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 $\fbox{\textbf{(A)}\ 44}$ .

~ Bxiao31415

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 $\triangle ACD$

$AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos \angle C$

$\cos \angle C = \frac{ AC^2 + CD^2 - AD^2 }{ 2 \cdot AC \cdot CD } = \frac{ 2^2 + \left( \frac{\sqrt{13}}{2} \right)^2 - \left( \frac{3}{2} \right)^2 }{ 2 \cdot 2 \cdot \frac{\sqrt{13}}{2} } = \frac{ 5 \sqrt{13} }{26}$

$5 + 13 + 26 = \boxed{\textbf{(A) }44}$

~isabelchen

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, $5+13+26=\boxed{\textbf{(A)}~44}$ .

~r00tsOfUnity

Solution 5 (Coordinate Bash)

Let $A$ be at $(0,0)$, with $E$ at $(1,0)$ and $G$ at $(\tfrac{1}{2},\tfrac{\sqrt{3}}{2})$. Because $\triangle AGE$ is equilateral, the equation of line $AD$ is $y=x\sqrt{3}$, and the equation for $BE$ is $y=-x\sqrt{3}+\sqrt{3}$. Because $BE$ is a median, we know that $C$ is at $(2,0)$.

Therefore, the equation for line $BC$ is $y=m(x-2)$, where $m$ is the slope of the line. Since $AD$ is also a median, $D$ is the midpoint of $\overline{BC}$.

To find $m$, solve for the coordinates of $D$ in terms of $m$:

$x\sqrt{3}=mx-2m$ $x(m-\sqrt{3})=2m$ $x=\frac{2m}{m-\sqrt{3}}$

The $y$-value of $D$ is then $y=x\sqrt{3}$:

$y=\frac{2m\sqrt{3}}{m-\sqrt{3}}$

Because $D$ is the midpoint of $\overline{BC}$ and $y_C=0$, the $y$-value of $B$ is double that of $D$:

$y_B=\frac{4m\sqrt{3}}{m-\sqrt{3}}$

Now use the equation for line $BE$ to express $B$:

$mx-2m=-x\sqrt{3}+\sqrt{3}$ $x(m+\sqrt{3})=2m+\sqrt{3}$ $x_B=\frac{2m+\sqrt{3}}{m+\sqrt{3}}$

Plug this back into $BE$ to find $y_B$:

$y_B=\frac{-2m-\sqrt{3}}{m+\sqrt{3}}+\sqrt{3}=\frac{-m\sqrt{3}}{m+\sqrt{3}}$

Equating the two expressions for $y_B$:

$-\frac{m\sqrt{3}}{m+\sqrt{3}}=\frac{4m\sqrt{3}}{m-\sqrt{3}}$ $-\sqrt{3}m^2+3m=4\sqrt{3}m^2+12m$ $5\sqrt{3}m^2+9m=0$ $m(5m\sqrt{3}+9)=0$

Since $m\neq0$, we have

$m=-\frac{3\sqrt{3}}{5}.$

Now find $B$:

$-x\sqrt{3}+\sqrt{3}=-\frac{3\sqrt{3}}{5}x+\frac{6\sqrt{3}}{5}$ $\frac{-2\sqrt{3}}{5}x=\frac{\sqrt{3}}{5}$ $x_B=-\frac{1}{2}$

Plug this into $BE$ to get

$y_B=\frac{3\sqrt{3}}{2}.$

Since we’re asked to find $\cos\angle C$, extend $\overline{AC}$ to $F$ where $F=(-\frac{1}{2},0)$, forming right triangle $\triangle BFC$. Using the Pythagorean theorem for $\overline{BC}$:

$\left(\frac{5}{2}\right)^2+\left(\frac{3\sqrt{3}}{2}\right)^2=BC^2,\quad BC=\sqrt{13}.$

Thus

$\cos\angle C=\frac{\frac{5}{2}}{\sqrt{13}}=\frac{5\sqrt{13}}{26}.$

The problem asks for $m+n+p$, so $5+13+26=\boxed{44}$.

~Voidling

Video Solution by MOP 2024

https://youtu.be/QNjvpYI1V5g

~r00tsOfUnity

Video Solution (Just 3 min!)

https://youtu.be/Q54sH65AJa4

~Education, the Study of Everything

Video Solution(Length & Angle Chasing)

https://youtu.be/JVDlHCSPF6k

~Hayabusa1

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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