Difference between revisions of "2022 AMC 12B Problems/Problem 19"
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==Solution 1 (Law of Cosines)== | ==Solution 1 (Law of Cosines)== | ||
− | Let <math>AG=AE=EG=2x</math>. Since <math>E</math> is the midpoint of <math>\overline{AC}</math>, we must have <math> | + | Let <math>AG=AE=EG=2x</math>. Since <math>E</math> is the midpoint of <math>\overline{AC}</math>, we must have <math>EC=2x</math>. |
Since the centroid splits the median in a <math>2:1</math> ratio, <math>GD=x</math> and <math>BG=4x</math>. | Since the centroid splits the median in a <math>2:1</math> ratio, <math>GD=x</math> and <math>BG=4x</math>. |
Revision as of 20:29, 11 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
.
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 |
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