Difference between revisions of "2022 AMC 12B Problems/Problem 19"

Revision as of 07:56, 18 November 2022

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

@@ Line 68: / Line 68: @@
 ==Solution 2 (Also Law of Cosines, but with one less computation) ==
-Let <math>AG = 1</math>. Since <math>\frac{BG}{GE}=2</math>, <math>BE = 3</math>. Also, <math>EC = 1</math> and <math>\angle{BEC} = 120^{\circ}</math>. By the law of cosines, <math>BC = \sqrt{13}</math>.
+Let <math>AG = 1</math>. Since <math>\frac{BG}{GE}=2</math> (as <math>G</math> is the centroid), <math>BE = 3</math>. Also, <math>EC = 1</math> and <math>\angle{BEC} = 120^{\circ}</math>. By the law of cosines, <math>BC = \sqrt{13}</math>.
 Applying the law of cosines again on <math>BEC</math> gives <math>\cos{\angle{C}} = \frac{1 + 13 - 9}{2\sqrt{13}} = \frac{5\sqrt{13}}{26}</math>, so the answer is <math>\fbox{\textbf{(A)}\ 44}</math>.

2022 AMC 12B (Problems • Answer Key • Resources)
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

Page

Toolbox

Search