The perfect nonagonal pancake

After you intended to make a perfectly circular pancake, the batter has spread out, filling every last corner of your square pan. (It is unclear why you were using a square pan in the first place.)

To salvage your breakfast, you plan to slice off the corners of your square pancake, giving you something closer to a circle. The image below shows one such slice you might make. Each slice must be straight, and no slice can pass through the inscribed blue circle that represents your original desired pancake.

Of course, there’s a catch. You can make at most five slices. If the blue circle has a radius of 1 unit, what is the minimum possible area your pancake can have after five slices?

By symmetry we can break the problem into quandrants. Three quadrants will have symmetric cuts, the fourth quadrant will have two cuts in it. Let's first focus on the three symmetric quadrants:

If we make a cut that is tangent to the blue circle at the point $(t, \sqrt{1-t^2})$ then the cut would be given by the line $xt + y\sqrt{1-t^2}=1.$ This line would intersect the line $y=1$ at the point $(\frac{1-\sqrt{1-t^2}}{t}, 1)$ and the line $x=1$ at the point $(1, \frac{\sqrt{1-t^2}}{1+t}).$ The remaining pancake area is then given by $$A(t) = \frac{1-\sqrt{1-t^2}}{t} + \frac{1}{2} \left( 1 - \frac{1 - \sqrt{1-t^2}}{t}\right) \left( 1 + \frac{\sqrt{1-t^2}}{1+t} \right) = \frac{1+t -\sqrt{1-t^2}}{t(1+t)}.$$ Differentiating with respect to $t$ we get $$A^\prime(t) = \frac{1+t-t^2 -(1+t)\sqrt{1-t^2}}{t^2(1+t)\sqrt{1-t^2}}.$$ Setting to zero and solving for $t$, we get $t^* = \frac{1}{\sqrt{2}},$ with minimal area $A(1/\sqrt{2}) = 2(\sqrt{2}-1).$

For the fourth quadrant, where we make two cuts assume that these cuts are tangent to $(r,\sqrt{1-r^2})$ and $(s,\sqrt{1-s^2}),$ respectively, which would be associated with cuts $$xr + y\sqrt{1-r^2}=1 \,\, \text{and} \,\, xs + y\sqrt{1-s^2} = 1.$$ Without loss of generality, if $r \leq s$, then the line $xr+y\sqrt{1-r^2}=1$ intersects the line $y=1$ at $(\frac{1-\sqrt{1-r^2}}{r},1),$ the line $xs+y\sqrt{1-s^2}=1$ intersects the line $x = 1$ at $(1,\frac{\sqrt{1-s^2}}{1+s}),$ and the two cuts intersect at $(\frac{\sqrt{1-r^2} - \sqrt{1-s^2}}{s-r}, \frac{s\sqrt{1-r^2} + r\sqrt{1-s^2}}{r+s}).$ The area of the the remaining pancake region is then given by \begin{align*}B(r,s) &= \frac{1-\sqrt{1-r^2}}{r}\\ &\,\,+ \frac{1}{2} \left(1 + \frac{s\sqrt{1-r^2} + r\sqrt{1-s^2}}{s+r}\right) \left(\frac{\sqrt{1-r^2} - \sqrt{1-s^2}}{s-r} - \frac{1 - \sqrt{1-r^2}}{r}\right)\\ &\,\,\,\, + \frac{1}{2} \left(\frac{s\sqrt{1-r^2} + r\sqrt{1-s^2}}{s+r} + \frac{\sqrt{1-s^2}}{1+s}\right) \left(1 - \frac{\sqrt{1-r^2} - \sqrt{1-s^2}}{s-r}\right)\end{align*} With some additional algebra and differentiation, we obtain optimal values of $r^* = \frac{1}{2}$ and $s^* = \frac{\sqrt{3}}{2},$ which gives $B(\frac{1}{2}, \frac{\sqrt{3}}{2}) = 3(2 - \sqrt{3}).$

Putting these together we get a minmal pancake area of $$3A(1/\sqrt{2}) + B(1/2,\sqrt{3}/2) = 3(2(\sqrt{2}-1)) + 3(2 - \sqrt{3}) = 3(2\sqrt{2}- \sqrt{3}).$$