Difference between revisions of "2000 AIME II Problems/Problem 8"
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=== Solution 4 === | === Solution 4 === | ||
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+ | Let <math>E</math> be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have | ||
+ | <cmath>\begin{align*} | ||
+ | BC^2&=BE^2+CE^2 \ | ||
+ | &=(AB^2-AE^2)+(CD^2-DE^2) \ | ||
+ | &=CD^2+\sqrt{11}^2-(AE^2+DE^2) \ | ||
+ | &=CD^2+11-AD^2 \ | ||
+ | &=CD^2-990 | ||
+ | \end{align*}</cmath> | ||
== See also == | == See also == |
Revision as of 18:26, 7 December 2019
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 .
Solution 3
Let and . From Pythagoras with , we obtain . Since and are perpendicular diagonals of a quadrilateral, then , so we have Substituting and simplifying yields and the quadratic formula gives . Then from , we plug in to find .
Solution 4
Let be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have
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.