Mock AIME 4 2006-2007 Problems/Problem 15
Triangle has sides , , and of length 43, 13, and 48, respectively. Let be the circle circumscribed around and let be the intersection of and the perpendicular bisector of that is not on the same side of as . The length of can be expressed as , where and are positive integers and is not divisible by the square of any prime. Find the greatest integer less than or equal to .
We set up a trivial coordinate bash. Let A = 0,0, C = 48,0, B = 83/2, 13sqrt3/2. We find the coordinates of the circumcenter to be 24, -11sqrt3/3. The radius is 43sqrt3.Then the coordinate of point D are 24, -18sqrt3. The answer is then 6 + sqrt43, which yields 12.
The perpendicular bisector of any chord of any circle passes through the center of that circle. Let be the midpoint of , and be the length of the radius of . By the Power of a Point Theorem, or . By the Pythagorean Theorem, .
Let's compute the circumradius : By the Law of Cosines, . By the Law of Sines, so .
Now we can use this to compute and thus . By the quadratic formula, . (We only take the positive sign because angle is obtuse so is the longer of the two segments into which the chord divides the diameter.) Then so , and so the answer is .
Let angle = , angle , and . Since ABCD is a cyclic quadrilateral, degrees. Using the Law of Cosines, , so . Since , then is . Using the Law of Cosines on triangle ADC, . Solving for , we get which is between and , so the answer is .