Difference between revisions of "2000 AIME II Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | {{ | + | 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=p</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 21:28, 27 January 2008
Problem
In trapezoid 
, leg 
is perpendicular to bases 
and 
, and diagonals 
and 
are perpendicular. Given that 
and 
, find 
.
Solution
Call the height of the trapezoid a. Triangles BAC and CBD are similar since AC is perpendicular to BD. Using similar triangles, 
. 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, 
. Letting 
(which is what we're trying to find) and simplifying yields the equation 
, 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 | ||
