2017 AMC 12B Problems/Problem 24

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 $AB/BC$ $\textbf{(A) } 1+\sqrt(2) \qquad \textbf{(B) } 2 + \sqrt(2) \qquad \textbf{(C) } \sqrt(17) \qquad \textbf{(D) } 2 + \sqrt(5) \qquad \textbf{(E) } 1 + 2\sqrt(3)$

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 $CF=x^3/(x^2+1)$ . Since $CF$ and $BF$ form altitudes of triangles $CED$ and $BEA$ , respectively, the areas of these triangles can be calculated. Additionally, the area of triangle $BEC$ can be calculated, as it is a right triangle. Solving for each of these yields: $Area(BEC)=Area(CED)=Area(BEA)=(x^3)/(2(x^2+1))$ $Area(ABCD)=Area(AED)+Area(DEC)+Area(CEB)+Area(BEA).$ $(AB+CD)(BC)/2= 17*Area(CEB)+ Area(CEB) + Area(CEB) + Area(CEB)$ $(x^3+x)/2=(20x^3)/(2(x^2+1))$ $(x)(x^2+1)=20x^3/(x^2+1)$ $(x^2+1)^2=20x^2$ $x^4-18x^2+1=0$ $x^2=9+4sqrt(5)=4+2(2sqrt(5))+5$ (Minus yields a negative value) $x=2+sqrt(5)$ Thus the answer is D: 2+sqrt(5)

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2017 AMC 12B Problems/Problem 24

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