2000 AIME II Problems/Problem 8
In trapezoid , leg is perpendicular to bases and , and diagonals and are perpendicular. Given that and , find .
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 .
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 .
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 .
Let be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have Followed by dropping the perpendicular like in solution 1, we obtain system of equation Rearrange the first equation yields Equating it with the second equation we have Which gives . Substituting into equation 1 obtains the quadratic in terms of Solving the quadratic to obtain .
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