Difference between revisions of "2022 AMC 12B Problems/Problem 19"
(→Problem) |
(→Problem) |
||
Line 8: | Line 8: | ||
\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 04:59, 18 November 2022
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
Solution 1: Law of Cosines
Note: can someone add the diagram here please, I don't know how to do that
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.