Assume you have two circles with radius $n$, the radius between the centers of these two circles are $a$. Where $0<a<n$.

Now remove the overlapping part between the two circles. Now let $B$ be the biggest triangle you can inscribe in the circles.

What is the ratio between the area of the circle sector and the triangle?

I actually gave this problem a shot. Here is my attempt at dealing with the specific case when $n=3$, $a=1$.


This gives out a ratio about 2.6. Now, I do not know if my drawing is 100% correct. Now how to actually show this in a easy way? After what I see one way would be to exploit the symmetry, and only look at the right side.

One could then assume the triangle has the greatest area when it is equilateral. Putting u a equation for the circle, then start messing with the intersection point between the top and the circle. Do anyone know a better way to do this problem ?

Also I am not able to perform the steps above, It is a tad too advanced for me. Hopefully someone is willing to help me!

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    $\begingroup$ The last editor was not too careful with the deletions, which resulted in the picture being removed. Hurray for "Edit History"! $\endgroup$ – J. M. isn't a mathematician Dec 19 '11 at 0:23
  • $\begingroup$ As the union $S$ of the two disks is not convex you have to specify what is meant by "inscribe": Do you only require that the vertices of the triangle $\Delta$ lie on the boundary of of $S$ or do you insist on $\Delta\subset S$? $\endgroup$ – Christian Blatter Dec 19 '11 at 11:24
  • $\begingroup$ By inscribed I mean that the edges of the triangle does not cross either of the circles. And it seems a more efficent method is obtained by turning the triangle 90degrees. =) $\endgroup$ – N3buchadnezzar Dec 19 '11 at 12:10

enter image description hereWith n=3 a=1 it can be done with an area of about 13.23, by making the base of the triangle go through an intersection point between the circles and making the other intersection point the top of the triangle. The area is then $\frac{(\frac{a^2}{2}+2an)\sqrt{4a^2-(\frac{b^2}{2}+2an)^2}}{2}$.

If a/n is smaller, a larger triangle can be inscribed with its base through the centre of one circle and its top where the other circle intersects the line through the cenctres of the circles. This will have an area of $2n(a+n)$.enter image description here


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