SOLUTION TO PROBLEM 2

There are many ways to solve this old problem - calculus, analytic geometry, trigonometry. The following clever decomposition, however, leads to a most elegant solution.

If the other regions that occur in the figure have areas B and C, as shown, then we immediately obtain the equations

$$\begin{array}{c} A+4B+4C=1,\\ A+3B+2C=\mbox{ one quadrant }=\frac{\pi}{4}. \end{array}$$

A third equation follows the neat alternative decomposition of the region TQR of the area A+2B+C, that is bounded by the arcs TQ, TR, and the side QR:

D is an equilateral triangle of unit side, and we have

$$D=\frac{1}{2}\cdot 1\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}:$$

and clearly D+E consitutes a $60^\circ$ sector of a unit circle, implying

$$D+E=\frac{\pi}{6}.$$

Hence

$$A+2B+C=D+2E=2(D+E)-D=\frac{\pi}{3}-\frac{\sqrt{3}}{4}$$

Solving we obtain

$$\begin{array}{lcl} \displaystyle A&\displaystyle =&\displayst... ...&\displaystyle 1-\frac{\sqrt{3}}{4}- \frac{\pi}{6}. \end{array}$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Thu Sep 6 16:55:46 CDT 2001