Hoops

Let $X = x^2+y^2, Y = x+y.$ If $\cos X = \sin Y,$ then there are two cases: either $\sin X = \cos Y$ or $\sin X = -\cos Y.$

In the first case, $\sin(X+Y) = \sin X \cos Y + \cos X \sin Y = \sin^2 X + \cos^2 X = 1,$ so $X+Y = \pi/2 + 2\pi k,$ $k$ an integer.

In the second case, $\sin(X-Y) = \sin X \cos Y - \cos X \sin Y = -\sin^2 X - \cos^2 X = -1,$ so $X-Y = 3\pi/2 +2\pi k,$ $k$ an integer.

The first case is of the form $x^2+y^2+x+y = \pi/2 + 2\pi k.$ These are concentric circles centered at $(-1/2,-1/2)$ with radius $\sqrt{\frac{\pi+1}2 + 2\pi k}$ , so $k \ge 0.$

The second case is of the form $x^2+y^2-x-y = 3\pi/2 + 2\pi k.$ These are concentric circles centered at $(1/2,1/2)$ with radius $\sqrt{\frac{3\pi+1}2 + 2\pi k}$ , so $k \ge 0.$

The smallest two circles are clearly when $k=0.$ The points of intersection are where $x_0^2+y_0^2+x_0+y_0 = \pi/2$ and $x_0^2+y_0^2-x_0-y_0 = 3\pi/2,$ so $x_0^2+y_0^2 = \pi$ and $x_0+y_0 = -\pi/2.$ The angle between the circles is the same as the angle between the radii. We can compute this by dot products. This is not too bad since we know the length of the radii, so we know the denominator of the formula $\cos \theta = \frac{{\bf x} \cdot {\bf y}}{\|x\| \|y\|}.$ $\cos \theta = \frac{((x_0,y_0) + (-1/2,-1/2)) \cdot ((x_0,y_0) + (1/2,1/2))}{\sqrt{\frac{\pi+1}2}\sqrt{\frac{3\pi +1}2}} = \frac{x_0^2+y_0^2 - 1/2}{\sqrt{\frac{\pi+1}2}\sqrt{\frac{3\pi +1}2}} = \frac{\pi - 1/2}{\sqrt{\frac{3\pi^2+4\pi + 1}4}} = \frac{2\pi-1}{\sqrt{3\pi^2+4\pi+1}}.$

So the answer is $2 \times 3 \times 4 = \fbox{24}.$