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

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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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[[Category:Area Ratio Problems]]

Revision as of 22:37, 11 April 2013

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 $\triangle ABE$?

AMC10 2004A 20.png

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

Solution

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

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 10 Problems and Solutions