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Difference between revisions of "2018 USAMO Problems/Problem 5"

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  /* end of picture */
 
  /* end of picture */
 
</asy>
 
</asy>
−
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----
−
<cmath>\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</cmath>
+
<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:
 
so <math>P,E,Q</math> are collinear. Furthermore, note that <math>DQBP</math> is cyclic because:
 
<cmath>\angle EDQ = \angle BAE = BPE.</cmath>
 
<cmath>\angle EDQ = \angle BAE = BPE.</cmath>
Line 71: Line 78:
 
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>
 
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
 
~AopsUser101
 +
 +
==Video Solution by MOP 2024==
 +
https://youtu.be/jORAIJDLzp4
 +
 +
~r00tsOfUnity

Latest revision as of 09:46, 27 August 2023

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), 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]

\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*} 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 $DQBP$.

Now define $X$ as the intersection of $BQ$ and $DP$. From Pappus's theorem on $BFPDGQ$ that $A,M,X$ are collinear. It’s a well known property of Miquel points that $\angle EAX = 90$, so it follows that $MA \perp AE$, as desired. $\blacksquare$ ~AopsUser101

Video Solution by MOP 2024

https://youtu.be/jORAIJDLzp4

~r00tsOfUnity

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