Question

Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $\overline{AD}.$ Let $F$ be the intersection of lines $EB$ and $AC.$ What is the ratio of the area of quadrilateral $CDEF$ to the area of triangle $CFB?$

$5 : 4$

$4 : 3$

$3 : 2$

$5 : 3$

$2 : 1$

Solution:

Area ratios don't change under an affine map, so drop in convenient coordinates: $A = (0,0),$ $B = (1,0),$ $C = (1,1),$ $D = (0,1),$ which makes $E = (0, \tfrac12).$ Line $AC$ is $y = x,$ and line $EB$ runs from $(0, \tfrac12)$ to $(1, 0);$ they cross at $F = (\tfrac13, \tfrac13).$ The shoelace formula gives quadrilateral $CDEF$ area $\tfrac{5}{12}$ and triangle $CFB$ area $\tfrac13.$ So the ratio is $\tfrac{5}{12} : \tfrac13 = 5 : 4.$ Therefore, the answer is A.

2024 AMC 10B Problem 10

Solution:

Problem 10 in Other Years