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

Revision as of 00:30, 10 January 2023

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 $\EC=2x$ (Error compiling LaTeX. Unknown error_msg).

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

@@ Line 64: / Line 64: @@
 ==Solution 2 (Law of Cosines: One Fewer Step) ==
-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 (applied on <math>\bigtriangleup BEC</math>), <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 (applied on <math>\triangle BEC</math>), <math>BC = \sqrt{13}</math>.
-Applying the law of cosines again on <math>\bigtriangleup 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>.
+Applying the law of cosines again on <math>\triangle 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>.
 ~[[User:Bxiao31415 | Bxiao31415]]

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