Eq. 2

Eq. 3

Eq. 4. Final Result

Equal Areas Solution

First we set up the geometry necessary to solve the "equal areas" problem. This geometry is shown in Fig. 1, to the right. The radius of the "fence" is small *r* and the radius of the "leash" is big *R*. The angle θ is measured from the horizontal axis to the intersection point of the two circles, as shown. The strategy will be to determine the value of θ necessary to make the intersection area equal to one-half the area of the smaller circle. Once we have θ we can easily determine the ratio *R/r* using the relationship shown in Eq. 1. Eq. 2 shows the θ equation that meets the equal-areas condition. This cannot be solved analytically. We thus resort to numerical analysis. Fig. 3 is a plot of the left side of Eq. 2. The intersection of this curve with the 0 horizontal yields the result for θ that meets the condition of Eq. 2. We know that θ is less than 90 deg thus the proper solution for θ is shown in Eq. 3, below the plot. Our final result for *R/r* is shown in Eq. 4 below. This answer has been rounded off to 5 decimal places. As expected *R/r* > 1. Fig. 3, below, shows the geometric solution with the proper proportions.

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Eq. 1