1970 AMC 12 Problem 28

Question

In triangle

ABC,

the median from vertex

A

is perpendicular to the median from vertex

B.

If the lengths of sides

AC

and

BC

are

6

and

7,

respectively, then the length of side

AB

is:

\sqrt{17}

4

4\tfrac12

2\sqrt5

4\tfrac14

Place the centroid at the origin and let the position vectors of

A,B

be perpendicular vectors

u,v

Then

C=-u-v,

so

AC^2=|2u+v|^2

and

BC^2=|u+2v|^2

Place the centroid at the origin. Let the position vectors of

A

and

B

be

u

and

v.

The two medians lie along

u

and

v,

so

u\cdot v=0.

Also the centroid condition gives

C=-u-v.

Let

U=|u|^2

and

V=|v|^2.

Then

AC^2=|2u+v|^2=4U+V=36

and

\begin{aligned} BC^2&=|u+2v|^2\\ &=U+4V=49. \end{aligned}

Adding appropriate multiples gives

U+V=17.

Finally,

AB^2=|u-v|^2=U+V=17,

so

AB=\sqrt{17}.

Therefore, the correct answer is A.

Problem 28 in other years