Difference between revisions of "2000 AIME II Problems/Problem 8"
Mathgeek2006 (talk | contribs) (→Solution) |
Mathgeek2006 (talk | contribs) m (→Solution 2) |
||
Line 24: | Line 24: | ||
<cmath>(x^2-11)^2=11(1001-x^2)</cmath> | <cmath>(x^2-11)^2=11(1001-x^2)</cmath> | ||
<cmath>x^4-11x^2-11\cdot 990=0.</cmath> | <cmath>x^4-11x^2-11\cdot 990=0.</cmath> | ||
− | As above, we can solve this quadratic to get the positve solution <math>BC=x^2=\boxed{110}</math>. | + | As above, we can solve this quadratic to get the positve solution <math>BC^2=x^2=\boxed{110}</math>. |
== See also == | == See also == |
Revision as of 10:51, 23 July 2013
Contents
[hide]Problem
In trapezoid , leg is perpendicular to bases and , and diagonals and are perpendicular. Given that and , find .
Solution
Solution 1
Let be the height of the trapezoid, and let . Since , it follows that , so .
Let be the foot of the altitude from to . Then , and is a right triangle. By the Pythagorean Theorem,
The positive solution to this quadratic equation is .
Solution 2
Let . Dropping the altitude from and using the Pythagorean Theorem tells us that . Therefore, we know that vector and vector . Now we know that these vectors are perpendicular, so their dot product is 0. As above, we can solve this quadratic to get the positve solution .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.