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2017 AMC 12B Problems/Problem 24

Revision as of 21:27, 16 February 2017 by Thedoge (talk | contribs) (→‎Solution)

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

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 $\triangle BCD ~ \triangle ABC ~ \triangle CEB$, 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 $CEB~CFE~EFB$, 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: \[[BEC]=[CED]=Area[BEA]=(x^3)/(2(x^2+1))\] \[Area(ABCD)=Area(AED)+Area(DEC)+Area(CEB)+Area(BEA).\] \[(AB+CD)(BC)/2= 17*Area(CEB)+ Area(CEB) + Area(CEB) + Area(CEB)\] \[(x^3+x)/2=(20x^3)/(2(x^2+1))\] \[(x)(x^2+1)=20x^3/(x^2+1)\] \[(x^2+1)^2=20x^2\] \[x^4-18x^2+1=0 \implies x^2=9+4sqrt(5)=4+2(2sqrt(5))+5\] Therefore, the answer is $\boxed{\textbf{(D) } 2+\sqrt{5}}$


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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