2021 AIME II Problems/Problem 14

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 $\angle XOY=17k.$

Note that:

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 $OG.$

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 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 (Reconstruction)

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 $\boxed{592}.$

Solution 4 (Easy and Simple)

Video Solution 1

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

Video Solution 2

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

~Osman Nal

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