The Art and Craft of Problem Solving

is an invariant, always equal to zero.

3.4 INVARIANTS 93

Example 3.4.2 The Power of a Point Theorem. Given a fixed point P and a fixed
circle, draw a line through P that intersects the circle at X and Y. The power of the
point P with respect to this circle is defined to be the quantity PX. PY.
The Power of a Point theorem (also known as POP) states that this quantity is
invariant; i.e., it does not depend on the line that is drawn. For example, in the picture
below,

PX ·PY =PX'·Py'.

Y'

You undoubtedly learned this theorem in elementary geometry, at least for the case
where the point P lies inside the circle. For a proof, see Example 8.3.11.

Example 3.4.3 Euler's Formula. You encountered this originally as Problem 2.2. 12
on page 37, which asked you to conjecture a relationship between the number of ver­
tices, edges and faces of any polyhedron. It turns out that if v, e , and f denote the
number of vertices, edges and faces of a polyhedron without "holes," then

v-e+f=2

always holds; i.e., the quantity v - e + f is an invariant. This is known as Euler's
Formula. See Problem 3.4.4 0 for some hints on how to prove this formula.

Example 3.4.4 Symmetry. Even though we devoted the first section of this chapter
to symmetry, this topic is logically contained within the concept of invariants. If a
particular object (geometric or otherwise) contains symmetry, that is just another way
of saying that the object itself is an invariant with respect to some transformation or set
of transformations. For example, a square is invariant with respect to rotations about
its center of 0, 90, 180 and 270 degrees.
On a deeper level, the substitution u := x + x-I, which helped solve

x^4 +x^3 +.l+x+l = 0

in Example 3.1.1 0 worked, because u is invariant with respect to a permutation of

some of the roots.^5

(^5) This idea is the germ of one of the greatest achievements of 19th-century mathematics, Galois theory, which
among other things develops a systematic way of determining which polynomials can be solved with radicals and
which cannot. For more information, consult Herstein's wonderful book [20].