jmc

Question

What is the sum of the squares of the lengths of the medians of a triangle whose side lengths are 10, 10, and 12?

Accepted Answer

Let us draw our triangle and medians and label our points of interest: We have made

A B = A C = 10

and

B C = 12.

We can notice a few useful things. Since

A B C

is isosceles, it follows that

A D

is an altitude as well as a median, which is useful for finding lengths, since it means we can use the Pythagorean Theorem. At this point, we can drop additional segments from

E

and

F

perpendicular to

B C,

meeting

B C

at

G

and

H,

respectively: Since

D C = 6

and

A C = 10,

we have a

3 : 4 : 5

Pythagorean triple and

A D = 8

. Since

△ B F H \sim △ B A D

and

B F = \frac{1}{2} \cdot A B

(since F is the midpoint of AB), we can see that

F H = \frac{1}{2} \cdot A D = 4

and

B H = \frac{1}{2} \cdot B D = \frac{1}{4} \cdot B C = 3.

H C = B C - B H = 12 - 3 = 9.

To find

C F^{2},

we simply use the Pythagorean Theorem:

C F^{2} = F H^{2} + H C^{2} = 16 + 81 = 97.

By symmetry, we can see that

B E^{2} = 97.

From before, we have that

A D^{2} = 8^{2} = 64.

Our answer is therefore

A D^{2} + B E^{2} + C F^{2} = 64 + 97 + 97 = 258 .