2004 AMC 10A Problems/Problem 20

Problem

Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$ ?

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

Solution

Since triangle $BEF$ is equilateral, $EA=FC$ , and $EAB$ and $FCB$ are $SAS$ congruent. Thus, triangle $DEF$ is an isosceles right triangle. So we let $DE=x$ . Thus $EF=EB=FB=x\sqrt{2}$ . If we go angle chasing, we find out that $\angle AEB=75^{\circ}$ , thus $\angle ABE=15^{\circ}$ . $\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}$ . Thus $\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$ , or $AE=\frac{x(\sqrt{3}-1)}{2}$ . Thus $AB=\frac{x(\sqrt{3}+1)}{2}$ , and $[AEB]=\frac{x^2}{4}$ , and $[DEF]=\frac{x^2}{2}$ . Thus the ratio of the areas is $\boxed{\mathrm{(D)}\ 2}$

Solution 2 (Non-trig)

Without loss of generality let the side length of $ABCD$ be 1. Let $DE = x$ and $AE = x$ . Then $<ABE = \frac{90^\circ - 60^\circ}{2} = 15^\circ$ , so $<AEB = 75^\circ$ , and $<DEF = 45^\circ$ ; i.e. that $DEF$ is 45-45-90. We find that $BE = EF = x \sqrt{2}$ , and so, by the Pythagorean Theorem, we have $(1 - x)^2 + 1 = 2x^2.$ This yields $x^2 + 2x = 2$ , so $x^2 = 2 - 2x$ . Thus, the desired ratio of areas is $\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = 2.$

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2004 AMC 10A Problems/Problem 20

Contents

Problem

Solution

Solution 2 (Non-trig)

See also