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

Let $CD=1$ , $BC=x$ , and $AB=x^2$ . Note that $AB/BC=x$ . By the Pythagorean Theorem, $BD=\sqrt{x^2+1}$ . Since $\triangle BCD ~ \triangle ABC ~ \triangle CEB$ , the ratios of side lengths must be equal. Since $BC=x$ , $CE=\frac{x^2}{\sqrt{x^2+1}}$ and $BE=\frac{x}{\sqrt{x^2+1}}$ . Let F be a point on $\overline{BC}$ such that $\overline{EF}$ is an altitude of triangle $CEB$ . Note that $CEB~CFE~EFB$ , so $BF$ and $CF$ can be calculated. Solving for these lengths gives $BF=\frac{x}{x^2+1}$ and $CF=\frac{x^3}{x^2+1}$ . Since $\overline{CF}$ and $\overline{BF}$ 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: $[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))$ $[ABCD]=[AED]+[DEC]+[CEB]+[BEA]$ $(AB+CD)(BC)/2= 17*[CEB]+ [CEB] + [CEB] + [CEB]$ $(x^3+x)/2=(20x^3)/(2(x^2+1))$ $(x)(x^2+1)=20x^3/(x^2+1)$ $(x^2+1)^2=20x^2$ $x^4-18x^2+1=0 \implies x^2=9+4sqrt(5)=4+2(2sqrt(5))+5$ Therefore, the answer is $\boxed{\textbf{(D) } 2+\sqrt{5}}$

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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