1965 AMC 12 Problem 37

Attempt Problem 37 of the 1965 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 1965 AMC 12 solutions, or check the answer key.

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37.

Point

E

is selected on side

AB

of triangle

ABC

in such a way that

AE:EB=1:3,

and point

D

is selected on side

BC

so that

CD:DB=1:2.

The point of intersection of

AD

and

CE

is

F.

Then

\frac{EF}{FC}+\frac{AF}{FD}

is:

\dfrac45

\dfrac54

\dfrac32

2

\dfrac52

Answer: C

Concepts:mass points ratio and proportion

Difficulty rating: 2170

Small hint:

Assign masses

m_A=3,

m_B=1,

and

m_C=2

Big hint:

Use the combined masses at

E

and

D

to read the two cevian ratios

Solution:

The side ratios are represented by masses

m_A=3,

m_B=1,

and

m_C=2.

Then

m_E=m_A+m_B=4

and

m_D=m_B+m_C=3.

Along

CE,

\frac{EF}{FC}=\frac{m_C}{m_E}=\frac12,

while along

AD,

\frac{AF}{FD}=\frac{m_D}{m_A}=1.

Their sum is

\frac{3}{2}.

Therefore, the correct answer is C.

Problem 37 in other years