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Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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Let <math>C = (0,0), D=(\frac1a, 0), B = (0,1), A = (a,1)</math> where <math>a>1</math>. Because <math>BC = 1, a = \frac{AB}{BC}</math>. Notice that the diagonals are perpendicular with slopes of <math>\frac1a</math> and <math>-a</math>. Let the intersection of <math>AC</math> and <math>BD</math> be <math>F</math>, then <math>\triangle BFC \sim \triangle ABC</math>. However, because <math>ABCD</math> is a trapezoid, <math>\triangle</math><math>BCF</math> and <math>\triangle ADF</math> share the same area, therefore <math>\triangle</math><math>BCE</math> is the reflection of <math>\triangle</math><math>BCF</math> over the perpendicular bisector of <math>BC</math>, which is <math>y=\frac12</math>. We use the linear equations of the diagonals, <math>y = -ax + 1, y = \frac1a x</math>, to find the coordinates of <math>F</math>. <cmath>-ax+1 = \frac1ax \Longrightarrow x = \frac{1}{a+\frac1a} = \frac{a}{a^2+1}</cmath> <cmath>y = \frac1ax = \frac{1}{a^2+1}</cmath>
 
Let <math>C = (0,0), D=(\frac1a, 0), B = (0,1), A = (a,1)</math> where <math>a>1</math>. Because <math>BC = 1, a = \frac{AB}{BC}</math>. Notice that the diagonals are perpendicular with slopes of <math>\frac1a</math> and <math>-a</math>. Let the intersection of <math>AC</math> and <math>BD</math> be <math>F</math>, then <math>\triangle BFC \sim \triangle ABC</math>. However, because <math>ABCD</math> is a trapezoid, <math>\triangle</math><math>BCF</math> and <math>\triangle ADF</math> share the same area, therefore <math>\triangle</math><math>BCE</math> is the reflection of <math>\triangle</math><math>BCF</math> over the perpendicular bisector of <math>BC</math>, which is <math>y=\frac12</math>. We use the linear equations of the diagonals, <math>y = -ax + 1, y = \frac1a x</math>, to find the coordinates of <math>F</math>. <cmath>-ax+1 = \frac1ax \Longrightarrow x = \frac{1}{a+\frac1a} = \frac{a}{a^2+1}</cmath> <cmath>y = \frac1ax = \frac{1}{a^2+1}</cmath>
 
The y-coordinate of <math>E</math> is simply <math>1-\frac{1}{a^2+1} = \frac{a^2}{a^2+1}</math>
 
The y-coordinate of <math>E</math> is simply <math>1-\frac{1}{a^2+1} = \frac{a^2}{a^2+1}</math>
−
The area of <math>\triangle BCE</math> is <math>\frac12 \frac{a}{a^2+1}</math>. We apply shoelace theorem to solve for the area of <math>\triangle ADE</math>. The coordinates of the triangle are <math>(\frac{a}{a^2+1}, \frac{a^2}{a^2+1}), (a,1), (\frac1a, 0)</math>, so the area is  
+
The area of <math>\triangle BCE</math> is <math>\frac12 \frac{a}{a^2+1}</math>. We apply shoelace theorem to solve for the area of <math>\triangle ADE</math>. The coordinates of the triangle are <math>\{(\frac{a}{a^2+1}, \frac{a^2}{a^2+1}), (a,1), (\frac1a, 0)\}</math>, so the area is  
  
 
<cmath>\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|</cmath>
 
<cmath>\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|</cmath>

Revision as of 01:17, 7 November 2021

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 $A=(-1,4a), B=(-1,0), C=(1,0), D=\bigg(1,\frac{1}{a}\bigg)$. 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 4

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

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).

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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