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Difference between revisions of "2000 AIME II Problems/Problem 8"

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== Problem ==
 
== Problem ==
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In trapezoid <math>ABCD</math>, leg <math>\overline{BC}</math> is perpendicular to bases <math>\overline{AB}</math> and <math>\overline{CD}</math>, and diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> are perpendicular. Given that <math>AB=\sqrt{11}</math> and <math>AD=\sqrt{1001}</math>, find <math>BC^2</math>.
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In [[trapezoid]] <math>ABCD</math>, leg <math>\overline{BC}</math> is [[perpendicular]] to bases <math>\overline{AB}</math> and <math>\overline{CD}</math>, and diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> are perpendicular. Given that <math>AB=\sqrt{11}</math> and <math>AD=\sqrt{1001}</math>, find <math>BC^2</math>.
  
 
== Solution ==
 
== Solution ==
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Call the height of the trapezoid a. Triangles BAC and CBD are similar since AC is perpendicular to BD. Using similar triangles, <math>CD=a^2/\sqrt{11}</math>. Call the foot of the altitude from A to CD H. We know AH=a, but AHD is a right triangle, so by the Pythagorean Theorem, <math>a=\sqrt{1001-(a^2/\sqrt{11}-\sqrt{11})^2}</math>. Letting <math>a^2=n</math> (which is what we're trying to find) and simplifying yields the equation <math>n^2-11n-10890=0</math>, and the positive solution is 110.
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Let <math>x = BC</math> be the height of the trapezoid, and let <math>y = CD</math>. Since <math>AC \perp BD</math>, it follows that <math>\triangle BAC \sim \triangle CBD</math>, so <math>\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}</math>.
  
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Let <math>E</math> be the foot of the altitude from <math>A</math> to <math>\overline{CD}</math>. Then <math>AE = x</math>, and <math>ADE</math> is a [[right triangle]]. By the [[Pythagorean Theorem]],
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<cmath>x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0</cmath>
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The positive solution to this [[quadratic equation]] is <math>x^2 = \boxed{110}</math>.
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<center><asy>
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size(200); pathpen = linewidth(0.7);
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pair C=(0,0),B=(0,110^.5),A=(11^.5,B.y),D=(10*11^.5,0),E=foot(A,C,D);
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D(MP("A",A,(2,.5))--MP("B",B,W)--MP("C",C)--MP("D",D)--cycle); D(A--C);D(B--D);D(A--E,linetype("4 4") + linewidth(0.7));
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MP("\sqrt{11}",(A+B)/2,N);MP("\sqrt{1001}",(A+D)/2,NE);MP("\sqrt{1001}",(A+D)/2,NE);MP("x",(B+C)/2,W);MP("y",(D+C)/2);D(rightanglemark(B,IP(A--C,B--D),C,20));
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</asy></center>
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== See also ==
 
{{AIME box|year=2000|n=II|num-b=7|num-a=9}}
 
{{AIME box|year=2000|n=II|num-b=7|num-a=9}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 10:16, 30 August 2008

Problem

In trapezoid $ABCD$, leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$, and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$, find $BC^2$.

Solution

Let $x = BC$ be the height of the trapezoid, and let $y = CD$. Since $AC \perp BD$, it follows that $\triangle BAC \sim \triangle CBD$, so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$.

Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$. Then $AE = x$, and $ADE$ is a right triangle. By the Pythagorean Theorem,

\[x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0\]

The positive solution to this quadratic equation is $x^2 = \boxed{110}$.

[asy] size(200); pathpen = linewidth(0.7); pair C=(0,0),B=(0,110^.5),A=(11^.5,B.y),D=(10*11^.5,0),E=foot(A,C,D); D(MP("A",A,(2,.5))--MP("B",B,W)--MP("C",C)--MP("D",D)--cycle); D(A--C);D(B--D);D(A--E,linetype("4 4") + linewidth(0.7)); MP("\sqrt{11}",(A+B)/2,N);MP("\sqrt{1001}",(A+D)/2,NE);MP("\sqrt{1001}",(A+D)/2,NE);MP("x",(B+C)/2,W);MP("y",(D+C)/2);D(rightanglemark(B,IP(A--C,B--D),C,20)); [/asy]

See also

2000 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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