# 2022 AMC 12B Problems/Problem 19

## Problem

In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$, where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$ $\textbf{(A) }44 \qquad \textbf{(B) }48 \qquad \textbf{(C) }52 \qquad \textbf{(D) }56 \qquad \textbf{(E) }60$

## Diagram $[asy] import geometry; unitsize(2cm); real arg(pair p) { return atan2(p.y, p.x) * 180/pi; } pair G=(0,0),E=(1,0),A=(1/2,sqrt(3)/2),D=1.5*G-0.5*A,C=2*E-A,B=2*D-C; pair t(pair p) { return rotate(-arg(dir(B--C)))*p; } path t(path p) { return rotate(-arg(dir(B--C)))*p; } void d(path p, pen q = black+linewidth(1.5)) { draw(t(p),q); } void o(pair p, pen q = 5+black) { dot(t(p),q); } void l(string s, pair p, pair d) { label(s, t(p),d); } d(A--B--C--cycle); d(A--D); d(B--E); o(A); o(B); o(C); o(D); o(E); o(G); l("A",A,N); l("B",B,SW); l("C",C,SE); l("D",D,S); l("E",E,NE); l("G",G,NW); [/asy]$

## Solution 1 (Law of Cosines)

Let $AG=AE=EG=2x$. Since $E$ is the midpoint of $\overline{AC}$, we must have $EC=2x$.

Since the centroid splits the median in a $2:1$ ratio, $GD=x$ and $BG=4x$.

Applying Law of Cosines on $\triangle ADC$ and $\triangle{}AGB$ yields $AB=\sqrt{28}x$ and $CD=BD=\sqrt{13}x$. Finally, applying Law of Cosines on $\triangle ABC$ yields $\cos(C)=\frac{5}{2\sqrt{13}}=\frac{5\sqrt{13}}{26}$. The requested sum is $5+13+26=44$.

## Solution 2 (Law of Cosines: One Fewer Step)

Let $AG = 1$. Since $\frac{BG}{GE}=2$ (as $G$ is the centroid), $BE = 3$. Also, $EC = 1$ and $\angle{BEC} = 120^{\circ}$. By the law of cosines (applied on $\triangle BEC$), $BC = \sqrt{13}$.

Applying the law of cosines again on $\triangle BEC$ gives $\cos{\angle{C}} = \frac{1 + 13 - 9}{2\sqrt{13}} = \frac{5\sqrt{13}}{26}$, so the answer is $\fbox{\textbf{(A)}\ 44}$.

## Solution 3 (Law of Cosine)

Let $AG = AE = GE = CE = 1$. Since $G$ is the centroid, $DG = \frac12$, $BG = 2$. $$\angle BGD = \angle AGE = 60^{\circ}$$

By the Law of Cosine in $\triangle BGD$ $$BD^2 = BG^2 + DG^2 - 2 \cdot BG \cdot DG \cdot \cos \angle BGD$$ $$BD = \sqrt {2^2 + \left( \frac{1}{2} \right)^2 - 2 \cdot 2 \cdot \frac12 \cdot \cos \angle BGD} = \frac{\sqrt{13}}{2}, \quad CD = \frac{\sqrt{13}}{2}$$

By the Law of Cosine in $\triangle ACD$ $$AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos \angle C$$ $$\cos \angle C = \frac{ AC^2 + CD^2 - AD^2 }{ 2 \cdot AC \cdot CD } = \frac{ 2^2 + \left( \frac{\sqrt{13}}{2} \right)^2 - \left( \frac{3}{2} \right)^2 }{ 2 \cdot 2 \cdot \frac{\sqrt{13}}{2} } = \frac{ 5 \sqrt{13} }{26}$$ $$5 + 13 + 26 = \boxed{\textbf{(A) }44}$$

## Solution 4 (Barycentric Coordinates)

Using reference triangle $\triangle AGE$, we can let $$A=(1,0,0),G=(0,1,0),E=(0,0,1),C=(-1,0,2),D=(-\tfrac{1}{2},\tfrac{3}{2},0),B=(0,3,-2).$$ If we move $A,B,C$ each over by $(1,0,-2)$, leaving $\angle C$ unchanged, we have $$A=(2,0,-2),B=(1,3,-4),C=(0,0,0).$$ The angle $\theta$ between vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ satisfies $$\cos\theta=\frac{(2)(1)+(0)(3)+(-2)(-4)}{\sqrt{\left[2^{2}+0^{2}+(-2)^{2}\right]\left[1^{2}+3^{2}+(-4)^{2}\right]}}=\frac{10}{\sqrt{8\cdot 26}}=\frac{10}{4\sqrt{13}}=\frac{5\sqrt{13}}{26},$$ giving the answer, $5+13+26=\boxed{\textbf{(A)}~44}$.

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## Video Solution (Just 3 min!)

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## Video Solution(Length & Angle Chasing)

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