2007 AIME I Problems/Problem 12

Revision as of 19:16, 15 March 2007 by Azjps (talk | contribs) (let's c if I can work this out now w/o careless mistakes)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


In isosceles triangle $ABC$, $A$ is located at the origin and $B$ is located at (20,0). Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$. If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$-axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$, where $p,q,r,s$ are integers. Find $(p-q+r-s)/2$.


This problem needs a solution. If you have a solution for it, please help us out by adding it.

The union of the area is equal to $2$ times the area of $\triangle ABC$, minus the union of the area of the two triangles.

See also

2007 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS