Determining conditions on two circles inscribed in square so that a tangent rectangle has integer area In the accompanying figure, all seven contact points are tangential. It can be verified that if the areas of the major and minor circles, noted $A$ and $a$, are respectively equal to $8\pi$ and $2\pi$ then the area of the blue rectangle is equal to $7$.

Can you find without much work necessary and sufficient conditions between the areas $A$ and $a$ so that the area of the blue rectangle is an integer? Or would it be too difficult?


 A: If the radii of the circles are $r$ and $R$, then the side of the big square is $s=(r+R)(1+\frac{\sqrt{2}}{2})$ and so the sides of the rectangle are $s-2r=R(1+\frac{\sqrt{2}}{2})+r(-1+\frac{\sqrt{2}}{2})$ and $s-2R=R(-1+\frac{\sqrt{2}}{2})+r(1+\frac{\sqrt{2}}{2})$, so (after the calculation) the area of the rectangle is:
$$-\frac{1}{2}R^2-\frac{1}{2}r^2+3Rr=\frac{-A-a+6\sqrt{Aa}}{2\pi}$$
Now, I presume you want $A=u\pi, a=v\pi$, so this area then becomes:
$$-\frac{u+v}{2}+3\sqrt{uv}$$
which gives you necessary and sufficient conditions: (a) $u,v$ are of the same parity; (b) $uv$ is a square.
A: After observing that the centers of the two circles are on the diagonal involved, everything comes out easy with the $45º$ angles in play. Thus we arrive at the relationship between the radii $R$ and $r$ which expresses the integer area $n$ of the blue rectangle.
$$\frac12(6rR-R^2-r^2)=n$$
Moving then to areas $A$ and $a$ of the problem we have the equation
$$6\sqrt{aA}-A-a=2n\pi$$ from this an easy partial solution is given by the mentioned conditions in Stinking Bishop's answer above.
However the solution is much more general. Expressing the areas as a function of the radii, the transcendent $\pi$ disappears and we return to the original expression, so we can do
$$R=3r\pm\sqrt{8r^2-2n}$$ This suggests that for each natural integer $n$ there exists an infinity of real pairs $(r, R)$ such that the corresponding blue rectangle has an area equal to $n$.
We end with an example for area $n=1$ (there is an infinity possible).
Example.-$(n,r,R)=(1,\sqrt3,\space 3\sqrt3+\sqrt{22})$.
In fact $$6\sqrt3(3\sqrt3+\sqrt{22})-(\sqrt3)^2-(3\sqrt3+\sqrt{22})^2=2\cdot1$$
