# Elementary geometry problem

A square field has side length $100$ metres. One horse is tethered to the midpoint of the bottom side of the field. Another is tethered to the top right corner of the field. A third horse is tethered to the midpoint of the left side of the field. Each horse can access an equal area of the field and none of the areas overlap.

What length of rope must each horse have to minimise the area that no horse can reach?

My progress:

Let $L$ be the length of the rope for the horses at the bottom and on the left. These horses are both able to access semicircular regions, each with area $\dfrac{1}{2}\pi L^2$. The horse at the top right is able to access a quarter circular region with area $\dfrac{1}{2}\pi L^2$, and therefore the length of rope this horse has is $L\sqrt{2}$.

For none of the areas to overlap, I think we will need the two semicircles to touch, but I can't see how to use this to find a suitable function to minimise. A hint would be greatly appreciated.

EDIT: This is an image. The areas of the yellow, blue and red regions are supposed to be the same. • Could you add an image to help visualization? My imaginative powers are flawed. – Mohammad Zuhair Khan Jan 5 '18 at 15:07
• The left and bottom circles touch first. – Paul Jan 5 '18 at 15:13
• Join the midpoint of the bottom side to the midpoint of the left side - these are the centres of two of the circles and the line passes through the point of tangency (the radius of each circle to the point of tangency is perpendicular to the common tangent). The joining line is twice the radius and half the diagonal of the whole square. – Mark Bennet Jan 5 '18 at 15:52

Let $O$ be the left bottom point of the square. The point of tangency $T$ must then on the line $y=x$ by symmetry.
The line from the midpoint of the blue circle through $T$ is $y=-x+50$, therefore intersecting $y=x$ with $y=-x+50$ gives $T=(25,25)$.
Hence, the asked radius of the circle is $\sqrt{(25-50)^2+(25-0)^2}=\color{red}{25\sqrt{2}}$.
• Thanks! I think there's a typo at the end - should be $25\sqrt{2}$ – A. Goodier Jan 5 '18 at 15:32
• FWIW, if we remove the restriction of equal areas for all 3 horses, and make the red circle tangent to the blue and yellow circles, then its radius is $50\sqrt5 - 25\sqrt2$ – PM 2Ring Jan 5 '18 at 15:44