2005 AIME II Problems/Problem 12
Solution 1 (trigonometry)
Let be the foot of the perpendicular from to . Denote and , and (since and ). Then , and .
By the tangent addition rule , we see that Since , this simplifies to . We know that , so we can substitute this to find that .
Substituting again, we know have . This is a quadratic with roots . Since , use the smaller root, .
Now, . The answer is .
Solution 2 (synthetic)
Label , so . Rotate about until lies on . Now we know that therefore also since is the center of the square. Label the new triangle that we created . Now we know that rotation preserves angles and side lengths, so and . Draw and . Notice that since rotations preserve the same angles so too. By SAS we know that so . Now we have a right with legs and and hypotenuse . By the Pythagorean Theorem,
and applying the quadratic formula we get that . Since we take the positive root, and our answer is .
Let the midpoint of be and let , so then and . Drawing , we have , so By the Pythagorean Theorem on , Setting these two expressions for equal and solving for (it is helpful to scale the problem down by a factor of 50 first), we get . Since , we want the value , and the answer is .
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