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

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==Solution==
 
==Solution==
  
Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD</math> ~ <math>\triangle ABC</math> ~ <math>\triangle CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=\frac{x^2}{\sqrt{x^2+1}}</math> and <math>BE=\frac{x}{\sqrt{x^2+1}}</math>. Let F be a point on <math>\overline{BC}</math> such that <math>\overline{EF}</math> is an altitude of triangle <math>CEB</math>. Note that <math>\triangle CEB</math> ~ <math>\triangle CFE</math> ~ <math>\triangle EFB</math>. Therefore, <math>BF=\frac{x}{x^2+1}</math> and <math>CF=\frac{x^3}{x^2+1}</math>. Since <math>\overline{CF}</math> and <math>\overline{BF}</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields:
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Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD \sim \triangle ABC \sim \triangle CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=\frac{x^2}{\sqrt{x^2+1}}</math> and <math>BE=\frac{x}{\sqrt{x^2+1}}</math>. Let F be a point on <math>\overline{BC}</math> such that <math>\overline{EF}</math> is an altitude of triangle <math>CEB</math>. Note that <math>\triangle CEB \sim \triangle CFE \sim \triangle EFB</math>. Therefore, <math>BF=\frac{x}{x^2+1}</math> and <math>CF=\frac{x^3}{x^2+1}</math>. Since <math>\overline{CF}</math> and <math>\overline{BF}</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields:
 
<cmath>[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))</cmath>
 
<cmath>[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))</cmath>
 
<cmath>[ABCD]=[AED]+[DEC]+[CEB]+[BEA]</cmath>
 
<cmath>[ABCD]=[AED]+[DEC]+[CEB]+[BEA]</cmath>
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<cmath>(x)(x^2+1)=20x^3/(x^2+1)</cmath>
 
<cmath>(x)(x^2+1)=20x^3/(x^2+1)</cmath>
 
<cmath>(x^2+1)^2=20x^2</cmath>
 
<cmath>(x^2+1)^2=20x^2</cmath>
<cmath>x^4-18x^2+1=0 \implies x^2=9+4sqrt(5)=4+2(2sqrt(5))+5</cmath>
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<cmath>x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5</cmath>
 
Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math>
 
Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math>
  

Revision as of 10:13, 17 February 2017

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 $\boxed{\textbf{(D) } 2+\sqrt{5}}$

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
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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