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

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Applying Law of Cosines on <math>\triangle{}ADC</math> and <math>\triangle{}AGB</math> yields <math>\overline{AB}=\sqrt{28}x</math> and <math>\overline{CD}=\overline{BD}=\sqrt{13}x</math>. Finally, applying Law of Cosines on <math>\triangle{}ABC</math> yields <math>\cos{C}=\frac{5}{2\sqrt{13}}=\frac{5\sqrt{13}}{26}</math>. The requested sum is <math>5+13+26=44</math>.
 
Applying Law of Cosines on <math>\triangle{}ADC</math> and <math>\triangle{}AGB</math> yields <math>\overline{AB}=\sqrt{28}x</math> and <math>\overline{CD}=\overline{BD}=\sqrt{13}x</math>. Finally, applying Law of Cosines on <math>\triangle{}ABC</math> yields <math>\cos{C}=\frac{5}{2\sqrt{13}}=\frac{5\sqrt{13}}{26}</math>. The requested sum is <math>5+13+26=44</math>.
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 +
==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>.
 +
 +
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{(A)44}</math>.
 +
 +
~[[User:Bxiao31415 | Bxiao31415]]
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2022|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2022|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 06:47, 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 $\frac{BG}{GE}=2$, $BE = 3$. 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 $\fbox{(A)44}$.

~ Bxiao31415

See Also

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

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