1966 AMC 12 Problem 11

Attempt Problem 11 of the 1966 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1966 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

The sides of triangle

BAC

are in the ratio

2:3:4.

BD

is the angle-bisector drawn to the shortest side

AC,

dividing it into segments

AD

and

CD.

If the length of

AC

10,

then the length of the longer segment of

AC

is:

3\dfrac12

5

5\dfrac57

6

7\dfrac12

Since

AC

is the shortest side, the adjacent sides are in ratio

3:4

Apply the angle-bisector theorem to

AD:DC

The sides adjacent to angle

B

are in ratio

3:4.

By the angle-bisector theorem,

AD:DC=3:4.

The longer of the two parts is therefore

\frac4{3+4}\cdot10=\frac{40}{7}=5\frac57.

Therefore, the correct answer is C.

Problem 11 in other years