Question

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.

1965 AMC 12 Problem 37

Answer: C

Difficulty rating: 2170

Small Hint:

Big Hint:

Solution:

Problem 37 in Other Years