Difference between revisions of "2022 AMC 12B Problems/Problem 19"
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==Solution 2 (Law of Cosines: One Fewer Step) == | ==Solution 2 (Law of Cosines: One Fewer Step) == | ||
− | Let <math>AG = 1</math>. Since <math>\frac{BG}{GE}=2</math> (as <math>G</math> is the centroid), <math>BE = 3</math>. Also, <math>EC = 1</math> and <math>\angle{BEC} = 120^{\circ}</math>. By the law of cosines (applied on <math>\ | + | Let <math>AG = 1</math>. Since <math>\frac{BG}{GE}=2</math> (as <math>G</math> is the centroid), <math>BE = 3</math>. Also, <math>EC = 1</math> and <math>\angle{BEC} = 120^{\circ}</math>. By the law of cosines (applied on <math>\triangle BEC</math>), <math>BC = \sqrt{13}</math>. |
− | Applying the law of cosines again on <math>\ | + | Applying the law of cosines again on <math>\triangle 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{\textbf{(A)}\ 44}</math>. |
~[[User:Bxiao31415 | Bxiao31415]] | ~[[User:Bxiao31415 | Bxiao31415]] |
Revision as of 01:30, 10 January 2023
Contents
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)
Let . Since
is the midpoint of
, we must have $\EC=2x$ (Error compiling LaTeX. Unknown error_msg).
Since the centroid splits the median in a ratio,
and
.
Applying Law of Cosines on and
yields
and
. Finally, applying Law of Cosines on
yields
. The requested sum is
.
Solution 2 (Law of Cosines: One Fewer Step)
Let . Since
(as
is the centroid),
. Also,
and
. By the law of cosines (applied on
),
.
Applying the law of cosines again on gives
, so the answer 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.