Difference between revisions of "2004 AMC 10A Problems/Problem 20"

Revision as of 21:11, 31 July 2018

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 1

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 $[ABE]=\frac{x^2}{4}$ , and $[DEF]=\frac{x^2}{2}$ . Thus the ratio of the areas is $\boxed{\mathrm{(D)}\ 2}$

Solution 2 (Non-trig)

WLOG, let the side length of $ABCD$ be 1. Let $DE = x$ . It suffices that $AE = 1 - x$ . Then triangles $ABE$ and $CBF$ are congruent by HL, so $CF = AE$ and $DE = DF$ . 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 $\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.$

Solution 3

$\bigtriangleup BEF$ is equilateral, so $\angle EBF = 60^{\circ}$ , and $\angle EBA = \angle FBC$ so they must each be $15^{\circ}$ . Then let $BE=EF=FB=1$ , which gives $EA=\sin{15^{\circ}}$ and $AB=\cos{15^{\circ}}$ . The area of $\bigtriangleup ABE$ is then $\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}$ . $\bigtriangleup DEF$ is an isosceles right triangle with hypotenuse 1, so $DE=DF=\frac{1}{\sqrt{2}}$ and therefore its area is $\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}$ . The ratio of areas is then $\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}$

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 ==Solution 2 (Non-trig) ==
-Without loss of generality, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math>. It suffices that <math>AE = 1 - x</math>. Then triangles <math>ABE</math> and <math>CBF</math> are congruent by HL, so <math>CF = AE</math> and <math>DE = DF</math>. We find that <math>BE = EF = x \sqrt{2}</math>, and so, by the Pythagorean Theorem, we have
+WLOG, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math>. It suffices that <math>AE = 1 - x</math>. Then triangles <math>ABE</math> and <math>CBF</math> are congruent by HL, so <math>CF = AE</math> and <math>DE = DF</math>. We find that <math>BE = EF = x \sqrt{2}</math>, and so, by the Pythagorean Theorem, we have
 <math>(1 - x)^2 + 1 = 2x^2.</math> This yields <math>x^2 + 2x = 2</math>, so <math>x^2 = 2 - 2x</math>. Thus, the desired ratio of areas is
 <cmath>\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.</cmath>

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