1974 AHSME Problems/Problem 19

Problem

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

$[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((.82,0)--(1,1)--(0,.76)--cycle); label("A", (0,0), S); label("B", (1,0), S); label("C", (1,1), N); label("D", (0,1), N); label("M", (0,.76), W); label("N", (.82,0), S);[/asy]$

$\mathrm{(A)\ } 2\sqrt{3}-3 \qquad \mathrm{(B) \ }1-\frac{\sqrt{3}}{3} \qquad \mathrm{(C) \ } \frac{\sqrt{3}}{4} \qquad \mathrm{(D) \ } \frac{\sqrt{2}}{3} \qquad \mathrm{(E) \ }4-2\sqrt{3}$

Solution

Let $BN=x$ so that $AN=1-x$ . From the Pythagorean Theorem on $\triangle NBC$ , we get $CN=\sqrt{x^2+1}$ , and from the Pythagorean Theorem on $\triangle AMN$ , we get $MN=(1-x)\sqrt{2}$ . Since $\triangle CMN$ is equilateral, we must have $\sqrt{x^2+1}=(1-x)\sqrt{2}\implies x^2+1=2x^2-4x+2\implies x^2-4x+1=0$ . From the Pythagorean Theorem, we get $x=2-\sqrt{3}$ , since we want the root that's less than $1$ .

Therefore, $CN^2=x^2+1=(2-\sqrt{3})^2+1=8-4\sqrt{3}$ . The area of an equilateral triangle with side length $x$ is equal to $\frac{x^2\sqrt{3}}{4}$ , so the area of $\triangle CMN$ is $\frac{(8-4\sqrt{3})(\sqrt{3})}{4}=2\sqrt{3}-3, \boxed{\text{A}}$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1974 AHSME Problems/Problem 19

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