Difference between revisions of "2018 USAMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
− | https:// | + | <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), N * labelscalefactor); | ||
+ | dot((-7.42,-1.22),dotstyle); | ||
+ | label("$B$", (-7.36,-1.0866666666666638), SW * 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> | ||
+ | ---- | ||
+ | <cmath>\begin{align*} | ||
+ | &\mathrel{\phantom{=}}\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 | ||
+ | \end{align*}</cmath> | ||
+ | so <math>P,E,Q</math> are collinear. Furthermore, note that <math>DQBP</math> is cyclic because: | ||
+ | <cmath>\angle EDQ = \angle BAE = BPE.</cmath> | ||
+ | Notice that since <math>A</math> is the intersection of <math>(EDQ)</math> and <math>(BPE)</math>, it is the Miquel point of <math>DQBP</math>. | ||
+ | |||
+ | Now define <math>X</math> as the intersection of <math>BQ</math> and <math>DP</math>. From Pappus's theorem on <math>BFPDGQ</math> that <math>A,M,X</math> are collinear. It’s a well known property of Miquel points that <math>\angle EAX = 90</math>, so it follows that <math>MA \perp AE</math>, as desired. <math>\blacksquare</math> | ||
+ | ~AopsUser101 | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/jORAIJDLzp4 | ||
+ | |||
+ | ~r00tsOfUnity |
Latest revision as of 10:46, 27 August 2023
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
so
are collinear. Furthermore, note that
is cyclic because:
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
Video Solution by MOP 2024
~r00tsOfUnity