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2004 AMC 10A Problems/Problem 20

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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 $\triangle ABE$ ?

Image:AMC10_2004A_20.png

$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qqua...$

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 $2$ . $\mathrm{(D)}$

See also

AoPS topic

2004 AMC 10A (Problems • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_10A_Problems/Problem_20"

Category: Introductory Geometry Problems

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

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