Math Core

Most AIMEs have at least one solid-geometry problem, and they scare people more than they should. The trick is almost always to reduce to two dimensions: take a well-chosen cross-section, unfold a surface flat, or drop in coordinates. This module gives you the formulas and the reduction strategies.

Volumes and the one-third

Prisms and cylinders have volume $Bh$ (base area times height). Pyramids and cones have

$V = \frac{1}{3}Bh.$

Why one-third? A unit cube splits into three congruent pyramids that share the vertex $(1, 1, 1)$ and have the three faces through the origin as bases; each has volume $\tfrac13$ . For a general pyramid, use Cavalieri's principle: two solids whose cross-sections at every height have equal areas have equal volumes. The cross-section of a pyramid at height $y$ is a scaled copy of the base, with area $B\left(1 - \frac{y}{h}\right)^2$ , which depends only on $B$ and $h$ . So every pyramid with the same base area and height has the same volume as the cube-pyramid, scaled: $\tfrac13 Bh$ .

The same cross-section idea gives the most important scaling fact:

Similar solids

If two solids are similar with length ratio $k$ , their surface areas are in ratio $k^2$ and their volumes in ratio $k^3$ . Cutting a pyramid or cone by a plane parallel to its base at the fraction $k$ of the way down from the apex leaves a small pyramid with $k^3$ of the volume.

Right-corner tetrahedra

A tetrahedron $OABC$ with right angles at $O$ between all three edges, with $OA = a$ , $OB = b$ , $OC = c$ , is a corner cut off a box. Put $O$ at the origin and the edges on the axes.

Volume: $V = \dfrac{abc}{6}$ (base $\tfrac12 ab$ , height $c$ ).
The plane $ABC$ is $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$ .
Distance from $O$ to plane $ABC$ : $\dfrac{1}{h^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}$ .
Area of face $ABC$ (a 3D Pythagorean theorem): $[ABC]^2 = [OAB]^2 + [OBC]^2 + [OCA]^2$ .

Proof of the distance formula. The distance from a point $(x_0, y_0, z_0)$ to the plane $px + qy + rz = d$ is $\dfrac{|px_0 + qy_0 + rz_0 - d|}{\sqrt{p^2 + q^2 + r^2}}$ , by the same argument as in 2D. For $O$ and the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ , that is $h = \dfrac{1}{\sqrt{1/a^2 + 1/b^2 + 1/c^2}}$ . $\blacksquare$

Proof of the area formula. $V = \tfrac13 [ABC]\,h$ , so $[ABC] = \dfrac{3V}{h} = \dfrac{abc}{2}\sqrt{\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}}$ . Squaring, $[ABC]^2 = \dfrac{b^2c^2 + a^2c^2 + a^2b^2}{4}$ , which is the sum of the squares of $\tfrac12 ab$ , $\tfrac12 bc$ , $\tfrac12 ca$ . $\blacksquare$

Inscribed spheres

Inradius of a solid

If a sphere of radius $r$ is tangent to every face of a polyhedron with volume $V$ and surface area $S$ , then

$r = \frac{3V}{S}.$

Connect the sphere's center to every vertex. This cuts the solid into pyramids, one per face, each with height $r$ . So $V = \sum \tfrac13 (\text{face area}) \cdot r = \tfrac13 S r$ . It's the 3D version of $K = rs$ .

For a cone or a symmetric pyramid, you can instead take the cross-section through the axis: the inscribed sphere becomes the inscribed circle of a triangle.

Unfolding

The shortest path along a surface becomes a straight line when you unfold the surface flat.

Boxes: unfold the faces the path crosses into one plane.
Cones: a cone with base radius $r$ and slant height $\ell$ unrolls into a sector of radius $\ell$ and arc length $2\pi r$ , so its angle is $\dfrac{r}{\ell} \cdot 360^\circ$ .

A cube cut by a plane through one corner. The cross-section is a parallelogram.

Worked example: A corner of a box

A tetrahedron has three mutually perpendicular edges of lengths $3$ , $4$ and $12$ meeting at a vertex $O$ . Find its volume and the distance from $O$ to the opposite face.

$V = \dfrac{3 \cdot 4 \cdot 12}{6} = 24$ . The distance satisfies $\dfrac{1}{h^2} = \dfrac{1}{9} + \dfrac{1}{16} + \dfrac{1}{144} = \dfrac{16 + 9 + 1}{144} = \dfrac{26}{144}$ , so $h = \dfrac{12}{\sqrt{26}}$ .

Worked example: A regular tetrahedron

Find the volume of a regular tetrahedron with edge $s$ .

The base is equilateral with area $\dfrac{\sqrt3}{4}s^2$ and circumradius $\dfrac{s}{\sqrt3}$ . The apex is directly above the base's center, so the height is $h = \sqrt{s^2 - \dfrac{s^2}{3}} = s\sqrt{\dfrac23}$ . Then

$V = \frac13 \cdot \frac{\sqrt3}{4}s^2 \cdot s\sqrt{\frac23} = \frac{s^3}{6\sqrt2} = \frac{\sqrt2}{12}s^3.$

Worked example: Around a cone

A cone has base radius $2$ and slant height $6$ . An ant starts at a point $A$ on the rim, crawls once around the cone, and returns to $A$ . Find the shortest possible length of its path.

Unrolled, the cone is a sector of radius $6$ with angle $\dfrac{2}{6} \cdot 360^\circ = 120^\circ$ . The two copies of $A$ are on the two edges of the sector, $6$ from the apex and $120^\circ$ apart. The straight path between them has length $\sqrt{36 + 36 - 72\cos 120^\circ} = \sqrt{108} = 6\sqrt3$ .

Common mistake

When the unrolled sector angle is $180^\circ$ or more, the straight segment between the two copies of $A$ passes through the apex (or outside the sector), and the "go around" path no longer beats going straight to the apex and back. Always check the sector angle before using the chord.

Tip

For a plane through three points in a cube or box, find the normal vector with a cross product. If the normal has integer coordinates with an integer length (like $(3, 2, 6)$ , of length $7$ ), the problem was built for it.

Practice

Practice 1

A rectangular box has faces with areas $30$ , $40$ and $48$ . Find the square of the length of its space diagonal.