Avoid Computations – Use Variables!

If you’re like me, you’re prone to computational error. One technique that helps minimize such errors is to use variables and delay substitution of specific values for as long as possible.

Problem #14 on the 1990 AIME can be used to illustrate this nicely. For your convenience, here’s the problem:

The rectangle ABCD has dimensions AB = $12 \sqrt{3}$ and BC = $13 \sqrt{3}$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at P. If triangle ABP is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segments $\overline{CP}$ and $\overline{DP}$, we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.

If you work this problem through using the specific dimensions provided from the outset, you make yourself vulnerable to computational error. You might even perform some unnecessary, wasteful computations. For example you might multiply by a quantity only to later divide by the same quantity, giving yourself two chances to err when there shouldn’t even be one. Even more, by plowing ahead with specific numbers, you may not even realize that you’ve done something like that! Instead, find a general volume formula for an arbitrary a by b rectangle, where a represents the length AB in the figure.

We’ll use Heron’s area formula and the formula $R = \frac{abc}{4A}$ for the radius R of the circumcircle of a triangle with area A and side lengths a, b, and c.

The volume of the tetrahedron is given by $V = \frac{1}{3}Ah$, where A is the area of the base (using the base as indicated in the figure) and h is the altitude with respect to that base. Because the 3 red segments are corresponding parts of 3 congruent right triangles, they have equal lengths and are radii of the circumcircle of the base.

By the Pythagorean theorem,

Since $AP^2 = \frac{1}{4}(a^2 + b^2)$, we find that

Thus,

By Heron’s formula, $A^2 = (b + \frac{a}{2})(b-\frac{a}{2})(\frac{a}{2})^2$, so that

At this point, one can substitute $a = 12\sqrt{3}$ and $b = 13\sqrt{3}$, to find $V = 18\sqrt{3^2169 - 3(144)} = 18(3)(11) = 593$.

Ha! Just kidding…$18(3)(11) = 594$.

There are many advantages to using variables instead of specific quantities. You’re far less prone to computational error. Mistakes are easier to find. You can test the sensibility of the formula by checking if it has units of length cubed or make sure it agrees with easy-to-compute special cases such as when a = b. And the formula leads to additional questions and enables one to answer others, such as: What happens when $3b^2 - a^2 = 0$? Can you ever get an integral volume starting with integral dimensions?