds); draw((-5.58,1.98)--(-7.42,-1.22), linewidth(1) + wvvxds); draw((-3.8846455730896308,1.9324397054436309)--(-4.06,-3.18), linewidth(1) + rvwvcq); draw((-3.8846455730896308,1.9324397054436309)--(-3.775365275873233,5.118495172394378), linewidth(1) + rvwvcq); draw((-13.275816730447621,2.195893092761112)--(-4.06,-3.18), linewidth(1) + rvwvcq); draw((-13.275816730447621,2.195893092761112)--(-3.775365275873233,5.118495172394378), linewidth(1) + rvwvcq); draw((-8.947688716232484,-0.3288482488643849)--(-0.2874232022466262,3.539053630345969), linewidth(1) + rvwvcq); draw((-7.42,-1.22)--(-0.2874232022466262,3.539053630345969), linewidth(1) + rvwvcq); draw((-0.2874232022466262,3.539053630345969)--(-4.06,-3.18), linewidth(1) + rvwvcq); draw(circle((-6.783982903277441,1.6253383695771872), 2.915556239332651), linewidth(1) + rvwvcq); draw((-5.58,1.98)--(-3.8846455730896308,1.9324397054436309), linewidth(1) + ccwwff); draw((-5.216225985226909,0.7450829498492438)--(-3.912501589632712,1.120300499610314), linewidth(1) + ccwwff); draw((-8.947688716232484,-0.3288482488643849)--(-5.58,1.98), linewidth(1) + ccwwff); draw((-8.947688716232484,-0.3288482488643849)--(-5.216225985226909,0.7450829498492438), linewidth(1) + ccwwff); draw((-5.58,1.98)--(-3.912501589632712,1.120300499610314), linewidth(1) + ccwwff);

/* dots and labels */

dot((-5.58,1.98),dotstyle); label(" $A$ ", (-5.52,2.113333333333337), NE * labelscalefactor); dot((-7.42,-1.22),dotstyle); label(" $B$ ", (-7.36,-1.0866666666666638), NE * labelscalefactor); dot((-4.06,-3.18),dotstyle); label(" $C$ ", (-4,-3.046666666666664), NE * labelscalefactor); dot((-3.8846455730896308,1.9324397054436309),dotstyle); label(" $D$ ", (-3.8266666666666733,2.06), NE * labelscalefactor); dot((-5.216225985226909,0.7450829498492438),linewidth(4pt) + dotstyle); label(" $E$ ", (-5.16,0.8466666666666699), NE * labelscalefactor); dot((-3.775365275873233,5.118495172394378),linewidth(4pt) + dotstyle); label(" $F$ ", (-3.72,5.22), NE * labelscalefactor); dot((-13.275816730447621,2.195893092761112),linewidth(4pt) + dotstyle); label(" $G$ ", (-13.226666666666675,2.3), NE * labelscalefactor); dot((-8.947688716232484,-0.3288482488643849),linewidth(4pt) + dotstyle); label(" $P$ ", (-8.893333333333342,-0.22), NE * labelscalefactor); dot((-3.912501589632712,1.120300499610314),linewidth(4pt) + dotstyle); label(" $Q$ ", (-3.88,1.18), NE * labelscalefactor); dot((-0.2874232022466262,3.539053630345969),linewidth(4pt) + dotstyle); label(" $H$ ", (-0.24,3.6466666666666705), NE * labelscalefactor); dot((-7.211833579631486,1.4993048370077748),linewidth(4pt) + dotstyle); label(" $M$ ", (-7.16,1.60666666666667), NE * labelscalefactor);

clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);

/* end of picture */

[/asy]

$\angle DEQ + \angle AED + \angle AEP = \angle DAQ + \angle AQD + \angle AEP = 180 - \angle ADC + \angle AEP = 180 - \angle ADC + \angle ABP = \angle ABP + \angle ABC = 180$ so $P,E,Q$ are collinear. Furthermore, note that $DQBP$ is cyclic because: $\angle EDQ = \angle BAE = BPE.$ Notice that since $A$ is the intersection of $(EDQ)$ and $(BPE)$ , it is the Miquel point of $DQBP$ . It follows that $A,M,H$ are collinear. It’s a well known property of Miquel points that $\angle EAH = 90$ , so it follows that $MA \perp AE$ , as desired. $\blacksquare$ ~AopsUser101

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2018 USAMO Problems/Problem 5

Problem 5

Solution