Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(no spiral sim)
Line 2: Line 2:
 
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

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 $\angle A = \frac{585}{7} \implies \boxed{592}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_14&oldid=149967"