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Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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==Problem==
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==Problem 24==
  
 
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\tfrac{AB}{BC}</math>?
 
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\tfrac{AB}{BC}</math>?
  
 
<math>\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}</math>
 
<math>\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}</math>
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==Solution 1==
 
==Solution 1==

Revision as of 08:45, 20 February 2017

Problem 24

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, $FE:EG=[\triangle CBE]:[\triangle ADE]=1:17$.

$FE=\frac{x^2}{1+x^2}$. 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.


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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