2022 AMC 12B Problems/Problem 19
Problem
In 
medians 
and 
intersect at 
and 
is equilateral. Then 
can be written as 
, where 
and 
are relatively prime positive integers and 
is a positive integer not divisible by the square of any prime. What is 
?
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]](http://latex.artofproblemsolving.com/9/8/6/986e7af100c4c0c6eca5727bbe1d7a64cdc47449.png)
Solution 1: Law of Cosines
Let 
. Since 
is the midpoint of 
, 
must also be 
.
Since the centroid splits the median in a 
ratio, 
must be equal to 
and 
must be equal to 
.
Applying Law of Cosines on 
and 
yields 
and 
. Finally, applying Law of Cosines on 
yields 
. The requested sum is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
