You and two friends all have March birthdays, so you’ve decided to celebrate together with one big cake that has delicious frosting around its perimeter. To share the cake fairly, you want to ensure that (1) each of you gets the same amount of cake, by area, and (2) each of you gets the same amount of frosting along the cake’s edge. What’s more, you want to cut the cake by starting at a single point inside of it, and then making three straight cuts to the edge from that point.
As shown in the Fiddler email, you know how to make these cuts for circular or square cakes. However, the cake you bought is rectangular, with a length of 20 inches and a width of 10 inches. Using the coordinate system of your choice, describe a way this particular cake can be cut fairly, so that all three of you get the same amount in terms of both area and the cake’s perimeter. Again, there should be three straight cuts emanating from a single point inside the cake.
Let's put the lower left corner of our sheet cake at the origin of the $xy$-plane with opposite corner located at the point $(20,10)$, so that the long 20in side of the cake is parallel to the $x$-axis. Assume that one of the straight cuts ends up hitting the point $(t,0)$ for some $0\leq t\leq 10$. Then since each birthday person needs one third of the perimeter, the other two cuts must hit the perimeter of the sheet cake at $(20,t)$ and $(10-t,10)$.
If start cutting at the central point within the cake $P=P(x,y)$, then the upper right quadrilateral has area \begin{align*}A_1 &= \frac{1}{2} (10+t) (10 - y) + \frac{1}{2} (10 - t) (20 - x)\\& = 150 - 5t -\frac{1}{2} (10-t) x -\frac{1}{2} (10+t) y,\end{align*} while the lower right quadrilateral has area \begin{align*}A_2 &= \frac{1}{2} (20-t) y + \frac{1}{2} t (20 - x) \\ &= 10t -\frac{1}{2} t x +\frac{1}{2} (20-t) y.\end{align*} Imposing the condition that all three pieces of cake have the same area is mathematically equivalent to ensuring that $A_1 = A_2 = \frac{200}{3}$ which yields the following system of equations \begin{align*} (10-t) x + (10+t) y & = \frac{500-30t}{3} \\ -t x + (20-t) y & = \frac{400-60t}{3}.\end{align*} Solving for $x=x(t)$ and $y=y(t),$ we get $$P(t)=\begin{pmatrix} x(t)\\ y(t) \end{pmatrix} = \begin{pmatrix} 10-t & 10 + t \\ -t & 20-t \end{pmatrix}^{-1} \begin{pmatrix} \frac{500 -30t}{3} \\ \frac{400-60t}{3}\end{pmatrix} = \begin{pmatrix} \frac{15t^2-150t+1000}{t^2-10t +100} \\ \frac{15t^2-200t+250}{3(t^2-10t+100)} \end{pmatrix}.$$ Though the points won't exactly be correct, we can extend for $10 \leq t \leq 20$ by appealing to symmetry, from which we can define $x(t) = 20 - x(20-t)$ and $y(t)=y(20-t)$ whenever $10\leq t \leq 20.$
One solution (for $t=0$) has the center point $P(0)=(10, 20/3),$ with cuts hitting the perimeter at points $(0,0),$ $(20,0)$ and $(10,10).$
The parametric curve $P(t)=(x(t), y(t))$ encloses an area that can be approximated by the Riemann sum through horizontal slices for a given partition $t_0 = 0 \lt t_1 \lt \cdots \lt t_{N-1} \lt t_N = 10,$ as \begin{align*}A_N &= \sum_{i=1}^N \left(x(20 - t_{i-1}) - x(t_{i-1}) \right) \left(y(t_{i-1}) - y(t_i) \right) \\ &= \sum_{i=1}^N 2 \left(10 - x(t_{i-1}) \right) \frac{y(t_{i-1}) - y(t_i)}{t_i - t_{i-1}} \left( t_i - t_{i-1} \right).\end{align*} As $N \to \infty,$ we get that the area enclosed within the parametric curve of center points for the equitable area and perimeter cuts is given by \begin{align*}A &= \lim_{N \to \infty} A_N = -2 \int_0^{10} ( 10 - x(t) ) \frac{d}{dt} y(t) dt\\ &= \frac{1000}{3} \int_0^{10} \frac{(t^2 - 10t) (t^2 - 10t - 50)}{(t^2 - 10t + 100)^3} \,dt \\ &= \frac{1000}{3} \int_0^{10} \frac{t^4 - 20t^3 + 50t^2 + 500t}{(t^2 - 10t + 100)^3} \,dt \\ &= \frac{200}{27} \left[ \sqrt{3}\arctan \left( \frac{x-5}{5\sqrt{3}} \right) - \frac{30(x-5)^3}{(x^2 - 10x + 100)^2} \right]^{x=10}_{x=0} \\ &= \frac{400}{27} \left( \sqrt{3} \arctan\left(\frac{1}{\sqrt{3}}\right) - \frac{3}{8} \right) = \frac{50 \left( 4 \sqrt{3} \pi - 9 \right)}{81}\\ &\approx 7.879995... \,\, \text{in}^2.\end{align*}
