# Sphere inscribed in Regular Tetrahedron

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=bh, & 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.

In two dimensions, the intersection of medians of a triangle split the medians in a $1:2$ ratio. In 3 dimensions they split in a $1:3$ ratio.

And with regular triangles the altitudes are the medians.

I think that this problem becomes a little bit easier when you assign coordinates to the system.

Suppose the vertexes of our tetrahedron are $(t,t,t), (-t,-t,t), (t,-t,-t), (-t,t,-t).$

Verify that this is a regular tetrahedron.

The center of the sphere will be at the origin.

The incenter of the faces is $(-\frac {t}{3}, -\frac {t}{3},-\frac {t}{3}),(\frac {t}{3}, \frac {t}{3},-\frac {t}{3}),(-\frac {t}{3}, \frac {t}{3},\frac {t}{3}),(\frac {t}{3}, -\frac {t}{3},\frac {t}{3})$

The distance from the the origin to the incenter of each face is $\frac {1}{3}\sqrt 3$

The distance from the origin to the vertex is $\sqrt 3 t$

The length of the altitude is $\frac {4}{3}\sqrt 3 t$

The ratio of the altitude to the radius is $\frac 14$

• I'm confused. I follow that we first assign coordinates to the vertices of the tetrahedron. But how can I verify this produces a tetrahedron or the center of the sphere without vectors? (I'm a high school freshman). Then, you find the in centers of the faces, and find distance from origin to each in-center, and vertex, but how do we know that the distance from the origin to the in center of each face is equal to the radius of the sphere? – DarkRunner May 21 '18 at 22:03
• Revision: I just realized, one can pick the vertices of a tetrahedron from a cube. – DarkRunner May 21 '18 at 22:07
• Is the center of mass of a given cube the center of the sphere inscribed in a tetrahedron? By all accounts, this seems to be so, for the ratio 1/4 still holds. However, I don't know if this can be rigorously proved, or if it's true at all. – DarkRunner May 21 '18 at 23:42
• The distance from the origin to the center of all of the faces is the same, which puts the origin at the center of the sphere. – Doug M May 21 '18 at 23:45
• Ah, I see, since all faces are equidistant from the origin, the origin is the center of both the sphere, and the center of mass of the cube, no? – DarkRunner May 21 '18 at 23:51

Take a cube in vertex to vertex diagonal alignment. Then you would see that all vertices fall in 4 consecutive parallel planes: one at the tip, one at the neighbouring three vertices, one for the next three vertices, and finally one at the opposite tip. When considering the squares, each as a join of 3 consecutive vertex layers, it becomes evident that all these planes of vertex layers are equally spaced, 1:1:1.

Next consider the tetrahedron when inscribed into a cube by alternation of vertices. WRT the above orientation, you chop off the base tip say, resulting in the plane of the neighbouring vertices. Thus the inradius of the tetrahedron is just half of the central segment. Or stated differently, 3 times the inradius of the tetrahedron is the circumradius of the encasing cube. - But WRT the opposite vertex of the cube, it is clear that this one will be maintained in the vertex alternation. Therefore the inradius of the tetrahedron is just 1/3 of the circumradius of the tetrahedron.

As both radii are collinear and pointing in opposite directions, you get the altitude to be the sum of the circumradius plus the inradius of the tetrahedron. And as those are in ratio 3:1 you would get the altitude of the tetrahedron within this same scaling to be 3+1=4.

--- rk