2018 AMC 10A Problem 24

Question

Triangle

ABC

with

AB=50

and

AC=10

has area

120.

Let

D

be the midpoint of

\overline{AB},

and let

E

be the midpoint of

\overline{AC}.

The angle bisector of

\angle BAC

intersects

\overline{DE}

and

\overline{BC}

at

F

and

G,

respectively. What is the area of quadrilateral

FDBG?

60

65

70

75

80

Use the angle bisector theorem to locate

G

The midsegment

DE

makes a similar half-height trapezoid

Let

BC = a, BG = x, GC = y,

and

h

be the length of the altitude through

A.

By the Angle Bisector Theorem,

BG:GC=AB:AC=5:1,

so

BG=\dfrac{5a}{6}.

Because

DE

is a midsegment,

DE=\dfrac a2.

Applying the same theorem in

\triangle ADE

gives

DF:FE=AD:AE=5:1,

so

DF=\dfrac56DE=\dfrac{5a}{12}.

The trapezoid’s height is

\frac{h}{2},

and

\frac{ah}{2}=120.

Its average base length is

\dfrac12\left(\dfrac{5a}{12}+\dfrac{5a}{6}\right)=\dfrac{5a}{8}.

Therefore, its area is

\dfrac{5a}{8}\cdot\dfrac h2=\dfrac58\cdot\dfrac{ah}{2}=75.

Thus, D is the correct answer.

Problem 24 in other years