Difference between revisions of "2021 AIME II Problems/Problem 14"
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Let <math>\Delta ABC</math> be an acute triangle with circumcenter <math>O</math> and centroid <math>G</math>. Let <math>X</math> be the intersection of the line tangent to the circumcircle of <math>\Delta ABC</math> at <math>A</math> and the line perpendicular to <math>GO</math> at <math>G</math>. Let <math>Y</math> be the intersection of lines <math>XG</math> and <math>BC</math>. Given that the measures of <math>\angle ABC, \angle BCA, </math> and <math>\angle XOY</math> are in the ratio <math>13 : 2 : 17, </math> the degree measure of <math>\angle BAC</math> can be written as <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Let <math>\Delta ABC</math> be an acute triangle with circumcenter <math>O</math> and centroid <math>G</math>. Let <math>X</math> be the intersection of the line tangent to the circumcircle of <math>\Delta ABC</math> at <math>A</math> and the line perpendicular to <math>GO</math> at <math>G</math>. Let <math>Y</math> be the intersection of lines <math>XG</math> and <math>BC</math>. Given that the measures of <math>\angle ABC, \angle BCA, </math> and <math>\angle XOY</math> are in the ratio <math>13 : 2 : 17, </math> the degree measure of <math>\angle BAC</math> can be written as <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>M</math> be the midpoint of <math>BC</math>. Because <math>\angle{OAX}=\angle{OGX}=\angle{OGY}=\angle{OMY}=90^o</math>, <math>AXOG</math> and <math>OMYG</math> are cyclic, so <math>O</math> is the center of the spiral similarity sending <math>AM</math> to <math>XY</math>, and <math>\angle{XOY}=\angle{AOM}</math>. Because <math>\angle{AOM}=2\angle{BCA}+\angle{BAC}</math>, it's easy to get <math>\frac{585}{7} \implies \boxed{592}</math> from here. | Let <math>M</math> be the midpoint of <math>BC</math>. Because <math>\angle{OAX}=\angle{OGX}=\angle{OGY}=\angle{OMY}=90^o</math>, <math>AXOG</math> and <math>OMYG</math> are cyclic, so <math>O</math> is the center of the spiral similarity sending <math>AM</math> to <math>XY</math>, and <math>\angle{XOY}=\angle{AOM}</math>. Because <math>\angle{AOM}=2\angle{BCA}+\angle{BAC}</math>, it's easy to get <math>\frac{585}{7} \implies \boxed{592}</math> from here. | ||
~Lcz | ~Lcz | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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>. | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=13|num-a=15}} | {{AIME box|year=2021|n=II|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:04, 23 March 2021
Contents
[hide]Problem
Let be an acute triangle with circumcenter and centroid . Let be the intersection of the line tangent to the circumcircle of at and the line perpendicular to at . Let be the intersection of lines and . Given that the measures of and are in the ratio the degree measure of can be written as where and are relatively prime positive integers. Find .
Solution 1
Let be the midpoint of . Because , and are cyclic, so is the center of the spiral similarity sending to , and . Because , it's easy to get from here.
~Lcz
Solution 2
Let be the midpoint of . Because we have cyclic and so ; likewise since we have cyclic and so . Now note that are collinear since is a median, so . But . Now letting we have and so .
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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