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

Revision as of 08:25, 25 March 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 $m+n$ .

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 $BC$ . Because $\angle OAX = \angle OGX = 90$ we have $AXOG$ cyclic 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 $\angle A = \frac{585}{7} \implies \boxed{592}$ .

Guessing Solution for 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 $\boxed{592}.$

@@ Line 10: / Line 10: @@
 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>.
-==Guessing Solution for last 3 minutes (unreliable)
+==Guessing Solution for last 3 minutes (unreliable)==
 Notice that <math>\triangle ABC</math> looks isosceles, so we assume it's isosceles. Then, let <math>\angle BAC = \angle ABC = 13x</math> and <math>\angle BCA = 2x.</math> Taking the sum of the angles in the triangle gives <math>28x=180,</math> so <math>13x = \frac{13}{28} \cdot 180 = \frac{585}{7}</math> so the answer is <math>\boxed{592}.</math>

2021 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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