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 .
Solution 3 (similar triangles)
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 .
Solution 4 (Abusing Stewart)
Let , so . Let , . Applying Stewart's Theorem on triangles twice, first using as the base point and then , we arrive at the equations and Now applying law of sines and law of cosines on yields and Solving for from the sines equation and plugging into the law of cosines equation yields . We now finish by adding the two original stewart equations and obtaining: This is a quadratic which only takes some patience to solve for
Solution 5 (Complex Numbers)
Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with , and . Since = 400, . From , we can deduce that the rotation of point 45 degrees counterclockwise, , and the origin are collinear. In other words, is a real number. Simplyfying using the fact that , clearing the denominator, and setting the imaginary part equal to , we eventually get the quadratic which has solutions . It is given that , so and
Let G be a point such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB. Since , , and by Angle Bisector Theorem, . Thus, . , and (90 degree rotation), and now we can bash on 2 similar triangles .
I hope you like expanding Quadratic formula gives us Since AE < BF Thus, So, our answer is .
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