2001 AIME I Problems/Problem 12
The center of the insphere must be located at where is the sphere's radius. must also be a distance from the plane
The signed distance between a plane and a point can be calculated as , where G is any point on the plane, and P is a vector perpendicular to ABC.
A vector perpendicular to plane can be found as
Thus where the negative comes from the fact that we want to be in the opposite direction of
Notice that we can split the tetrahedron into smaller tetrahedrons such that the height of each tetrahedron is and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be and surface area be , using the volume formula for each pyramid(base times height divided by 3) we have . The surface area of the pyramid is . We know triangle ABC's side lengths, and , so using the expanded form of heron's formula, . Therefore, the surface area is , and the volume is , and using the formula above that , we have and thus , so the desired answer is .
(Solution by Shaddoll)
The intercept form equation of the plane is Its normal form is (square sum of the coefficients equals 1). The distance from to the plane is . Since and are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in ). Therefore we have So which solves the problem.
Additionally, if is on the other side of , we have , which yields corresponding an "ex-sphere" that is tangent to face as well as the extensions of the other 3 faces.
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