Trinity

For the regular trigonal hosohedron , we can think of three planes intersecting the ball $x^2+y^2+z^2 \leq 1$ :

• Plane 1: y=z
• Plane 2: x=z
• Plane 3: x=y.

Alternatively, or maybe complementary to that, we can also think of sets of points that are in each of the three parts:

• Part X consists of all points of the ball that have x>y and x>z
• Part Y has all points where y>x and y>z
• Part Z has all points where z>x and z>y

So each of the 3 parts has spherical area 4πr^2/3 (one third of the total sphere exterior) and planar area πr^2 (two half disks: one half on each of the two interior faces).

So the total area for each of the three parts is $\frac{7π}{3}$ . The volume of such a part is $\frac{4π}{9}$ . And the ratio $A = \frac{\text{Area}}{\text{Volume}} = \frac{63}{12}=\boxed{5.25}$ , which is the requested answer already. Note that it is an exact answer.

So what about the calotte and the number B?

I will assume that the cuts are parallel to the x-y- plane. For the spherical area we want to integrate $\int_{z=z_1}^{z_2}dA$ . The integrand dA is a thin ring at height z. Let φ be the angle between the z-axis and the line connecting the origin to the ring, then the radius of the ring is $r=\sin φ$ . The surface dA of the ring is at an angle 90-φ with the z-axis and therefore has area $dA=\frac{2πrdz}{\cos (90-φ)}=\frac{2π\sinφ dz}{\sin φ} =2πdz$ . So spherical area is just equal to $\int_{z_1}^{z_2}2πdz=2π(z_2-z_1)$ .

A cap of height h (h is the distance from the intersecting plane to the top of the unit sphere) has spherical area 2πh and planar area $(1-(1-h)^2)π=(2h-h^2)π$ , so $\text{Area}_{cap} = (4h-h^2)π$

The mid section between the caps has spherical area $4π(1-h)$ and planar area $(4h-2h^2)π$ so $\text{Area}_{mid} = (4-2h^2)π$

The volume of a cap is found easily from $\int_{1-h}^1 πr^2dx$ with $r^2=1-x^2$ . and is $\text{Volume}_{cap} = π(h^2-\frac13h^3)$

The volume of the mid section is $\text{Volume}_{mid} =\frac13π(4 - 6h^2+2h^3)$

The area to volume ratio of a cap is $\text{Ratio}_{cap}=\frac{4h-h^2}{h^2-\frac13h^3}$ The area to volume ratio of the mid section is $\text{Ratio}_{mid}=\frac{3(4-2h^2)}{4 - 6h^2+2h^3}$

Setting these ratios equal and solving numerically gives the value $h^*=0.7459$ , for which the area-to-volume for each of the three slices is equal. This ratio $B$ is found by setting $h=h^*$ in either ratio expression:

$\text{Ratio}_{cap}(h^*)=\text{Ratio}_{mid}(h^*)=B = 5.806$

Note that $h^*$ and $B$ are approximations.

And indeed, it turns out that $B - h^2 = 5.25 = A$