Difference between revisions of "2017 AMC 12B Problems/Problem 24"

Revision as of 20:24, 16 February 2017

Problem

Quadrilateral $ABCD$ has right angles at $B$ and $C$ , Triangle $ABC$ ~ Triangle $BCD$ , and $AB > BC$ . There is a point $E$ in the interior of $ABCD$ such that Triangle $ABC$ ~ Triangle $CEB$ and the area of Triangle $AED$ is $17$ times the area of Triangle $CEB$ . What is $AB/BC$ $\textbf{(A) } 1+sqrt(2) \qquad \textbf{(B) } 2 + sqrt(2) \qquad \textbf{(C) } sqrt(17) \qquad \textbf{(D) } 2 + sqrt(5) \qquad \textbf{(E) } 1 + 2sqrt(3)$

Solution

Solution by TorrTar

Let $CD=1$ , $BC=x$ , $AB=x^2$ . Note that $AB/BC=x$ . The Pythagorean theorem states that $BD=sqrt(x^2+1)$ . Since $BCD~ABC~CEB$ , the ratios of side lengths must be equal. Since $BC=x$ , $CE=x^2/sqrt(x^2+1)$ and $BE=x/sqrt(x^2+1)$ . Let Point F be a point on $BC$ such that $EF$ is an altitude of triangle $CEB$ . Note that $CEB~CFE~EFB$ , so $BF$ and $CF$ can be calculated. Solving for these lengths gives $BF=x/(x^2+1)$ and $CF=x^3/(x^2+1)$ .

2017 AMC 12B (Problems • Answer Key • Resources)
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Difference between revisions of "2017 AMC 12B Problems/Problem 24"

Revision as of 20:24, 16 February 2017

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Solution

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