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2004 AMC 10A Problems/Problem 20: Difference between revisions

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==Problem==
==Problem==
Points <math>E</math> and <math>F</math> are located on square <math>ABCD</math> so that <math>\triangle BEF</math> is equilateral. What is the ratio of the area of <math>\triangle DEF</math> to that of <math>\triangle ABE</math>?
Points <imath>E</imath> and <imath>F</imath> are located on square <imath>ABCD</imath> so that <imath>\triangle BEF</imath> is equilateral. What is the ratio of the area of <imath>\triangle DEF</imath> to that of <imath>\triangle ABE</imath>?


<center>[[Image:AMC10_2004A_20.png]]</center>
<center>
<asy>
unitsize(3 cm);


<math> \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} </math>
pair A, B, C, D, E, F;


==Solution==
A = (0,0);
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>
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>
</center>
 
<imath> \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} </imath>
 
==Solution 1==
Since triangle <math>BEF</math> is equilateral, <math>EA=FC</math>, and <math>EAB</math> and <math>FCB</math> are <math>HL</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>
 
z


==Solution 2 (Non-trig) ==
==Solution 2 (Non-trig) ==
Without loss of generality, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math> and <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
<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} = 2.</cmath>
<cmath>\dfrac{\dfrac{x^2}{2}}{\dfrac{1-x}{2}} = \dfrac{x^2}{1 - x} = \boxed{\text{(D) }2}.</cmath>
 
==Solution 3 (System of Equations)==
Assume <math>AB=1</math>. Then, <math>FC</math> is <math>x</math> and <math>ED</math> is <math>1-x</math>. We see that using <math>HL</math>, <math>FCB</math> is congruent to EAB. Using Pythagoras of triangles <math>FCB</math> and <math>FDE</math> we get <math>2{(1-x)}^2=x^2+1</math>. Expanding, we get <math>2x^2-4x+2=x^2+1</math>. Simplifying gives <math>x^2-4x+1=0</math> solving using completing the square (or other methods) gives 2 answers: <math>2-\sqrt{3}</math> and <math>2+\sqrt{3}</math>. Because <math>x < 1</math>, <math>x=2-\sqrt{3}</math>. Using the areas, the answer is <math>\boxed{\text{(D) }2}</math>
 
==Solution 4==
First, since <imath>\bigtriangleup BEF</imath> is equilateral and <imath>ABCD</imath> is a square, by the Hypothenuse Leg Theorem, <imath>\bigtriangleup ABE</imath> is congruent to <imath>\bigtriangleup CBF</imath>. Then, assume length <imath>AB = BC = x</imath> and length <imath>DE = DF = y</imath>, then <imath>AE = FC = x - y</imath>. <imath>\bigtriangleup BEF</imath> is equilateral, so <imath>EF = EB</imath> and <imath>EB^2 = EF^2</imath>, it is given that <imath>ABCD</imath> is a square and <imath>\bigtriangleup DEF</imath> and <imath>\bigtriangleup ABE</imath> are right triangles. Then we use the Pythagorean theorem to prove that <imath>AB^2 + AE^2 = EB^2</imath> and since we know that <imath>EB^2 = EF^2</imath> and <imath>EF^2 = DE^2 + DF^2</imath>, which means <imath>AB^2 + AE^2 = DE^2 + DF^2</imath>. Now we plug in the variables and the equation becomes <imath>x^2 + (x-y)^2 = 2y^2</imath>, expand and simplify and you get <imath>2x^2 - 2xy = y^2</imath>. We want the ratio of area of <imath>\bigtriangleup DEF</imath> to <imath>\bigtriangleup ABE</imath>. Expressed in our variables, the ratio of the area is <imath>\frac{y^2}{x^2 - xy}</imath> and we know <imath>2x^2 - 2xy = y^2</imath>, so the ratio must be <imath>2</imath>. So, the answer is <imath>\boxed{\text{(D) }2}</imath>
 
==solution 5:==
First, assume that <imath>AB=1</imath>, and let <imath>ED = DF = x</imath>. Then, we have <imath>[DEF] = \frac{x^2}{2}</imath> and <cmath>[ABE] = \frac{(AE)(AB)}{2} = \frac{(1-x)(1)}{2},</cmath>so <cmath>\frac{[DEF]}{[ABE]} = \frac{x^2}{1-x} .</cmath>By the Pythagorean Theorem applied to <imath>\triangle DEF</imath>, we have <cmath>EF^2 = DE^2 + DF^2 = 2x^2.</cmath>Applying the Pythagorean Theorem to <imath>\triangle AEB</imath>, we have <cmath>EB^2 = AB^2 + AE^2 = 1 + (1-x)^2 = 2 - 2x + x^2.</cmath>Since <imath>\triangle EFB</imath> is equilateral, we have <imath>EF = EB</imath>, so <cmath>2x^2 = 2-2x + x^2,</cmath>or <imath>x^2 = 2-2x= 2(1-x)</imath>. Hence the desired ratio of the areas is <cmath>\frac{[DEF]}{[ABE]} = \frac{x^2}{1-x} = \boxed{2}.</cmath>
 
==Video Solution==
https://youtu.be/hwHIHRukYMk
 
Education, the Study of Everything
 
==Video Solution by TheBeautyofMath==
https://youtu.be/BFKo9h8GhLY
 
~IceMatrix


==See also==
==See also==
*[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131332 AoPS topic]
{{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}
{{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}


[[Category:Introductory Geometry Problems]]
[[Category:Introductory Geometry Problems]]
[[Category:Area Ratio Problems]]
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 17:55, 8 November 2025

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 $HL$ 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 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 \[\dfrac{\dfrac{x^2}{2}}{\dfrac{1-x}{2}} = \dfrac{x^2}{1 - x} = \boxed{\text{(D) }2}.\]

Solution 3 (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 4

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

solution 5:

First, assume that $AB=1$, and let $ED = DF = x$. Then, we have $[DEF] = \frac{x^2}{2}$ and \[[ABE] = \frac{(AE)(AB)}{2} = \frac{(1-x)(1)}{2},\]so \[\frac{[DEF]}{[ABE]} = \frac{x^2}{1-x} .\]By the Pythagorean Theorem applied to $\triangle DEF$, we have \[EF^2 = DE^2 + DF^2 = 2x^2.\]Applying the Pythagorean Theorem to $\triangle AEB$, we have \[EB^2 = AB^2 + AE^2 = 1 + (1-x)^2 = 2 - 2x + x^2.\]Since $\triangle EFB$ is equilateral, we have $EF = EB$, so \[2x^2 = 2-2x + x^2,\]or $x^2 = 2-2x= 2(1-x)$. Hence the desired ratio of the areas is \[\frac{[DEF]}{[ABE]} = \frac{x^2}{1-x} = \boxed{2}.\]

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
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