Question

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.

1963 AMC 12 Problem 39

Answer: D

Difficulty rating: 2030

Small Hint:

Big Hint:

Solution:

Problem 39 in Other Years