Sliding Square

Labeling the final position for $A$ as $(a,b)$ , $a$ is the solution of the equation

$18\pi - 72*\sin^{-1}(\frac{\sqrt{2}(a + 2)}{12}) - (a + 2)*(\sqrt{72 - (a + 2)^{2}} - (a + 2)) = 8$ ,

which has solution $a = 1.897774......$ . Now since $b = a + 1$ , we have

$a + b = 2a + 1 = \boxed{4.796}$ to $3$ decimal places.

I will provide more details in the morning, but I just wanted to get something posted now so that others can make general comments and queries.

Note that the desired fractional coverage is $\frac{1}{2}$ , (I read it initially as $\frac{1}{3}$ , leading to an incorrect first attempt), and that the radius of the circle is $6\sqrt{2}$ .

EDIT: O.k., now for some details.

The circle has equation $(x + 2)^{2} + (y + 1)^{2} = 72$ , and $OE$ lies on the line $y = x + 1$ . So, as noted above, once we find $(a,b)$ the desired solution will be $2a + 1$ .

With the square in its final position, we will have $A$ at $(a,b)$ and $B$ at $(a,Y)$ where $Y$ is the point of intersection of the circle and the line $x = a$ . Plugging $x = a$ into the circle equation yields

$(a + 2)^{2} + (Y + 1)^{2} = 72 \Longrightarrow Y = \sqrt{72 - (a + 2)^{2}} - 1$ ,

as we are looking for $Y$ in the first quadrant. Now the area of the portion of the square lying within the circle will be the area of sector $OBD$ minus the combined areas of triangles $\Delta OBA$ and $\Delta ODA$ . Now by symmetry these two triangles have the same area.

Now let $\angle BOA = \theta$ and $\angle OBA = \alpha$ . We have that $\angle OAB = \frac{3\pi}{4}$ , and that $OA$ has length

$\sqrt{(a + 2)^{2} + (b + 1)^{2}} = \sqrt{(a + 2)^{2} + (a + 2)^{2}} = \sqrt{2} * (a + 2)$ .

So using the Sine Law on $\Delta OBA$ , we see that

$\dfrac{\sin(\frac{3\pi}{4})}{6\sqrt{2}} = \dfrac{\sin(\alpha)}{\sqrt{2}*(a + 2)} \Longrightarrow \alpha = \sin^{-1}(\frac{\sqrt{2}*(a + 2)}{12})$ .

Now $\angle BOD = 2\theta = 2(\frac{\pi}{4} - \alpha) = \frac{\pi}{2} - 2\alpha$ , and so the area of sector $OBD$ is

$(\frac{1}{2})*(6\sqrt{2})^{2}*(2\theta) = 18\pi - 72*\sin^{-1}(\frac{\sqrt{2}*(a + 2)}{12})$ .

The area of $\Delta OBA$ is

$(\frac{1}{2})*(a - (-2))*(Y - b) = (\frac{1}{2})*(a + 2)*(\sqrt{72 - (a + 2)^{2}} - 1 - (a + 1)))$ ,

and thus the combined area of $\Delta OBA$ and $\Delta ODA$ is

$(a + 2)*(\sqrt{72 - (a + 2)^{2}} - (a + 2))$ .

Given that we need to find $a$ such that half the box lies inside the circle, we are thus led to the equation presented at the beginning of this post, which is then solved using numerical methods.

Note that this problem can also be solved using calculus, (and probably more easily that way), but since Guiseppi tagged this as a geometry problem I chose to take the above approach.