1978 AMC 12 Problem 23

Attempt Problem 23 of the 1978 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 1978 AMC 12 solutions, or check the answer key.

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23.

Vertex

E

of equilateral triangle

ABE

is in the interior of square

ABCD,

and

F

is the point of intersection of diagonal

BD

and line segment

AE.

If length

AB

\sqrt{1+\sqrt3},

then the area of

\triangle ABF

1

\frac{\sqrt2}{2}

\frac{\sqrt3}{2}

4-2\sqrt3

\frac12+\frac{\sqrt3}{4}

Place

A=(0,0)

and

B=(s,0),

where

s=AB

Find the intersection of

AE:\ y=\sqrt3x

and

BD:\ y=s-x

Let

A=(0,0),

B=(s,0),

and

D=(0,s),

where

s=\sqrt{1+\sqrt3}.

The two lines containing

F

have equations

\begin{aligned} AE:\quad y&=\sqrt3x,\\ BD:\quad y&=s-x. \end{aligned}

Hence the altitude of

F

above

AB

\frac{s\sqrt3}{1+\sqrt3}.

Therefore

[\triangle ABF] =\frac{s^2\sqrt3}{2(1+\sqrt3)} =\frac{\sqrt3}{2}.

Therefore, the correct answer is C.

Problem 23 in other years