2005 AIME I Problems/Problem 7
In quadrilateral and Given that where and are positive integers, find
Draw line segment such that line is concurrent with line . Then, is an isosceles trapezoid so , and and . We are given that . Since , using Law of Cosines on gives which gives . Adding to both sides gives , so . and are both , so and . , and therefore .
Draw the perpendiculars from and to , labeling the intersection points as and . This forms 2 right triangles, so and . Also, if we draw the horizontal line extending from to a point on the line , we find another right triangle . . The Pythagorean Theorem yields that , so . Therefore, , and .
Extend and to an intersection at point . We get an equilateral triangle . We denote the length of a side of as and solve for it using the Law of Cosines: This simplifies to ; the quadratic formula yields the (discard the negative result) same result of .
Extend and to meet at point , forming an equilateral triangle . Draw a line from parallel to so that it intersects at point . Then, apply Stewart's Theorem on . Let . By the quadratic formula (discarding the negative result), , giving for a final answer of .
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