Intersecting Circles I

Let there be given a pasture with circular shape and diameter $r$ and let a sheep be attached to a point on the perimeter by a rope of length $l$. Then, how long does $l$ need to be so that the sheep can reach exactly half the pasture?

The answer is $$ \frac{l}{r} = {\verb+1.15872847301812+} $$ and the value can only be achieved by numerical means. Everything needed to derive this result is depicted and named in the following sketch.

We start with stating $$ m = 2r\cdot \sin \frac{\alpha}{2} = 2l\cdot \sin \frac{\beta}{2}~. $$ Moreover, for the quadrilateral $CBDA$ there holds $$ l^2 = \frac{m^2}{2} - 2pq $$ and since $$ p = r \cdot \cos \frac{\alpha}{2} \quad \mbox{and} \quad q = r \cdot ( 1 - \cos\frac{\alpha}{2}\,)~, $$ we’re left with the compact formula $$ l^2 = 2 r^2 \cdot (1 - \cos \frac{\alpha}{2}\,)~. $$ We also note that, because the triangle $ACB$ is isosceles, $$ \beta + \frac{\alpha}{2} = \pi~. $$ We now take into account that by requirement $$ \frac{l^2}{2} \, (\beta - \sin \beta) + \frac{r^2}{2} \, (\alpha - \sin \alpha) = \frac{r^2 \pi}{2} $$ in which we insert the last two representations for $l$ and $\beta$ respectively, so that finally, $$ 2 \cdot \sin \frac{\alpha}{2} + (2 \pi - \alpha ) \cdot\cos \frac{\alpha}{2} = \pi~. $$ The solution $\alpha$ of this nonlinear equation turns out to be $$ \alpha = {\verb+2.47179384855981+}~, $$ giving the above ratio for $r$ and $l$.

While we’re at it, let’s turn our attention to $\gamma = \displaystyle\frac{\beta}{2}\,$ . From the above relationships between the angles and the fact that $ACD$ is isosceles, we deduce $$ 2 \, \gamma - \delta = \frac{\pi}{2}\,~. $$ Now, if we claim that $$ \gamma = \frac{3 \pi}{10} = 54^°~, $$ then $$ \delta = \frac{\pi}{10} = 18^° =\displaystyle \frac{\gamma}{3} \,~. $$

Therefore, the above construction allows us to trisect an angle of $54^°$. This is the solution of level $7.10$ of the famous app Euclidea (available for Android, iOS, and online).

To find the ratio of $r$ and $l\,$ for this particular case, we calculate $$ \alpha = \frac{4 \pi}{5} \quad \mbox{and} \quad \beta = \frac{3 \pi}{5} $$ and see from the first formula above $$ \frac{l}{r} = \frac{\sin \displaystyle\frac{4 \pi}{10}}{\sin \displaystyle\frac{3 \pi}{10}} = 2 \sin \frac{\pi}{5} = \left( \frac{1}{2} + \frac{\sqrt{5}}{10} \right)^{-1 / 2}~. $$

We’re now going to construct a pentagon inscribed to the smaller circle. We observe that for the triangle $BAG$ $$ \overline{GA}^{\,2} = l^2 - r^2 = r^2 \cdot \frac{ \displaystyle\frac{1}{2} - \displaystyle\frac{\sqrt{5}}{10}}{ \displaystyle\frac{1}{2} + \displaystyle\frac{\sqrt{5}}{10}} = \frac{r^2}{2} \cdot (3 - \sqrt{5}) $$ and now it turns out that $$ \overline{GA} = \frac{r}{2} \cdot (\sqrt{5} - 1 \, ) $$ so that, with $H$ bisecting $\overline{AF}$, $$ \overline{GH} = \overline{GA} + \frac{1}{2} \cdot r = \frac{\sqrt{5}}{2}\cdot r~, $$ which, together with $\overline{BH}^{\,2} = \displaystyle\frac{5}{4} \cdot r^2$, finally gives $\overline{BH} =\overline{GH}$.