The Art and Craft of Problem Solving

is the annulus

v-u:::; Izl:::; v+u

4.2 COMPLEX NUMBERS 129

centered at the origin, i.e. the set of points whose distance to the origin is between

v - u and v+u.

Now it is a simple matter to wrap up the problem. Since

a + h ueit + veis

M=

• 2 -

+

2

our locus is an annulus centered at the midpoint of the line segment joining the two

centers, i.e., the point (a + h) /2. Call this point c. Then the locus is the set of points

whose distance from c is between (v - u) /2 and (v + u) /2. (In general, the annulus

need not be tangent to the circle on the right, as shown below.) _

G

h -----i

Example 4.2.15 (Putnam 2003 ) Let A, Band C be equidistant points on the circum­

ference of a circle of unit radius centered at 0, and let P be any point in the circle's

interior. Let a, h, c be the distances from P to A, B, C respectively. Show that there is

a triangle with side lengths a, h, c, and that the area of this triangle depends only on

the distance from P to O.

Solution: This problem has a nice, elementary solution using complex numbers.
The equilateral triangle ABC compels us to try cube roots of unity. We will just sketch
the argument. Please fill in the details.

1. We need to show that there actually is a triangle with side lengths a, b, c. Note

that this may not be possible in general; for example, suppose a = h = 1, c =

10. Given three line segments of lengths a, b, c, one could test to see if they

formed a triangle by physically moving the line segments around and placing

them end-to-end to see ifthey could form a triangle. Use this physical intuition

(and pictures, most likely) to prove the following lemma.

Let ZI ,Z 2 , Z 3 he complex numbers with

IZII =a, IZ (^21) = b, IZ (^31) = c.

Suppose there exist real numbers a, f3 with 0 :::; a, f3 < 2n such that

ZI + eiaZ 2 + eif3 Z 3 = O.