Determining conditions on two circles inscribed in square so that a tangent rectangle has integer area

Question

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?

Answer 1

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.

Answer 2

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 $$6\sqrt3(3\sqrt3+\sqrt{22})-(\sqrt3)^2-(3\sqrt3+\sqrt{22})^2=2\cdot1$$