Mock AIME I 2015 Problems/Problem 7
For all points in the coordinate plane, let denote the reflection of across the line . For example, if , then . Define a function such that for all points , denotes the area of the triangle with vertices , , and . Determine the number of lattice points in the first quadrant such that .
Let the point have the coordinates . We can then represent as the point with the coordinates . We can assume that and then multiply our answer by at the end to account for the case where . We can safely ignore the case where , since this leads to a degenerate triangle between , , and .
We can then devise an expression that represents the area of a triangle with the points , , and . We can see that the distance from to is the same as the distance from to . Therefore, the triangle is isosceles, with as the base.
The length of the base can be found with the distance formula:
Next, to find the distance from the origin to , we note that since the triangle is isosceles, the altitude is from the origin to the midpoint of . Therefore, we can find the midpoint of as . Then, we can use the distance formula to find the distance from the origin to the midpoint of as .
Then, we can find the area of the triangle by
We set this equal to , and we get the equation . We take the prime factorization of 8! to get . In order to get integer solutions for and , both and have to have the same parity. Otherwise, and would not be integers. Since is clearly even, both and have to be even.
We can set to and to in order to represent this. Since both and are positive, we have that . Each solution to , or , we have a solution for . We can find these solutions by counting the number of factors of . We calculate this with .
We then divide this by to get the number of solutions where . We then multiply the answer by to account for our original assumption of . As a result, our final result is