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

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== 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^2/11</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.
+
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.
  
 
== See also ==
 
== 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}}

Revision as of 21:29, 27 January 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

Call the height of the trapezoid a. Triangles BAC and CBD are similar since AC is perpendicular to BD. Using similar triangles, $CD=a^2/\sqrt{11}$. 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, $a=\sqrt{1001-(a^2/\sqrt{11}-\sqrt{11})^2}$. Letting $a^2=n$ (which is what we're trying to find) and simplifying yields the equation $n^2-11n-10890=0$, and the positive solution is 110.

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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