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

(Created page with "== Problem == In <math>\triangle ABC</math> medians <math>\overline{\rm AD}</math> and <math>\overline{\rm BE}</math> intersect at <math>G</math> and <math>\triangle AGE</mat...")
 
(→‎Problem)
Line 8: Line 8:
 
\textbf{(D)}\ 56 \qquad
 
\textbf{(D)}\ 56 \qquad
 
\textbf{(E)}\ 60 \qquad</math>
 
\textbf{(E)}\ 60 \qquad</math>
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==Solution 1: Law of Cosines==
 +
Note: can someone add the diagram here please, I don't know how to do that
 +
 +
Let <math>\overline{AG}=\overline{AE}=\overline{EG}=2x</math>. Since <math>E</math> is the midpoint of <math>\overline{AC}</math>, <math>\overline{EC}</math> must also be <math>2x</math>.
 +
 +
Since the centroid splits the median in a <math>2:1</math> ratio, <math>\overline{GD}</math> must be equal to <math>x</math> and <math>\overline{BG}</math> must be equal to <math>4x</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>.
  
 
== 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 04:04, 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 $m+n+p$?

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

Solution 1: Law of Cosines

Note: can someone add the diagram here please, I don't know how to do that

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 $5+13+26=44$.

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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