Topping Tetrahedral Ten

You can derive the result from the relationship between the radii of inshere and circumsphere of a tetrahedron.

The height of the vertex of the circumscribed tetrahedron above the top sphere does not depend on the existence of the other spheres below it. It depends only on the angles typical of the tetrahedron, the fact that it is tangent to this sphere, and the sphere's radius, $r$ . In other words the distance from the center of that sphere to the vertex is the same as if the tetrahedron enclosed only that one sphere. So we can simplify the problem to that of a single sphere inscribed in a tetrahedron. The distance from the center of this sphere (and of the tetrahedron enclosing it) to the vertex is the radius of a sphere circumscribed around a tetrahedron, $R$ . It is known that $r=\frac{R}{3}$ , therefore the distance from the center (of the top sphere) to the vertex of the enclosing tetrahedron can accommodate exactly three radii of the sphere. This can be the radius of the original sphere plus the two radii of the additional sphere on top of it.

Or you can get the answer using only elementary geometry, namely similar triangles, Pythagorean theorem, and the formula for the area of a triangle. Let’s set the radius of each sphere to $\frac{1}{4}$ . Then the centers of the ten spheres will then lay on the ends or the centers of the edges of a tetrahedron with edge equal 1. Let’s take a cross-section of it positioned so that $AB$ is an edge and $C$ is located in the middle of the opposite edge. This means that $AC=CB=\frac{\sqrt{3}}{2}$ .

If $D$ is the center of $AB$ , we can get $CD$ from the right triangle $BDC$ as $CD=\sqrt{BC^2-BD^2}=\sqrt{\frac{3}{4}-\frac{1}{4}}=\frac{1}{\sqrt{2}}$ .

Area of the triangle $ABC$ does not depend on which base and which height we pick, therefore $AB\times CD=AC\times EB$ . This can be used to calculate $EB=\frac{AB\times CD}{AC}=$ $\frac{1\times\frac{1}{\sqrt{2} } }{\frac{\sqrt{3}}{2}}$ $=\frac{\sqrt{2}}{\sqrt{3}}$ .

From a right triangle $CEB$ we can now find $EC=\sqrt{BC^2-EB^2}=\sqrt{\frac{3}{4}-\frac{2}{3}}=\frac{1}{2\sqrt{3}}$ .

Triangles $ADC$ and $OEC$ are similar. This means $\frac{OE}{EC}=\frac{AD}{DC}$ . This can be used to calculate

$OE=\frac{AD\times EC}{DC}=$ $\frac {\frac{1}{2}\times\frac{1}{2\sqrt{3}}} {\frac{1}{\sqrt{2}}}$ $=\frac{1}{2\sqrt{6}}$ .

Distance $OB=EB-OE=\frac{\sqrt{2}}{\sqrt{3}}-\frac{1}{2\sqrt{6}}=\frac{3}{2\sqrt{6}}$ .

So we have $OA=OB=\frac{3}{2\sqrt{6}}$ and $OE=\frac{1}{2\sqrt{6}}$ , or $OA=3\times OE$ .

All of this still applied only to the tetrahedron generated by the centers of the spheres. So to finish the calculation, we need to move out to the tetrahedron encasing the spheres, that is go outward one radius of one sphere, which we set at $\frac{1}{4}$ , in a direction perpendicular to the face of tetrahedron. Line $AC$ is a side view of such a face. Line $AF$ is perpendicular to it. Triangle $OEA$ is similar to triangle $AFG$ , therefore $AG=\frac{OA}{EO}\times AF=\frac{3}{4}$ . That is three times the radius of one of the spheres.

Consider that the spheres all fit perfectly into the tetrahedron, the edges touching.

It is reasonable to say, then, that every time you go down a layer, another layer's worth of spheres may fit exactly inside the tetrahedron.

Similarly, we can conclude that every time you go up a layer, one less layer's worth of spheres may fit exactly inside the tetrahedron.

So, from the top level with 1 sphere to an imaginary level with 0 spheres, the same vertical distance is traveled as from level 1 to level 2.

Additionally, this distance is half the distance from level 1 to level 3.

We notice that level 3 is just level 1 with a sphere added onto the side, (from the face closest to the tetrahedron). The slope between levels 3 and 1, therefore, is exactly 2 sides-of-spheres sideways to 1 sphere-heights up.

The use of different units here is because the spheres are aligned in such a way that the distances are not the same. However, it will cancel out later so it doesn't matter.

We then use this rate of 1 sphere-height/ 2 sphere-sides from the top of the level 1 sphere to the top of the nonexistant level 0 sphere, but because we only need to go half a sphere-side towards the center, (both edges are converging) we can say that the tip of the pyramid is exactly 1 sphere-height higher than the top of the sphere.