1959 AMC 12 Problem 28

Attempt Problem 28 of the 1959 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 1959 AMC 12 solutions, or check the answer key.

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

28.

In triangle

ABC,

\overline{AL}

bisects angle

A

and

\overline{CM}

bisects angle

C.

Points

L

and

M

are on

\overline{BC}

and

\overline{AB},

respectively. The sides of triangle

ABC

are

a,

b,

and

c.

Then

\dfrac{AM}{MB}=k\dfrac{CL}{LB},

where

k

is:

1

\dfrac{bc}{a^2}

\dfrac{a^2}{bc}

\dfrac cb

\dfrac ca

Answer: E

Concepts:angle bisector theorem ratio and proportion

Difficulty rating: 1300

Small hint:

Apply the angle bisector theorem separately to

CM

and

AL

Big hint:

Use the standard notation

a=BC,\ b=CA,\ c=AB

Solution:

By the angle bisector theorem,

\begin{aligned} \frac{AM}{MB}&=\frac{AC}{CB}=\frac ba,\\ \frac{CL}{LB}&=\frac{AC}{AB}=\frac bc. \end{aligned}

Therefore

k=\frac{\frac{b}{a}}{\frac{b}{c}}=\frac ca.

Thus, the correct answer is E.