1962 AMC 12 Problem 39

Question

The medians

AN

and

BP

of a triangle with unequal sides are, respectively,

3

inches and

6

inches long. Its area is

3\sqrt{15}

square inches. The length of the third median, in inches, is:

4

3\sqrt3

3\sqrt6

6\sqrt3

6\sqrt6

The three medians form the side lengths of a triangle whose area is three-fourths the original area

Use the two known median lengths and that area to find the two possible included angles, then reject the case that makes two medians equal

The triangle whose sides are the three medians has area

\frac34\left(3\sqrt{15}\right)=\frac{9\sqrt{15}}4.

If

\theta

is the included angle between its sides

3

and

6,

then

9\sin\theta=\frac{9\sqrt{15}}4,

so

\cos\theta=\pm\frac{1}{4}.

By the law of cosines, the third median

m

satisfies

m^2=3^2+6^2-2(3)(6)\cos\theta,

giving

m^2=36

or

54.

The value

m=6

would make two medians, and hence two sides, equal. Because the triangle has unequal sides,

m=\sqrt{54}=3\sqrt6.

Therefore, the correct answer is C.

Problem 39 in other years