1991 AMC 12 Problem 23

Attempt Problem 23 of the 1991 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 1991 AMC 12 solutions, or check the answer key.

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

23.

ABCD

is a

2\times2

square,

E

is the midpoint of

AB,

F

is the midpoint of

BC,

AF

and

DE

intersect at

I,

and

BD

and

AF

intersect at

H,

then the area of quadrilateral

BEIH

\frac13

\frac25

\frac7{15}

\frac8{15}

\frac35

Assign coordinates to the square and write equations for

AF,

DE,

and

BD

Find

I

and

H,

then use the shoelace formula on

B,E,I,H

Set

B=(0,0),

C=(2,0),

D=(2,2),

and

A=(0,2).

Then

E=(0,1),

F=(1,0),

and the line intersections are

\begin{aligned} I&=AF\cap DE =\left(\frac25,\frac65\right),\\ H&=AF\cap BD =\left(\frac23,\frac23\right). \end{aligned}

The shoelace formula gives

\begin{aligned} [BEIH] &=\frac12\left| \frac4{15}-\frac25-\frac45 \right|\\ &=\frac7{15}. \end{aligned}

Thus the correct answer is C.

Problem 23 in other years