2017 AMC 12B Problems/Problem 24

Problem

Quadrilateral $ABCD$ has right angles at $B$ and $C$, $\triangle ABC \sim \triangle BCD$, and $AB > BC$. There is a point $E$ in the interior of $ABCD$ such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is $17$ times the area of $\triangle CEB$. What is $\tfrac{AB}{BC}$?

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

Solution 1

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 $AD$, $FG \parallel AB$. WLOG let $CD=1$, $CB=x$, $AB=x^2$. By weighted average $FG=\frac{1+x^4}{1+x^2}$.

Meanwhile, $FE:EG=[\triangle CBE]:[\triangle ADE]=1:17$. This follows from comparing the ratios of triangle DEG to CFE and triangle AEG to FEB, both pairs in which the two triangles share a height perpendicular to FG, and have base ratio $EG:FE$.

$FE=\frac{x^2}{1+x^2}$. We obtain $\frac{1+x^4}{1+x^2}=\frac{18x^2}{1+x^2}$, namely $x^4-18x^2+1=0$.

The rest is the same as Solution 1.

Solution 3

Let $AB = a$, $BC = b$, $AC = \sqrt{a^2+b^2}$

Note that $E$ cannot be the intersection of $AC$ and $BD$, as that would mean $[AED] = [CEB]$

$$\because \triangle BCD \sim \triangle ABC, \quad \therefore \frac{CD}{BC} = \frac{BC}{AB}, \quad CD = BC \cdot \frac{BC}{AB} = b \cdot \frac{b}{a} = \frac{b^2}{a}$$

$$[CEB] = ( \frac{BC}{AC} )^2 \cdot [ABC] = ( \frac{b}{ \sqrt{a^2+b^2} } )^2 \cdot \frac{ab}{2} = \frac{ab^3}{2(a^2+b^2)}$$

$$BF = \frac{ 2[CEB] }{BC} = \frac{ 2 \cdot \frac{ab^3}{ 2(a^2+b^2) } }{b} = \frac{ab^2}{a^2+b^2}$$

$$\because \triangle BFE \sim \triangle ABC, \quad \therefore \frac{EF}{BF} = \frac{BC}{AB}, \quad EF = BF \cdot \frac{BC}{AB} = \frac{ab^2}{a^2+b^2} \cdot \frac{b}{a} = \frac{b^3}{a^2+b^2}$$

$$EG = FG - EF = b - \frac{b^3}{a^2+b^2} = \frac{a^2b}{a^2+b^2}$$

$$[ABCD] = \frac{AB + CD}{2} \cdot BC = \frac{a + \frac{b^2}{a} }{2} \cdot b = \frac{ b(a^2+b^2) }{ 2a }$$

$$[ABE] = \frac12 \cdot AB \cdot EF = \frac12 \cdot a \cdot \frac{b^3}{a^2+b^2} = \frac{ab^3}{ 2(a^2+b^2) }$$

$$[CDE] = \frac12 \cdot CD \cdot EG = \frac12 \cdot \frac{a^2b}{a^2+b^2} \cdot \frac{a^2b}{a^2+b^2} = \frac{ab^3}{ 2(a^2+b^2) }$$

$$ADE = [ABCD] - [ABE] - [CEB] - [CDE] = \frac{ b(a^2+b^2) }{ 2a } - \frac{ab^3}{ 2(a^2+b^2) } - \frac{ab^3}{ 2(a^2+b^2) }- \frac{ab^3}{ 2(a^2+b^2) } = \frac{ b(a^2+b^2)^2 - 3a^2b^3 }{ 2a(a^2+b^2) }$$

$$\frac{ [ADE] }{ [CEB] } = \frac { \frac{ b(a^2+b^2)^2 - 3a^2b^3 }{ 2a(a^2+b^2) } }{ \frac{ab^3}{2(a^2+b^2)} } = \frac{ (a^2 + b^2)^2 - 3a^2b^2 }{ a^2b^2 } = \frac{ a^4 - a^2b^2 + b^4 }{ a^2b^2 } = 17$$

$$a^4 - a^2b^2 + b^4 = 17 a^2b^2, \quad a^4 + b^4 = 18 a^2b^2, \quad \frac{a^2}{b^2} + \frac{b^2}{a^2} = 18$$

Let $x = \frac{a}{b}$,

$$x^2 + \frac{1}{x^2} = 18, \quad x^4 - 18x^2 + 1 = 0, \quad x^2 = \frac{18 + \sqrt{324-4} }{2} = 9+ 4\sqrt{5}$$

$$\frac{a}{b} = \sqrt{ 9+ 4\sqrt{5} } = \sqrt{ 4+ 4\sqrt{5}+5 } = \boxed{\textbf{(D) } 2+ \sqrt{5}}$$

Solution 4

Let $A=(-1,4a), B=(-1,0), C=(1,0), D=\bigg(1,\frac{1}{a}\bigg)$. Then from the similar triangles condition, we compute $CE=\frac{4a}{\sqrt{4a^2+1}}$ and $BE=\frac{2}{\sqrt{4a^2+1}}$. Hence, the $y$-coordinate of $E$ is just $\frac{BE\cdot CE}{BC}=\frac{4a}{4a^2+1}$. Since $E$ lies on the unit circle, we can compute the $x$ coordinate as $\frac{1-4a^2}{4a^2+1}$. By Shoelace, we want $$\frac{1}{2}\det\begin{bmatrix} -1 & 4a & 1\\ \frac{1-4a^2}{4a^2+1} & \frac{4a}{4a^2+1} & 1\\ 1 & \frac{1}{a} & 1 \end{bmatrix}=17\cdot\frac{1}{2}\cdot 2 \cdot \frac{4a}{4a^2+1}$$Factoring out denominators and expanding by minors, this is equivalent to $$\frac{32a^4-8a^2+2}{2a(4a^2+1)}=\frac{68a}{4a^2+1} \Longrightarrow 16a^4-72a^2+1=0$$This factors as $(4a^2-8a-1)(4a^2+8a-1)=0$, so $a=1+\frac{\sqrt{5}}{2}$ and so the answer is $\textbf{(D) \ }$.

Solution 5

Let $C = (0,0), D=(\frac1a, 0), B = (0,1), A = (a,1)$ where $a>1$. Because $BC = 1, a = \frac{AB}{BC}$. Notice that the diagonals are perpendicular with slopes of $\frac1a$ and $-a$. Let the intersection of $AC$ and $BD$ be $F$, then $\triangle BFC \sim \triangle ABC$. However, because $ABCD$ is a trapezoid, $\triangle$$BCF$ and $\triangle ADF$ share the same area, therefore $\triangle$$BCE$ is the reflection of $\triangle$$BCF$ over the perpendicular bisector of $BC$, which is $y=\frac12$. We use the linear equations of the diagonals, $y = -ax + 1, y = \frac1a x$, to find the coordinates of $F$. $$-ax+1 = \frac1ax \Longrightarrow x = \frac{1}{a+\frac1a} = \frac{a}{a^2+1}$$ $$y = \frac1ax = \frac{1}{a^2+1}$$ The y-coordinate of $E$ is simply $1-\frac{1}{a^2+1} = \frac{a^2}{a^2+1}$ The area of $\triangle BCE$ is $\frac12 \frac{a}{a^2+1}$. We apply shoelace theorem to solve for the area of $\triangle ADE$. The coordinates of the triangle are $\{(\frac{a}{a^2+1}, \frac{a^2}{a^2+1}), (a,1), (\frac1a, 0)\}$, so the area is

$$\frac12 |\frac{a^3}{a^2+1} + \frac1a - \frac{a}{a^2+1} - \frac{a}{a^2+1}| = \frac12 |\frac{a^3-2a}{a^2+1} + \frac1a|$$ $$= \frac12 |\frac{a^4-2a^2}{a(a^2+1)} + \frac{a^2+1}{a(a^2+1)}| = \frac12 \frac{a^4-a^2+1}{a(a^2+1)}$$ Finally, we use the property that the ratio of areas equals $17$ $$\frac{\frac12}{\frac12} \frac{\frac{a^4-a^2+1}{a(a^2+1)}}{\frac{a}{a^2+1}} = 17 \Rightarrow \frac{a^4-a^2+1}{a^2} = 17 \Rightarrow a^4-18a^2+1 = 0$$ $$a^2 = 9+4\sqrt{5} = (2+\sqrt{5})^2 \Longrightarrow a = \boxed{\textbf{(D) } 2+\sqrt{5}}$$

~Zeric

~r00tsOfUnity

Notes

1) $\sqrt{17}$ is the most relevant answer choice because it shares numbers with the givens of the problem.

2) It's a very good guess to replace finding the area of triangle AED with the area of the triangle DAF, where F is the projection of D onto AB(then find the closest answer choice).