1966 AMC 12 Problem 14

Attempt Problem 14 of the 1966 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 1966 AMC 12 solutions, or check the answer key.

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14.

The length of rectangle

ABCD

5

inches and its width is

3

inches. Diagonal

AC

is divided into three equal segments by points

E

and

F.

The area of triangle

BEF,

expressed in square inches, is:

\dfrac32

\dfrac53

\dfrac52

\dfrac13\sqrt{34}

\dfrac13\sqrt{68}

Place the rectangle at

(0,0),

(5,0),

(5,3),

and

(0,3)

The trisection points of diagonal

AC

have easy coordinates

Take

A=(0,0),

B=(5,0),

and

C=(5,3).

Then

E=(\frac{5}{3},1)

and

F=(\frac{10}{3},2).

The determinant formula gives

\begin{aligned} [BEF] &=\frac12\left| \left(-\frac{10}{3}\right)2 -\left(-\frac53\right)1 \right|\\ &=\frac52. \end{aligned}

Therefore, the correct answer is C.

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