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

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

Revision as of 17:22, 25 January 2024

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

z

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 $x$ and $ED$ is $1-x$. We see that using $HL$, $FCB$ is congruent to 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 $x < 1$, $x=2-\sqrt{3}$. Using the areas, the answer is $\boxed{\text{(D) }2}$

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 $2$. So, the answer is $\boxed{\text{(D) }2}$

Video Solution

https://youtu.be/hwHIHRukYMk

Education, the Study of Everything

Video Solution by TheBeautyofMath

https://youtu.be/BFKo9h8GhLY

~IceMatrix

See also

2004 AMC 10A (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

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