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2022 AMC 12B Problems/Problem 19: Difference between revisions

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\textbf{(D)}\ 56 \qquad
\textbf{(D)}\ 56 \qquad
\textbf{(E)}\ 60 \qquad</math>
\textbf{(E)}\ 60 \qquad</math>
== 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==
==Solution 1: Law of Cosines==

Revision as of 03:59, 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

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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