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

## Problem

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

## 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 $\triangle BCD \sim \triangle ABC \sim \triangle CEB$, 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 $\triangle CEB \sim \triangle CFE \sim \triangle EFB$. Therefore, $BF=\frac{x}{x^2+1}$ 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: $$[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))$$ $$[ABCD]=[AED]+[DEC]+[CEB]+[BEA]$$ $$(AB+CD)(BC)/2= 17*[CEB]+ [CEB] + [CEB] + [CEB]$$ $$(x^3+x)/2=(20x^3)/(2(x^2+1))$$ $$(x)(x^2+1)=20x^3/(x^2+1)$$ $$(x^2+1)^2=20x^2$$ $$x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5$$ Therefore, the answer is $\boxed{\textbf{(D) } 2+\sqrt{5}}$

[SOLUTION 2] Draw line FG through E, with F on BC and G on AD, FG//AB. WOLG let CD=1, CB=x, AB=x^2. By weighted average FG=(1+x^4)/(1+x^2).

Meanwhile, FE:EG=[CBE]:[ADE]=1:17. FE=x^2/(1+x^2). We obtain (1+x^4)/(1+x^2)=18x^2/(1+x^2), Namely x^4-18x^2+1=0.

The rest is the same with solution 1.

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