Grazing In A Circular Pen

This question is similar to the famous Goat problem . Consider a unit circle where a goat is tied to a point on the perimeter of the circle with a tether length of $L =r$ . By circle-circle intersection , we can see that the area that the goat can graze is

$A = r^2 \arccos^{-1} \left( \dfrac{d^2+r^2-R^2}{2dr} \right) + R^2 \cos^{-1} \left( \dfrac{d^2+R^2-r^2}{2dR} \right) - \dfrac12 \sqrt{(-d + r+ R)(d+r-R)(d-r+R)(d+r+R)} \; .$

In this case, $R=d=1$ gives

$A(r) = -\dfrac12 \sqrt{4-r^2} + r^2 \cos^{-1} \left( \dfrac r2 \right) + \cos^{-1} \left( 1- \dfrac {r^2}2 \right) \; .$

Since the question states that the goat can graze half of the pen, then the area, $A(r)$ is simply half of the area of a unit circle,

$A(r) = A\left( \dfrac12 \right) = \pi \left(\dfrac 12\right)^2 = -\dfrac12 r\sqrt{4-r^2} + r^2 \cos^{-1} \left( \dfrac r2\right) + \cos^{-1} \left( 1 - \dfrac {r^2}2 \right) \; .$

The equation above can only be solved using numerical methods. If we consider using Newton-Raphson method with initial point of $x_0 = \dfrac{141 + 100}2 = 120.5$ , a couple of iterations gives $r = 1.159\cdots$ . Note that we only care about the first 3 decimal places, because we just want to find $100 r$ to the nearest integer.

In this case, we have $L = \lfloor 100r \rceil = \boxed{116}$ .

The condition that must be satisfied is to have the 'lime colored area' equal the 'rose colored area'. The points A, B, C, E, F and G are easily determined. The areas can be found by integration.

In the image above, the length BF is 115 m and 'rose area' exceeds 'lime area' by 96.28 $m^2$ .

In the second image, the length BF is 116 m and 'rose area' exceeds 'lime area' by 14.04 $m^2$ . Since L is to be determined to the nearest integer, we choose $L= \boxed {116}$