2022 AMC 12B Problems/Problem 19

Revision as of 06:56, 18 November 2022 by Bxiao31415 (talk | contribs) (Solution 2 (Also Law of Cosines, but with one less computation))

Problem

In $\triangle ABC$ medians $\overline{\rm AD}$ and $\overline{\rm BE}$ 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 $m+n+p$?

$\textbf{(A)}\ 44 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 52 \qquad \textbf{(D)}\ 56 \qquad \textbf{(E)}\ 60 \qquad$

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

Since the centroid splits the median in a $2:1$ ratio, $\overline{GD}$ must be equal to $x$ and $\overline{BG}$ must be equal to $4x$.

Applying Law of Cosines on $\triangle{}ADC$ and $\triangle{}AGB$ yields $\overline{AB}=\sqrt{28}x$ and $\overline{CD}=\overline{BD}=\sqrt{13}x$. 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 (Also Law of Cosines, but with one less computation)

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, $BC = \sqrt{13}$.

Applying the law of cosines again on $BEC$ 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

See Also

2022 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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