Difference between revisions of "2018 USAMO Problems/Problem 5"
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label("$Q$", (-3.88,1.18), NE * labelscalefactor); | label("$Q$", (-3.88,1.18), NE * labelscalefactor); | ||
dot((-0.2874232022466262,3.539053630345969),linewidth(4pt) + dotstyle); | dot((-0.2874232022466262,3.539053630345969),linewidth(4pt) + dotstyle); | ||
| − | label("$ | + | label("$X$", (-0.24,3.6466666666666705), NE * labelscalefactor); |
dot((-7.211833579631486,1.4993048370077748),linewidth(4pt) + dotstyle); | dot((-7.211833579631486,1.4993048370077748),linewidth(4pt) + dotstyle); | ||
label("$M$", (-7.16,1.60666666666667), NE * labelscalefactor); | label("$M$", (-7.16,1.60666666666667), NE * labelscalefactor); | ||
Revision as of 18:54, 15 March 2021
Problem 5
In convex cyclic quadrilateral 
we know that lines 
and 
intersect at 
lines 
and 
intersect at 
and lines 
and 
intersect at 
Suppose that the circumcircle of 
intersects line 
at 
and 
, and the circumcircle of 
intersects line at 
and 
, where 
and 
are collinear in that order. Prove that if lines 
and 
intersect at 
, then 
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("$X$", (-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]](http://latex.artofproblemsolving.com/4/7/6/47613dab00f8c03e9ab66b4d0a1951fab020e73a.png)
![\[\angle DEQ + \angle AED + \angle AEP = \angle DAQ + \angle AQD + \angle AEP = 180 - \angle ADC + \angle AEP = 180 - \angle ADC + \angle ABP = \angle ABP + \angle ABC = 180\]](http://latex.artofproblemsolving.com/1/d/7/1d7956974acc14bf0af31d432dfc8f91cfd73f4c.png)
so 
are collinear. Furthermore, note that 
is cyclic because:
![\[\angle EDQ = \angle BAE = BPE.\]](http://latex.artofproblemsolving.com/c/5/6/c569e46c4a7583f9181abf802abcb3164fa998f6.png)
Notice that since 
is the intersection of 
and 
, it is the Miquel point of .
Now define 
as the intersection of 
and 
. From Pappus's theorem on 
that 
are collinear. It’s a well known property of Miquel points that 
, so it follows that 
, as desired. 
~AopsUser101
