1959 AMC 12 Problem 40

Attempt Problem 40 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).

40.

In triangle

ABC,

\overline{BD}

is a median.

\overline{CF}

intersects

\overline{BD}

at

E

so that

BE=ED.

Point

F

is on

\overline{AB}.

Then, if

BF=5,

BA

equals:

10

12

15

20

none of these

Answer: C

Concepts:median (geometry)vector mass points

Difficulty rating: 1590

Small hint:

Use coordinates or masses, noting that

D

is the midpoint of

AC

and

E

is the midpoint of

BD

Big hint:

Write

E

both as

\frac{B+D}{2}

and as a point on line

CF

Solution:

Use vectors with

A=\mathbf0,\ B=\mathbf b,

and

C=\mathbf c.

Since

D=\frac{\mathbf c}{2}

and

E

is the midpoint of

BD,

E=\frac12\mathbf b+\frac14\mathbf c.

A point

F

on

AB

has the form

F=t\mathbf b.

Since

E

lies on

CF,

comparison of the

\mathbf c

-coefficient shows that

E=\tfrac14C+\tfrac34F.

Hence

\frac34t=\frac12,

so

t=\frac{2}{3}.

Thus

\frac{AF}{AB}=\frac{2}{3},

and

\frac{BF}{BA}=\frac{1}{3}.

Since

BF=5,

BA=15.

Therefore, the correct answer is C.

Problem 40 in other years