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

m (→‎Solution 2)
m (→‎Solution 2)
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==Solution 2==
 
==Solution 2==
 +
In this solution, all angle measures are in degrees.
 +
 
Let <math>M</math> be the midpoint of <math>\overline{BC}</math> so that <math>A,G,</math> and <math>M</math> are collinear. Let <math>\angle ABC=13k,\angle BCA=2k</math> and <math>\angle XOY=17k.</math>  
 
Let <math>M</math> be the midpoint of <math>\overline{BC}</math> so that <math>A,G,</math> and <math>M</math> are collinear. Let <math>\angle ABC=13k,\angle BCA=2k</math> and <math>\angle XOY=17k.</math>  
  
 
We note that:
 
We note that:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
−
   <li>Since <math>\angle OGX = \angle OAX = 90^\circ,</math> quadrilateral <math>OGAX</math> is cyclic by the Converse of the Inscribed Angle Theorem.<p>It follows that <math>\angle OAG = \angle OXG,</math> as they share the same intercepted arc <math>OG.</math></li><p>
+
   <li>Since <math>\angle OGX = \angle OAX = 90,</math> quadrilateral <math>OGAX</math> is cyclic by the Converse of the Inscribed Angle Theorem.<p>It follows that <math>\angle OAG = \angle OXG,</math> as they share the same intercepted arc <math>OG.</math></li><p>
−
   <li>Since <math>\angle OGY = \angle OMY = 90^\circ,</math> quadrilateral <math>OGYM</math> is cyclic by the supplementary opposite angles.<p>It follows that <math>\angle OMG = \angle OYG,</math> as they share the same intercepted arc <math>OG.</math></li><p>
+
   <li>Since <math>\angle OGY = \angle OMY = 90,</math> quadrilateral <math>OGYM</math> is cyclic by the supplementary opposite angles.<p>It follows that <math>\angle OMG = \angle OYG,</math> as they share the same intercepted arc <math>OG.</math></li><p>
 
</ol>
 
</ol>
 
Together, we conclude that <math>\triangle OAM \sim \triangle OXY</math> by AA, so <math>\angle AOM = \angle XOY = 17k.</math>
 
Together, we conclude that <math>\triangle OAM \sim \triangle OXY</math> by AA, so <math>\angle AOM = \angle XOY = 17k.</math>

Revision as of 14:37, 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

In this solution, all angle measures are in degrees.

Let $M$ be the midpoint of $\overline{BC}$ so that $A,G,$ and $M$ are collinear. Let $\angle ABC=13k,\angle BCA=2k$ and $\angle XOY=17k.$

We note that:

  1. Since $\angle OGX = \angle OAX = 90,$ quadrilateral $OGAX$ is cyclic by the Converse of the Inscribed Angle Theorem.

    It follows that $\angle OAG = \angle OXG,$ as they share the same intercepted arc $OG.$

  2. Since $\angle OGY = \angle OMY = 90,$ quadrilateral $OGYM$ is cyclic by the supplementary opposite angles.

    It follows that $\angle OMG = \angle OYG,$ as they share the same intercepted arc $OG.$

Together, we conclude that $\triangle OAM \sim \triangle OXY$ by AA, so $\angle AOM = \angle XOY = 17k.$

Next, we express $\angle BAC$ in terms of $k.$ By angle addition, we have \begin{align*} \angle AOM &= \angle AOB + \angle BOM \\ &= \underbrace{2\angle BCA}_{\substack{\text{Inscribed} \\ \text{Angle} \\ \text{Theorem}}} + \underbrace{\frac12\angle BOC}_{\substack{\text{Perpendicular} \\ \text{Bisector} \\ \text{Property}}} \\ &= 2\angle BCA + \underbrace{\angle BAC}_{\substack{\text{Inscribed} \\ \text{Angle} \\ \text{Theorem}}}. \end{align*} Substituting back gives $17k=2(2k)+\angle BAC,$ from which $\angle BAC=13k.$

For the sum of the interior angles (in degrees) of $\triangle ABC,$ we get \begin{align*} \angle ABC + \angle BCA + \angle BAC &= 180 \\ 13k+2k+13k&=180 \\ 28k&=180 \\ k&=\frac{45}{7}. \end{align*} Finally, we obtain $\angle BAC=13k=\frac{585}{7},$ from which the answer is $585+7=\boxed{592}.$

~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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