A sphere is inscribed in a regular tetrahedron. If the length of an altitude of the tetrahedron is 36, what is the length of a radius of the sphere?
I understand the solution to this problem is outlined @Radius of inscribed sphere within regular tetrahedron, however, I have trouble understanding why my own method is faulty.
My thought process:
If the incenter of every face of the tetrahedron was found, and perpendicularly projected into space, their point of concurrency would lie on the altitude of the tetrahedron. Thus, since the radius of the sphere lay on the altitude of the tetrahedron, all I needed to find was the ratio of the altitude of the tetrahedron to the radius of the sphere.
To do so, I simplified the problem into 2 dimensions, and used the formulas
A=rs to solve for the in-radius,
r. I then divided the in-radius
r, over the altitude of the equilateral triangle (keeping in mind the area of an equilateral triangle is always
s^2sqrt(3)/4). I then ended up with a ratio of 1:3 (length of in-radius to altitude in equilateral triangle). I then applied this ratio to the tetrahedron, getting a final answer of 1/3*36=12. However, the final answer turned out to be 9 (as in 1/4*36=9, since apparently the ratio of the altitude to the in-radius of the sphere is 1:4).
Is there any faulty logic in my reasoning? If not, why is my answer wrong? Any help is appreciated.