A RELATIONSHIP BETWEEN CHORDS AND THE RADIUS OF A CIRCLE

Russell Euler

Let a, b, and c represent the distinct lengths of the chords subtending arcs AB, BC, and CD, respectively. Assume that arc AD forms a semicircle (see Figure 1).

Figure 1.

The purpose of this paper is to determine the length of the radius of the semicircle as a function of a, b, and c. The level of the paper is for undergraduate students.

Let r ≠ 0 represent the length of the radius of the semicircle. Ptolemy’s Theorem [2] asserts that in a cyclic quadrilateral, the product of the lengths of the diagonals is equal to the sum of the products of the lengths of the two pairs of opposite sides. Hence, (AC)(BD) = (AB)(CD) + (AD)(BC) and so

(1)  (AC)(BD) = ac + 2rb.

Since triangles ACD and ABD are right triangles,

AC = \sqrt{4r^{2} − c^{2}} and BD = \sqrt{4r^{2} − a^{2}}.

So, identity (1) becomes

\sqrt{(4r^{2} − c^{2})(4r^{2} − a^{2})} = ac + 2rb.

Squaring both sides of this equation and simplifying gives

(2)  r^{3} − (a^{2} + b^{2} + c^{2})r/4 − abc/4 = 0.

By Descartes rule of sign, equation (2) has one positive root and that is the solution that is desired. Before determining this root, a brief review of Cardano’s formula [1] will be given.

The roots of a cubic equation of the form

(3)  x^{3} + 3px + q = 0

depend upon the sign of 4p^{3} + q^{2}. If 4p^{3} + q^{2} ≥ 0, the solution set of (3) is

\{\sqrt[3]{m} + \sqrt[3]{n}, ω\sqrt[3]{m} + ω^{2} \sqrt[3]{n}, ω^{2} \sqrt[3]{m} + ω\sqrt[3]{n}\}

where

ω = (−1 + i\sqrt{3})/2, m = (−q + \sqrt{4p^{3} + q^{2}})/2,

and

n = (−q − \sqrt{4p^{3} + q^{2}})/2.

If 4p^{3} + q^{2} < 0, then write m in the polar form m = |m| e^{iθ} (and n = \overline{m}). The solution set of (3) is

\{2\sqrt[3]{|m|} cos(θ/3), 2\sqrt[3]{|m|} cos[(θ + 2π)/3], 2\sqrt[3]{|m|} cos[(θ + 4π)/3]\}.

To relate these results to (2), let p = −(a^{2} + b^{2} + c^{2})/12 and q = −abc/4. By the arithmetic mean–geometric mean inequality,

(a^{2} + b^{2} + c^{2})/3 > \sqrt[3]{a^{2}b^{2}c^{2}},

which is equivalent to

(a^{2} + b^{2} + c^{2})^{3} > 27a^{2}b^{2}c^{2}.

That is, 4p^{3} + q^{2} < 0. Strict inequality was used above since a, b, and c are not equal.

Now,

m = [3abc + i\sqrt{[(a^{2} + b^{2} + c^{2})^{3} − 27a^{2}b^{2}c^{2}]/3]}/24

and so

|m| = \frac{\sqrt{(a^{2} + b^{2} + c^{2})^{3}/3}}{24}

and

θ = Arctan [\frac{\sqrt{[(a^{2} + b^{2} + c^{2})^{3} − 27a^{2}b^{2}c^{2}]/3}}{3abc}].

Therefore, the positive root of (2) is

(4)  r = \sqrt[3]{\frac{1}{3}\sqrt{(a^{2} + b^{2} + c^{2})^{3}/3}}

· cos \{\frac{1}{3}Arctan [\frac{\sqrt{[(a^{2} + b^{2} + c^{2})^{3} − 27a^{2}b^{2}c^{2}]/3}}{3abc}]\}.

If the restriction of distinctness is removed and one lets a = b = c, then equation (2) becomes r^{3} − 3a^{2}r/4 − a^{3}/4 = 0 and so (r − a)(r + a/2)^{2} = 0. The positive root is r = a which agrees with (4) when a = b = c. Geometrically, this means that to inscribe a regular hexagon in a circle, use the radius of the circle to mark the vertices on the circle.

\underline{References}

1. CRC Standard Mathematical Tables, 21st ed., 103–104.

2. D. C. Kay, College Geometry – A Discovery Approach, Harper Collins, (1994), 31.

Russell Euler Department of Mathematics and Statistics Northwest Missouri State University Maryville, MO 64468-6001 email: 0100120@acad.nwmissouri.edu