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

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Solution by TorrTar
 
Solution by TorrTar
  
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Let <math>CD=1</math>, <math>BC=x</math>, <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. The Pythagorean theorem states that <math>BD=sqrt(x^2+1)</math>. Since <math>BCD~ABC~CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=x^2/sqrt(x^2+1)</math> and <math>BE=x/sqrt(x^2+1)</math>. Let Point F be a point on <math>BC</math> such that <math>EF</math> is an altitude of triangle <math>CEB</math>. Note that <math>CEB~CFE~EFB</math>, so <math>BF</math> and <math>CF</math> can be calculated. Solving for these lengths gives <math>BF=x/(x^2+1)</math> and <math>CF=x^3/(x^2+1)</math>.
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Let <math>CD=1</math>, <math>BC=x</math>, <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. The Pythagorean theorem states that <math>BD=sqrt(x^2+1)</math>. Since <math>BCD~ABC~CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=x^2/sqrt(x^2+1)</math> and <math>BE=x/sqrt(x^2+1)</math>. Let Point F be a point on <math>BC</math> such that <math>EF</math> is an altitude of triangle <math>CEB</math>. Note that <math>CEB~CFE~EFB</math>, so <math>BF</math> and <math>CF</math> can be calculated. Solving for these lengths gives <math>BF=x/(x^2+1)</math> and <math>CF=x^3/(x^2+1)</math>. Since <math>CF</math> and <math>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:
 +
<math>Area(BEC)=Area(CED)=Area(BEA)=(x^3)/(2(x^2+1))</math>
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<math>Area(ABCD)=Area(AED)+Area(DEC)+Area(CEB)+Area(BEA).</math>
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<math>(AB+CD)(BC)/2= 17*Area(CEB)+ Area(CEB) + Area(CEB) + Area(CEB)</math>
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<math>(x^3+x)/2=(20x^3)/(2(x^2+1))</math>
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<math>(x)(x^2+1)=20x^3/(x^2+1)</math>
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<math>(x^2+1)^2=20x^2</math>
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<math>x^4-18x^2+1=0</math>
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<math>x^2=9+4sqrt(5)=4+2(2sqrt(5))+5</math> (Minus yields a negative value)
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<math>x=2+sqrt(5)</math>
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Thus the answer is D: 2+sqrt(5)
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==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:41, 16 February 2017

Problem

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

Solution

Solution by TorrTar

Let $CD=1$, $BC=x$, $AB=x^2$. Note that $AB/BC=x$. The Pythagorean theorem states that $BD=sqrt(x^2+1)$. Since $BCD~ABC~CEB$, the ratios of side lengths must be equal. Since $BC=x$, $CE=x^2/sqrt(x^2+1)$ and $BE=x/sqrt(x^2+1)$. Let Point F be a point on $BC$ such that $EF$ is an altitude of triangle $CEB$. Note that $CEB~CFE~EFB$, so $BF$ and $CF$ can be calculated. Solving for these lengths gives $BF=x/(x^2+1)$ and $CF=x^3/(x^2+1)$. Since $CF$ and $BF$ 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: $Area(BEC)=Area(CED)=Area(BEA)=(x^3)/(2(x^2+1))$ $Area(ABCD)=Area(AED)+Area(DEC)+Area(CEB)+Area(BEA).$ $(AB+CD)(BC)/2= 17*Area(CEB)+ Area(CEB) + Area(CEB) + Area(CEB)$ $(x^3+x)/2=(20x^3)/(2(x^2+1))$ $(x)(x^2+1)=20x^3/(x^2+1)$ $(x^2+1)^2=20x^2$ $x^4-18x^2+1=0$ $x^2=9+4sqrt(5)=4+2(2sqrt(5))+5$ (Minus yields a negative value) $x=2+sqrt(5)$ Thus the answer is D: 2+sqrt(5)

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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