1974 AMC 12 Problem 19

Question

In the adjoining figure

ABCD

is a square and

CMN

is an equilateral triangle. If the area of

ABCD

is one square inch, then the area of

CMN

in square inches is

2\sqrt3-3

1-\frac{\sqrt3}{3}

\frac{\sqrt3}{4}

\frac{\sqrt2}{3}

4-2\sqrt3

The square has side

1;

set

DM=NB=x

by comparing the equal sides

CM

and

CN

Equate

CM^2=1+x^2

with

MN^2=2(1-x)^2

The square has side

1.

Since

CM=CN,

symmetry of their squared lengths gives

DM=NB=x.

Also,

\begin{aligned} CM^2&=1+x^2,\\ MN^2&=2(1-x)^2. \end{aligned}

Equating these because

CMN

is equilateral gives

x^2-4x+1=0.

The root in

(0,1)

is

x=2-\sqrt3.

The triangle’s side has square

1+x^2,

so its area is

\begin{aligned} A_\triangle&=\frac{\sqrt3}{4}(1+x^2)\\ &=\frac{\sqrt3}{4} \bigl(1+(2-\sqrt3)^2\bigr)\\ &=2\sqrt3-3. \end{aligned}

Therefore, the correct answer is A.

Problem 19 in other years