# Area of intersection of a square and an equilateral triangle inscribed in a circle

A square and an equilateral triangle are both inscribed in a circle. Let $$A$$ be the area of intersection of these two inscribed polygons. How much greater is the maximum value of $$A$$ than its minimum value?

$$%$$

Note: I'm not really sure what the best method is to tackle this problem. Working out the area of intersection is not too difficult for certain positions of the square relative to the triangle. However, figuring out the maximum and minimum values of $$A$$ is another story. Am I missing something very obvious here?

• Have you tried drawing a picture? Commented Aug 24, 2022 at 20:03
• Draw the square aligned with the x and y axis, with the circle centered at origin. Then parametrize the positions of the triangle vertices based on the angle that the first vertex makes with $x$ axis Commented Aug 24, 2022 at 20:07
• I have indeed. In fact I've experimented different positions of the square relative to the triangle.
– JCr
Commented Aug 24, 2022 at 20:07
• I think we'll get just 1 such case possible and so Min(A)=Max(A). Commented Aug 24, 2022 at 20:25
• Could you show me how to prove it?
– JCr
Commented Aug 24, 2022 at 20:29

Let $$AB=R$$, and $$\angle BAG=\alpha$$. You can use $$R=1$$ in your calculations, then if you want, multiply all lengths by $$R$$, and all areas by $$R^2$$. Write positions for all the points in the figure, and the positions of the intersections of the triangle with the square. You can now either calculate the areas of the triangle outside the square (three small triangles) or the area of the square outside the equilateral triangle (two small triangles and a right angle trapezoid). This will yield the area of the intersection as a function of $$\alpha$$. It should be a simple function, linear in $$\sin\alpha$$ and $$\cos\alpha$$. If it's not a constant, calculate the derivative and set it to $$0$$.
Note: due to symmetry of the problem, you can reduce the domain of $$\alpha$$ in the interval $$[0,\pi/12]$$