1963 AMC 12 Problem 39

Attempt Problem 39 of the 1963 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 1963 AMC 12 solutions, or check the answer key.

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

39.

In triangle

ABC

lines

CE

and

AD

are drawn so that

\dfrac{CD}{DB}=\dfrac31

and

\dfrac{AE}{EB}=\dfrac32.

Let

r=\dfrac{CP}{PE},

where

P

is the intersection point of

CE

and

AD.

Then

r

equals:

3

\dfrac32

4

5

\dfrac52

Answer: D

Concepts:mass points ratio and proportion

Difficulty rating: 2030

Small hint:

Assign endpoint masses inversely proportional to the given side ratios

Big hint:

Choose masses

m_A=2,

m_B=3,

and

m_C=1

, then find the mass at

E

Solution:

The ratio

AE:EB=3:2

is represented by masses

m_A=2

and

m_B=3.

The ratio

CD:DB=3:1

then gives

m_C=1.

Point

E

has mass

m_A+m_B=5,

so along cevian

CE,

\frac{CP}{PE}=\frac{m_E}{m_C}=\frac51=5.

Thus, the correct answer is D.

Problem 39 in other years