Square Pyramid

A square pyramid is a Pyramid with a Square base. If the top of the pyramid is cut off by a Plane, a square Pyramidal Frustum is obtained. If the four Triangles of the square pyramid are Equilateral, the square pyramid is the ``regular'' Polyhedron known as Johnson Solid $J_1$ and, for side length $a$, has height

$$h={\textstyle{1\over 2}}\sqrt{2}\,a.$$ (1)

Using the equation for a general Pyramid, the Volume of the ``regular'' is therefore

$$V={\textstyle{1\over 3}}hA_b={\textstyle{1\over 6}}\sqrt{2}\,a^3.$$ (2)

If the apex of the pyramid does not lie atop the center of the base, then the Slant Height is given by

$$s=\sqrt{h^2+{\textstyle{1\over 2}}a^2},$$ (3)

where $h$ is the height and $a$ is the length of a side of the base.

Consider a Hemisphere placed on the base of a square pyramid (having side lengths $a$ and height $h$). Further, let the hemisphere be tangent to the four apex edges. Then what is the volume of the Hemisphere which is interior the pyramid (Cipra 1993)?

From Fig. (a), the Circumradius of the base is $a/\sqrt{2}$. Now find $h$ in terms of $r$ and $a$. Fig. (b) shows a Cross-Section cut by the plane through the pyramid's apex, one of the base's vertices, and the base center. This figure gives

$$b = \sqrt{{\textstyle{1\over 2}}a^2-r^2}$$ (4)

$$c = \sqrt{h^2-r^2}\,,$$ (5)

so the Slant Height is

$$s=\sqrt{h^2+{\textstyle{1\over 2}}a^2}=b+c=\sqrt{{\textstyle{1\over 2}}a^2-r^2}+\sqrt{h^2-r^2}\,.$$ (6)

Solving for $h$ gives

$$h={ra\over \sqrt{a^2-2r^2}}.$$ (7)

We know, however, that the Hemisphere must be tangent to the sides, so $r=a/2$, and

$$h={{\textstyle{1\over 2}}a\over\sqrt{a^2-{\textstyle{1\over ... ...\textstyle{1\over 2}}}} a = {\textstyle{1\over 2}}\sqrt{2}\,a.$$ (8)

Fig. (c) shows a Cross-Section through the center, apex, and midpoints of opposite sides. The Pythagorean Theorem once again gives

$$l=\sqrt{{\textstyle{1\over 4}}a^2+h^2} = \sqrt{{\textstyle{1... ...\textstyle{1\over 2}}a^2} = {\textstyle{1\over 2}}\sqrt{3}\,a.$$ (9)

We now need to find $x$ and $y$.

$$\sqrt{{\textstyle{1\over 4}}a^2-x^2}+d=l.$$ (10)

But we know $l$ and $h$, and $d$ is given by

$$d=\sqrt{h^2-x^2}\,,$$ (11)

so

$$\sqrt{{\textstyle{1\over 4}}a^2-x^2}+\sqrt{{\textstyle{1\over 2}}a^2-x^2}={\textstyle{1\over 2}}\sqrt{3}\, a.$$ (12)

Solving gives

$$x={\textstyle{1\over 6}}\sqrt{6}\,a,$$ (13)

so

$$y=\sqrt{r^2-x^2}=\sqrt{{\textstyle{1\over 4}}-{\textstyle{1\over 6}}} \,a = \sqrt{3-2\over 12}\,a = {a\over 2\sqrt{3}}.$$ (14)

We can now find the Area of the Spherical Cap as

$$V_{\rm cap}={\textstyle{1\over 6}}\pi H(3A^2+H^2),$$ (15)

where

$$A \equiv y={a\over 2\sqrt{3}}$$ (16)

$$H \equiv r-x = {\textstyle{1\over 2}}a-{a\over\sqrt{6}} = a\left({{1\over 2}-{1\over\sqrt{6}}}\right),$$ (17)

so

$$V_{\rm cap} = {\textstyle{1\over 6}} \pi a^3 \left[{3\left({1\over 12}\right)+\... ...}-{1\over\sqrt{6}}}\right)^2}\right] \left({{1\over 2}-{1\over\sqrt{6}}}\right)$$

$$= {\textstyle{1\over 6}}\pi a^3\left[{{1\over 4}+\left({{1\over 4}+... ... 6}-{1\over\sqrt{6}}}\right)}\right] \left({{1\over 2}-{1\over\sqrt{6}}}\right)$$

$$= {\textstyle{1\over 6}}\pi a^3 \left({{2\over 3}-{1\over\sqrt{6}}}\right)\left({{1\over 2}-{1\over\sqrt{6}}}\right)$$

$$= {\textstyle{1\over 6}}\pi a^3\left({{1\over 3}-{1\over 2\sqrt{6}}-{2\over 3\sqrt{6}}+{1\over 6}}\right)$$

$$= {\textstyle{1\over 6}}\pi a^3\left({{1\over 2}-{7\over 6\sqrt{6}}}\right).$$ (18)

Therefore, the volume within the pyramid is

$$V_{\rm inside} = {\textstyle{2\over 3}}\pi r^3-4V_{\rm cap} = {\textstyle{2\over 3... ... a^3 -{\textstyle{2\over 3}}\pi a^3\left({{1\over 2}-{7\over 6\sqrt{6}}}\right)$$

$$= {\textstyle{2\over 3}}\pi a^3\left({{1\over 8}-{1\over 2}+{7\over... ...) = {\textstyle{2\over 3}} \pi a^3 \left({{7\over 6\sqrt{6}}-{3\over 8}}\right)$$

$$= \pi a^3\left({{7\over 9\sqrt{6}}-{1\over 4}}\right).$$ (19)

This problem appeared in the Japanese scholastic aptitude test (Cipra 1993).

See also Square Pyramidal Number

References

Cipra, B. ``An Awesome Look at Japan Math SAT.'' Science 259, 22, 1993.