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Difference between revisions of "2021 AIME II Problems/Problem 14"

m (→‎Guessing Solution for last 3 minutes (unreliable): Changed title so it is "Solution 3".)
(→‎Solution 2: This solution is GREAT, but some notations (like angle A) are ambiguous. I will reformat it a little.)
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==Solution 2==
 
==Solution 2==
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Let <math>M</math> be the midpoint of <math>BC</math>. Because <math>\angle OAX = \angle OGX = 90</math> we have <math>AXOG</math> cyclic and so <math>\angle GXO = \angle OAG</math>; likewise since <math>\angle OMY = \angle OGY = 90</math> we have <math>OMYG</math> cyclic and so <math>\angle OYG = \angle OMG</math>. Now note that <math>A, G, M</math> are collinear since <math>\overline{AM}</math> is a median, so <math>\triangle AOM \sim \triangle XOY</math>. But <math>\angle AOM = \angle AOB + \angle BOM = 2 \angle C + \angle A</math>. Now letting <math>\angle C = 2k, \angle B = 13k, \angle AOM = \angle XOY = 17k</math> we have <math>\angle A = 13k</math> and so <math>\angle A = \frac{585}{7} \implies \boxed{592}</math>.
+
Let <math>M</math> be the midpoint of <math>\overline{BC}.</math> We note that:
 +
<ol style="margin-left: 1.5em;">
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  <li>Since <math>\angle OGX = \angle OAX = 90^\circ,</math> we conclude that <math>OGAX</math> is cyclic by the Converse of the Inscribed Angle Theorem.</li><p>
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  <li>Since <math>\angle OGY = \angle OMY = 90^\circ,</math> we conclude that <math>OGYM</math> is cyclic by the supplementary opposite angles.</li><p>
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</ol>
 +
 +
 
 +
and so <math>\angle GXO = \angle OAG</math>; likewise since <math>\angle OMY = \angle OGY = 90</math> we have <math>OMYG</math> cyclic and so <math>\angle OYG = \angle OMG</math>. Now note that <math>A, G, M</math> are collinear since <math>\overline{AM}</math> is a median, so <math>\triangle AOM \sim \triangle XOY</math>. But <math>\angle AOM = \angle AOB + \angle BOM = 2 \angle C + \angle A</math>. Now letting <math>\angle C = 2k, \angle B = 13k, \angle AOM = \angle XOY = 17k</math> we have <math>\angle A = 13k</math> and so <math>\angle A = \frac{585}{7} \implies \boxed{592}</math>.
 +
 
 +
~Constance-variance (Fundamental Logic)
 +
 
 +
~MRENTHUSIASM (Reformatting)
  
 
==Solution 3 (Guessing in the Last 3 Minutes, Unreliable)==
 
==Solution 3 (Guessing in the Last 3 Minutes, Unreliable)==

Revision as of 01:21, 1 June 2021

Problem

Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$. Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$. Let $Y$ be the intersection of lines $XG$ and $BC$. Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Diagram

File:2021 AIME II Problem 14 Diagram.png

~MRENTHUSIASM (by Geometry Expressions)

Solution 1

Let $M$ be the midpoint of $BC$. Because $\angle{OAX}=\angle{OGX}=\angle{OGY}=\angle{OMY}=90^o$, $AXOG$ and $OMYG$ are cyclic, so $O$ is the center of the spiral similarity sending $AM$ to $XY$, and $\angle{XOY}=\angle{AOM}$. Because $\angle{AOM}=2\angle{BCA}+\angle{BAC}$, it's easy to get $\frac{585}{7} \implies \boxed{592}$ from here.

~Lcz

Solution 2

Let $M$ be the midpoint of $\overline{BC}.$ We note that:

  1. Since $\angle OGX = \angle OAX = 90^\circ,$ we conclude that $OGAX$ is cyclic by the Converse of the Inscribed Angle Theorem.

  2. Since $\angle OGY = \angle OMY = 90^\circ,$ we conclude that $OGYM$ is cyclic by the supplementary opposite angles.


and so $\angle GXO = \angle OAG$; likewise since $\angle OMY = \angle OGY = 90$ we have $OMYG$ cyclic and so $\angle OYG = \angle OMG$. Now note that $A, G, M$ are collinear since $\overline{AM}$ is a median, so $\triangle AOM \sim \triangle XOY$. But $\angle AOM = \angle AOB + \angle BOM = 2 \angle C + \angle A$. Now letting $\angle C = 2k, \angle B = 13k, \angle AOM = \angle XOY = 17k$ we have $\angle A = 13k$ and so $\angle A = \frac{585}{7} \implies \boxed{592}$.

~Constance-variance (Fundamental Logic)

~MRENTHUSIASM (Reformatting)

Solution 3 (Guessing in the Last 3 Minutes, Unreliable)

Notice that $\triangle ABC$ looks isosceles, so we assume it's isosceles. Then, let $\angle BAC = \angle ABC = 13x$ and $\angle BCA = 2x.$ Taking the sum of the angles in the triangle gives $28x=180,$ so $13x = \frac{13}{28} \cdot 180 = \frac{585}{7}$ so the answer is $\boxed{592}.$

Video Solution 1

https://www.youtube.com/watch?v=zFH1Z7Ydq1s

Video Solution 2

https://www.youtube.com/watch?v=7Bxr2h4btWo

~Osman Nal

See also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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