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 $\triangle ABE$ ?

$[asy] unitsize(3 cm); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (1,1); D = (0,1); E = (0,Tan(15)); F = (1 - Tan(15),1); draw(A--B--C--D--cycle); draw(B--E--F--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, W); label("$F$", F, N); [/asy]$

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

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 $\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}$

Solution 4(system of equations)

Assume AB=1 then FC is x ED is $1-x$ then we see that using HL FCB is congruent is EAB. Using Pythagoras of triangles FCB and FDE we get $2{(1-x)}^2=x^2+1$ . Expanding we get $2x^2-4x+2=x^2+1$ . Simplifying gives $x^2-4x+1=0$ solving using completing the square(or other methods) gives 2 answers $2-\sqrt{3}$ and $2+\sqrt{3}$ because $x < 1$ then $x=2-\sqrt{3}$ then using the areas we get the answer to be D

Solution 5

First, since $\bigtriangleup BEF$ is equilateral and $ABCD$ is a square, by the Hypothenuse Leg Theorem, $\bigtriangleup ABE$ is congruent to $\bigtriangleup CBF$ . Then, assume length $AB = BC = x$ and length $DE = DF = y$ , then $AE = FC = x - y$ . $\bigtriangleup BEF$ is equilateral, so $EF = EB$ and $EB^2 = EF^2$ , it is given that $ABCD$ is a square and $\bigtriangleup DEF$ and $\bigtriangleup ABE$ are right triangles. Then we use the Pythagorean theorem to prove that $AB^2 + AE^2 = EB^2$ and since we know that $EB^2 = EF^2$ and $EF^2 = DE^2 + DF^2$ , which means $AB^2 + AE^2 = DE^2 + DF^2$ . Now we plug in the variables and the equation becomes $x^2 + (x+y)^2 = 2y^2$ , expand and simplify and you get $2x^2 - 2xy = y^2$ . We want the ratio of area of $\bigtriangleup DEF$ to $\bigtriangleup ABE$ . Expressed in our variables, the ratio of the area is $\frac{y^2}{x^2 - xy}$ and we know $2x^2 - 2xy = y^2$ , so the ratio must be 2. Choice D

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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
All AMC 10 Problems and Solutions

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