2022 AMC 12B Problem 19

Question

In

\triangle ABC

medians

\overline{AD}

and

\overline{BE}

intersect at

G

and

\triangle AGE

is equilateral. Then

\cos(C)

can be written as

\dfrac{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?

44

48

52

56

60

Answer: A

Concepts:median (geometry)centroid law of cosines

Difficulty rating: 2020

Small hint:

The centroid splits

AD

so

AG = \tfrac23 AD,

and

GE = \tfrac13 BE;

also

AE = \tfrac12 AC

Big hint:

Set all three equal and use the median-length formula to relate the sides, then apply the law of cosines

Solution:

Let

a = BC,

b = CA,

c = AB.

Since

E

is the midpoint of

AC,

AE = \tfrac{b}{2}.

The centroid gives

AG = \tfrac23 m_a

and

GE = \tfrac13 m_b,

where

m_a, m_b

are the medians from

A

and

B.

Equilateral

\triangle AGE

means

AG = GE = AE.

From

\tfrac23 m_a = \tfrac{b}{2}

we get

m_a = \tfrac34 b,

which with

m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}

gives

2c^2 - a^2 = \tfrac{b^2}{4}.

From

\tfrac13 m_b = \tfrac{b}{2}

we get

m_b = \tfrac32 b,

giving

a^2 + c^2 = 5b^2.

Solving,

c^2 = \tfrac{7b^2}{4}

and

a^2 = \tfrac{13b^2}{4}.

Taking

b = 2