Leading the logarithmic pack

You’re doing a $30$-minute workout on your stationary bike. There’s a live leaderboard that tracks your progress, along with the progress of everyone else who is currently riding, measured in units of energy called kilojoules. Once someone completes their ride, they are removed from the leaderboard.

Suppose many riders are doing the $30$-minute workout right now, and that they all begin at random times. Further suppose that they are burning kilojoules at different constant rates (i.e., everyone is riding at constant power) that are uniformly distributed between $0$ and $200$ Watts.

Halfway through (i.e., $15$ minutes into) your workout, you notice that you’re exactly halfway up the leaderboard. How far up the leaderboard can you expect to be as you’re finishing your workout?

Let's start by determining the distribution of the random other riders' outputs at any one time. At any particular time, say $\tau$, the only riders still on the leaderboard would have started at times $t \in (\tau - 1800, \tau).$ Let's assume that riders join the $30$ minute class uniformly randomly such that the probability that any join between the time $t$ and $t+dt$ is proportional to $dt,$ that is, at time \tau, the riders would have already completed $T \sim U(0,1800)$ seconds of the rider. These riders would also have uniformly distribution constant powers, $P \sim U(0,200).$ The total output is $O = P \cdot T,$ which we can get the distribution of by directly computing the \begin{align*}\mathbb{P} \{ O \leq \theta \} &= \frac{1}{360000}\int_0^{200} \int_0^{1800} \chi \{ p t \leq \theta \} \,dt \,dp \\ &= \frac{1}{360000} \int_0^{200} \int_0^{\min \{ 1800, \theta / p \}} \,dt \,dp \\ &= \frac{1}{360000} \int_0^{200} \min \left\{1800, \frac{\theta}{p} \right\} \,dp \\ &= \frac{1}{360000} \left( \int_0^{\theta/1800} 1800 \,dp + \int_{\theta / 1800}^200 \frac{\theta}{p} \,dp \right) \\ &= \frac{1}{360000} \left( \theta + \theta \left( \ln 200 - \ln (\theta / 1800) \right) \right) \\ &= \frac{\theta}{360000} \left( 1- \ln \left( \frac{\theta}{360000} \right) \right).\end{align*}

So let's simplify slightly and focus on the function $\Phi(t) = t ( 1 - \ln t ).$ So in particular, if at some point I find myself half way up the leaderboard, then that would mean that my output $\tilde{O} = 360000 \tilde{\theta}$ where $\Phi(\tilde{\theta}) = \frac{1}{2}.$ Of course, $\Phi$ does not have a neat and tidy inverse function, so we would have to implicitly solve for $\tilde{\theta} = \Phi^{-1}(0.5),$ but more on this later.

So since the distribution of random riders is time invariant, if halfway through my ride I have output $\tilde{O} = 360000 \Phi^{-1}(0.5),$ then I can expect my output at end of my ride I have output $2\tilde{O}.$ In this case, the proportion of riders that I will be ahead of the leaderboard is \begin{align*}\Phi(2 \Phi^{-1}(0.5)) &= 2\Phi^{-1}(0.5) \left( 1- \ln (2\Phi^{-1}(0.5))\right) \\ &= 2 \Phi^{-1}(0.5) \left( 1- \ln 2 - \ln \Phi^{-1}(0.5) \right) \\ &= -2 \Phi^{-1}(0.5) \ln 2 + 2 \Phi^{-1} (0.5) \left( 1 - \ln \Phi^{-1}(0.5) \right) \\ &= -2 \Phi^{-1}(0.5) \ln 2 + 2 \Phi \left( \Phi^{-1}(0.5) \right) \\ &= 1 - 2 \Phi^{-1}(0.5) \ln 2 = 1 - 2 \tilde{\theta} \ln 2 .\end{align*} So, since we can analytically solve the inverse function to find that $\tilde{\theta} = \Phi^{-1}(0.5) = 0.1866823,$ I can expect to be ahead of about $$1-2 \tilde{\theta} \ln 2 \approx 74.1203380189...\%$$ of the riders at the end of my ride.

As an added bonus problem (though not quite Extra Credit), what’s the highest up the leaderboard you could expect to be $15$ minutes into your workout?

If I am killing it at 200 Watts for the first 15 minutes, then I would have an output of $\hat{O} = 200 \cdot 900 = 180000$ kJ, which would put me ahead of about $$\Phi\left(\frac{\hat{O}}{360000}\right) = \Phi(0.5) = 0.5 (1 + \ln 2) \approx 84.657359028...\%$$ of the riders after $15$ minutes.

The arc of high jumping bends towards ... 2?

In the high jump, an athlete’s entire body must clear the bar. However, not every part of their body has to clear the bar at the same time. As a result, athletes arc their bodies over the bar, so that only a fraction of their mass is above the bar at any given time. In fact, athletes can theoretically clear the bar despite their center of mass remaining below the bar throughout the jump.

Let’s model the athlete’s jump as an arc of angle $2\phi$ that is centered over the bar, as shown below. For simplicity, assume that their mass is uniformly distributed across the length of their body. The dot in the diagram represents the athlete’s center of mass. Let $a$ represent the vertical distance between the athlete’s center of mass and their lowest points (presumably their outstretched fingers and toes), and let $b$ represent the vertical distance between the athlete’s center of mass and their highest point (presumably their waist).

If $\phi = \frac{\pi}{2}$ then the arc would be a complete semicircle. In that case, what is the ratio $a/b$?

As the angle $\phi$ gets very, very small (i.e., in the limit as $\phi$ goes to zero), what value does the ratio $a/b$ approach?

Let's solve the generic case for $\phi$ so that we can then do the two cases, both when $\phi = \frac{\pi}{2}$ and when $\phi \downarrow 0.$ Since we'll be taking ratios of everything anyway, let's just go ahead and assume that the circle on which the high jumper's arc is lying in the figure has radius $1$ and is centered at the origin. Then, with some trigonometric know-how, we see that the lowest point of the arc will be sitting at a height of $\cos \phi.$ So we must have $a + b = 1 - \cos \phi.$ Additionally we see that the jumper's body has a length of $2\phi,$ the length of this arc. So to find $a,$ we need only take the average of the jumper's height along this arc by taking the integral $$a + \cos \phi = \frac{1}{2\phi} \int_{\frac{\pi}{2} - \phi}^{\frac{\pi}{2} + \phi} \sin t \,dt = \frac{1}{2\phi} \left( -\cos (\frac{\pi}{2} + \phi) + \cos (\frac{\pi}{2} - \phi) \right) = \frac{\cos (\frac{\pi}{2} - \phi) }{\phi} = \frac{\sin \phi}{\phi}.$$ That is, $$a = \frac{\sin \phi}{\phi} - \cos \phi$$ and $$b = 1 - \cos \phi - a = 1 - \frac{\sin \phi}{\phi}.$$ Therefore for any value of \phi, the ratio of $a / b$ is $$r(\phi) = \frac{\frac{\sin \phi}{\phi} - \cos \phi}{1 - \frac{\sin \phi}{\phi}} = \frac{ \sin \phi - \phi \cos \phi }{\phi - \sin \phi}.$$

So in particular, if $\phi = \frac{\pi}{2}$ and the jumper's body makes a complete semicircle then the ratio $a/b$ is $$r(\frac{\pi}{2}) = \frac{\sin \frac{\pi}{2} - \frac{\pi}{2} \cos \frac{\pi}{2} }{\frac{\pi}{2} - \sin \frac{\pi}{2}} = \frac{1}{\frac{\pi}{2} - 1} = \frac{2}{\pi - 2} \approx 1.75193839388....$$

On the other hand, if we use some rough Taylor approximations around $\phi \approx 0$ we see that we have $$r(\phi) = \frac{\left(\phi - \frac{\phi^3}{6} + O(\phi^5)\right) - \phi \left( 1 - \frac{\phi^2}{2} + O(\phi^4)\right)}{\phi - \left(\phi - \frac{\phi^3}{6} + O(\phi^5)\right)} = \frac{ \frac{\phi^3}{3} + O(\phi^5) }{\frac{\phi^3}{6} + O(\phi^2)} = 2 \frac{1 + O(\phi^2)}{1 + O(\phi^2)}.$$ So, as $\phi \downarrow 0,$ we have $r(\phi) \to 2.$

Tour de Fiddler fun

This time around, a lone rider in the Tour de Fiddler is being pursued by a group of four riders. The four riders have an advantage—they take equal turns being in the lead position, while the other three riders draft behind. At any given speed, being in the lead position (as well as riding solo) requires twice as much power as drafting.

Assume that every rider must maintain the exact same average power over time, whether they are the lone rider or in the pack of four. To be clear, their power can change over time, but the time-averaged value must be the same for every rider. Also, when leading a pursuing group or riding solo, one’s speed is directly proportional to one’s power. When drafting, one's speed matches that of the leader (again, at half the power output).

The pursuers just passed under a banner indicating there are 10 kilometers left in the stage. How far back of the lone rider can they afford to be, such that they still catch them at the finish line?

Let's first start with some preliminaries. Assume that all riders have average power output $\bar{P}$, such that, since speed is directly proportional to ones power, then the average speed of a single rider who has no one to draft off of is $v_1 = C\bar{P},$ for some constant of proportionality $C \gt 0.$

Let's now analyze what would happen if $k$ riders work together, taking equal turns being in the lead position. Since the average power must be $\bar{P},$ but any of the riders in this group spends only $\frac{1}{k}$ amount of time leading and the rest drafting at half-power, the amount of power spent while leading a pack of $k$ riders, say $P_k$, can be found in the equation $$\bar{P} = \frac{1}{k} P_k + \frac{k-1}{k} \frac{1}{2} P_k = \frac{k+1}{2k} P_k,$$ that is, $$P_k = \frac{2k}{k+1} \bar{P}.$$ So that means that the average velocity of a group of $k$ riders working together must be $$v_k = C P_k = C \frac{2k}{k+1} \hat{P} = \frac{2k}{k+1} v_1.$$

Ok, so with that in mind, the pursuers will finish in $T_4 = \frac{10}{v_4} = \frac{25}{4v_1},$ while if the leader has a lead of $x$ kilometers, then he will finish in $T_1 = \frac{10-x}{v_1}.$ Since we want $$T_4 = \frac{25}{4v_1} \leq \frac{10-x}{v_1} = T_1$$ in order for the pursuers to catch the leader, then the leader can be no more than $x \leq \frac{15}{4} = 3.75$ kilometers ahead of the pursuers if they want to catch him.

In today’s stage of the Tour de Fiddler, there are 176 total riders. Some riders are grouped together in a single breakaway, while the remainder are grouped together in the peloton. The breakaway group is 10 kilometers from the finish, while the peloton is one kilometer behind (i.e., 11 kilometers from the finish). What is the smallest number of riders that the breakaway needs to reach the finish line before the pursuing peloton?

Let's say there are $B$ riders in the breakaway group and $P$ riders in the peleton, where here $B+P = 176.$ The average speed of the peleton will be $$v_P = C\frac{2P}{P+1} v_1,$$ while the average speed of the breakaway is $$v_B = C\frac{2B}{B+1} v_1,$$ so the finishing times are $$T_P = \frac{11}{v_P} = \frac{11(P+1)}{2Pv_1}$$ and $$T_B = \frac{10}{v_B} = \frac{10(B+1)}{2Bv_1},$$ respectively. Since we want the breakaway group to win the stage, we must have $T_B \leq T_P,$ or equivalently $$\frac{10B+10}{B} \leq \frac{11P+11}{P},$$ which is again equivalent to $$BP+11B-10P \geq 0.$$ Since we have restriction that $B + P = 176,$ we are now looking for the minimum integer $B$ such that $$B(176-B) + 11B -10(176-B) = -B^2 + 197B - 1760 \geq 0.$$ Since the quadratic inequality holds for all (real) values of $$ B \in \left[ \frac{197 -\sqrt{31769}}{2} , \frac{197 + \sqrt{31769}}{2} \right],$$ we see that the breakaway must have at least $$B_\min = \left\lceil \frac{197 - \sqrt{31769}}{2} \right\rceil = \lceil 9.3806979381.... \rceil = 10$$ riders in order to beat the peleton.

If this is what the job entails, then I'm not sure I want it ....

Starting with a line segment of length $1$, randomly split it somewhere along its length into two parts. Compute the product of these two lengths. Then take each of the two resulting segments and repeat the process. That is, for each one, randomly split it somewhere along its length into two parts and compute the product. Then do this for all four resulting segments, then the eight after that, and the $16$ after that, and so on.

After doing this (forever), you add up all the products you computed throughout. On average, what value would you expect this sum to approach?

There is nothing magical about the initial value of the line segment, that is, $1$, so let's instead generalize and say that the function $f : \mathbb{R}_+ \to \mathbb{R}_+$ gives the expected value of the sum of the products of randomly splitting a line segment of initial length $\ell \in \mathbb{R}_+$ into two parts, then breaking each of those segments into two parts, etc. Then to restate the question, we want the value of $f(1).$

We can approximate $f$ by not allowing for infinite recursion, but instead just a finite number of breaking into smaller pieces. If we just have one splitting then we get the expected value of the product is $$f_1(\ell) = \int_0^\ell x (\ell-x) \, dx = \left[ \ell \frac{x^2}{2} - \frac{x^3}{3} \right]^\ell_0 = \frac{\ell^3}{6}.$$ Going one step further, we then see that after two splits, we add in the expected product of two uniformly random numbers that add up to $x$ and the expected product of two uniformly random numbers that add up to $\ell - x.$ That is, we get $$f_2(\ell) = \int_0^\ell x(\ell - x) + f_1(x) + f_1(\ell-x) \,dx = f_1(\ell) + 2 \int_0^\ell f_1(x) \,dx.$$ Similarly, if we keep going with more recursion, we get that the expected sum of the products after $n+1$ rounds of splitting is given by $$f_{n+1} (\ell) = f_1 (\ell) + 2\int_0^\ell f_n(x) \,dx,$$ for $n \geq 1.$ As $n \to \infty,$ we get that we must have $$f(\ell) = f_1(\ell) + 2\int_0^\ell f(x)\,dx = \frac{\ell^3}{6} + \int_0^\ell f(x) \,dx.$$ Differentiating both sides with respect to $\ell,$ we get the differential equation $$f^\prime (\ell) = \frac{t^2}{2} + 2f(\ell),$$ with initial conditions $f(0) = 0.$ This is solved by the generic form $f(\ell) = Ce^{2\ell} + p(\ell),$ where $p$ is a polynomial, say $p(t) = at^2 + bt+ c.$ Plugging back into the differential equation we get that $$2at + b = \frac{t^2}{2} + 2at^2 + 2bt + 2c,$$ for all $t.$ By gathering like terms and setting equal to zero, we get \begin{align*}2a + \frac{1}{2} &= 0\\ 2b - 2a &= 0 \\ 2c - b &= 0, \end{align*} that is $a = b = -\frac{1}{4}$ and $c = -\frac{1}{8}.$ Additionally, we have $f(0) = C - \frac{1}{8} = 0,$ so $C = \frac{1}{8}.$ So finally, we see that $$f(\ell) = \frac{e^{2\ell} - 2\ell^2 -2\ell -1}{8},$$ and so the expected value of the sum of the products of random splitting into two subparts is $$f(1) = \frac{e^2 - 5}{8} \approx 0.29863201236.....$$

But wait ... there's more, what if as opposed to two pieces, each time we randomly split, we split into three pieces. Let $g: \mathbb{R}_+ \to \mathbb{R}_+$ be the expected value of the sum of the random triple products of splitting a line segement of initial length $\ell \in \mathbb{R}_+.$ Here too we can demonstrate the recurrence with a few small number of splits.

We see that after one splitting we get $$g_1(\ell) = \int_0^\ell \int_0^{\ell - x} xy(\ell - x -y) \,dy \,dx = \int_0^\ell x f_1(\ell - x) \,dx = \int_0^\ell x \frac{(\ell - x)^3}{6} \,dx = \frac{\ell^5}{120}.$$ If we again split each piece into three then after two splittings we have the original value of original splitting, followed by the expected product of three uniformly random numbers that add up to $x$, the expected product of three random numbers that add up to $y$, and the expected product of three random numbers that add up to $\ell - x - y.$ That is, \begin{align*}g_2(\ell) &= \int_0^\ell \int_0^{\ell - x} xy(\ell - x - y) + g_1(x) + g_1(y) + g_1(\ell - x - y) \,dy\,dx\\ &= g_1(\ell) + 3 \int_0^\ell (\ell - x) g_1(x) \,dx.\end{align*} In fact, this relation holds for any arbitrary number of splits, $n \geq 1,$ that is $$g_{n+1}(\ell) = g_1(\ell) + 3 \int_0^\ell (\ell-x) g_n(x) \,dx$$ and in particular as $n \to \infty$ we get $$g(\ell) = \frac{\ell^5}{120} + 3\int_0^\ell (\ell - x) g(x) \, dx.$$

Let's define the integral of $g$ to be $G(\ell) = \int_0^\ell g(x)\,dx.$ Differentiating under the integral sign with respect to $\ell,$ we get $$g^\prime(\ell) = \frac{\ell^4}{24} + 3\int_0^\ell g(x) \,dx = \frac{\ell^4}{24} + 3G(\ell).$$ Differentiating once more gives the second order inhomogeneous differential equation $$g^{\prime\prime}(\ell) - 3g(\ell) = \frac{\ell^3}{6},$$ where $g(0) = g^\prime(0) = 0.$ This is solved by a function of the form $$g(\ell) = C \sinh (\sqrt{3} \ell) + p(\ell),$$ where $p$ is a polynoial, say $p(t) = at^3 + bt^2 + ct + d.$ Plugging back into the differential equation we get $$6at + 2b - 3(at^3 + bt^2 + ct+d) = \frac{t^3}{6},$$ for all $t.$ By gathering terms and setting equal to zero, we get \begin{align*} -3a &= \frac{1}{6} \\ b &= 0 \\ 6a-3c &= 0 \\ 2b -3d &= 0, \end{align*} that is $a = -\frac{1}{18}, $$b = d = 0,$ and $c = -\frac{1}{9}.$ Additionally, we have $g^\prime(0) = \sqrt{3} C - \frac{1}{9} = 0,$ so $C = \frac{\sqrt{3}}{27}.$ So finally, we see that $$g(\ell) = \frac{2\sqrt{3} \sinh (\sqrt{3} \ell) - 3\ell^3 - 6\ell}{54},$$ and so the expected value of the sum of the triple products of random splittings into three subparts is $$g(1) = \frac{2\sqrt{3} \sinh(\sqrt{3}) - 9}{54} \approx 0.00895406261852....$$

The Great Ellipse Eclipse

[S]uppose you have two congruent ellipses that, together, cover a unit circle (i.e., a circle with radius $1$). These ellipses can have any eccentricity you like, but they must be congruent to each other. What is the smallest possible area one of these ellipses can have, such that they completely cover the circle?

By symmetry (both rotational and optimization-al?), let's assume that we have two congruent ellipses that have parallel major axis along the $y$-axis and are centered at points $(c,0)$ and $(-c,0)$ for some $c > 0.$ That is, let the equations for the ellipses be \begin{align*} \frac{(x-c)^2}{a^2} + \frac{y^2}{b^2} &= 1 \\ \frac{(x+c)^2}{a^2} + \frac{y^2}{b^2} &= 1.\end{align*}

In order to ensure that these two ellipses cover the entire unit circle we would need to make sure that they cover the extreme points, $(\pm 1, 0)$ and $(0, \pm 1)$ (and for good measure that the origin is contained at least one of them). In order for the origin to be contained we can plug in $x=y=0$ and get $$\frac{c^2}{(1-c)^2} \leq 1,$$ or equivalently $2c - 1 \leq 0,$ or $c \leq \frac{1}{2}.$ In order for $(1,0)$ to be contained in the ellipse centered at $(c,0)$ we must have $$\frac{(1-c)^2}{a^2} + \frac{0}{b^2} = 1,$$ or equivalently $a = 1 - c.$ In order for (0,1) to be contained in either ellipse we must have $$\frac{c^2}{(1-c)^2} + \frac{1}{b^2} = 1$$ or equivalently $$b^2 = \frac{1}{1 - \frac{c^2}{(1-c)^2}} = \frac{(1-c)^2}{(1-c)^2 - c^2} = \frac{(1-c)^2}{1-2c},$$ that is, $b = \frac{1-c}{\sqrt{1-2c}}.$

The area of one of the ellipses is $$A(c) = \pi a(c)b(c) = \pi \frac{(1-c)^2}{\sqrt{1-2c}}.$$ Differentiating we get \begin{align*}\frac{\partial}{\partial c} A(c) &= \pi \left(\frac{(1-c)^2}{(1-2c)^{3/2}} + \frac{2(c-1)}{\sqrt{1-2c}} \right)\\ &= \pi\frac{ (1-c)^2 - 2 (1-2c) (1-c) }{(1-2c)^{3/2}}\\ &= \frac{\pi(1-c)(3c-1)}{(1-2c)^{3/2}}.\end{align*} So the unconstrained critical points of $A$ are $c = 1$ and $c = \frac{1}{3}.$ Since we want $c \leq \frac{1}{2},$ we can eliminate the critical point $c = 1,$ and instead need to just test the critical points $c \in \left\{ 0, \frac{1}{3}, \frac{1}{2} \right\}.$ We have $A(0) = \pi,$ and $\lim_{c \uparrow \frac{1}{2}} A(c) = +\infty$ and $$A\left(\frac{1}{3}\right) = \pi \frac{ \frac{4}{9} }{ \sqrt{\frac{1}{3} }} = \frac{4\pi \sqrt{3}}{9} \approx 2.418399....,$$ which is the smallest area of one ellipse of a congruent pair that can cover the unit circle. For completeness, those ellipses are \begin{align*} \frac{ (3x - 1)^2 }{4} + \frac{3y^2}{4} &= 1 \\ \frac{(3x+1)^2}{4} + \frac{3y^2}{4} &= 1\end{align*}

Let them eat equitable cake slices

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*}

I come to bury the Niners, not to praise them

Football is complicated, so let’s assume a simplified scoring model. Every time your team is on offense, suppose there’s a 1-in-3 chance they score a touchdown (which we’ll say is worth a total of 7 points, as we won’t bother with 2-point conversions here), a 1-in-3 chance they score a field goal (worth 3 points), and a 1-in-3 chance they don’t score any points (i.e., they punt or turn the ball over on downs). After any of these three things happens, your team will then be on defense and the other team will have the same scoring probabilities.

Now, here’s how overtime will work: Your team is on offense first. No matter how many points your team does or does not score, the other team then gets a chance at offense. If the game is still tied beyond this point, the teams will continue alternating between offense and defense. Whichever team scores next wins immediately. Again, your team is on offense first. What is your team’s probability of winning?

In this first model, each team has a probability $p_0$ of scoring no points, $p_3$ of scoring 3 points, and $p_7$ of scoring 7 points, with $p_0 + p_3 + p_7 = 1.$ Though, I suppose, you don't hope to get there, let's first analyze what happens if you end up getting to sudden death.

The probability of you winning in sudden death in your first possession of sudden death is $p_3 + p_7 = 1-p_0.$ Of course the only way that you could get to a second possession in sudden death is to first score nothing (with probability $p_0$) and have you opponent also score nothing (with probability $p_0$), and then to score with probability $1-p_0$. That is, the probability of scoring in sudden death on your second possession is $p_0^2 (1-p_0).$ Similarly, we can reason that the probability of winning on your $k$th possession in sudden death must by $(k-1)$ rounds of both teams not scoring, then you scoring in the $k$th possession, for a probability of $p_0^{2(k-1)}(1-p_0).$ Adding this all up, the probability of you winning in sudden death conditional on making it to sudden death is $$p_{SD} = \mathbb{P} \{ W \mid SD \} = \sum_{k=1}^\infty p_0^{2(k-1)} (1-p_0) = \frac{1-p_0}{1-p_0^2} = \frac{1}{1+p_0}.$$

If your first drive ends in scoring $7$, then there are two outcomes based on your opponent's outcomes, either (a) your opponent scores less than $7$, with probability $p_0 + p_3$, and you win; or (b) you enter sudden death, with probability $p_7$. So the probability of winning conditional on scoring a TD in your first drive is $$\Pi_7 = \mathbb{P} \{ W \mid 7 \} = p_0 + p_3 + p_7 p_{SD} = p_0 + p_3 + \frac{p_7}{1+p_0}.$$

If your first drive ends in scoring $3$, then there are three outcomes based on your opponent's outcomes, either (a) your opponent scores zero, with probability $p_0$, and you win; or (b) you enter sudden death, with probability $p_3$; or (c) your opponent scores 7, with probability $p_7,$ and you lose. So the probability of you winning conditional on scoring a FG in your first drive is $$\Pi_3 = \mathbb{P} \{ W \mid 3 \} = p_0 + p_3 p_{SD} = p_0 + \frac{p_3}{1 + p_0}.$$

If you don't score in your first drive, then you either (a) lose, with probability $p_3 + p_7$; or (b) enter into sudden death, with probability $p_0$. So the probability of you winning conditional on not scoring in your first drive is $$\Pi_0 = \mathbb{P} \{ W \mid 0 \} = p_0 p_{SD} = \frac{p_0}{1+p_0}.$$

Putting these all together, you have a total probability of winning if you go first of $$\Pi (p_0, p_3, p_7) = p_0 \Pi_0 + p_3 \Pi_3 + p_7 \Pi_7 = p_0p_3 + p_0p_7 + p_3p_7 + \frac{p_0^2 + p_3^2 + p_7^2}{1 + p_0}.$$ In this case, if, as in our toy model, we have $p_0 = p_3 = p_7 = \frac{1}{3}$, then the probability of winning if you go first is $$\Pi = \Pi(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) = \frac{1}{3^2} + \frac{1}{3^2} + \frac{1}{3^2} + \frac{\frac{1}{3^2} + \frac{1}{3^2} + \frac{1}{3^2}}{1 + \frac{1}{3}} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12}.$$

Let's add this complexity to the model: When either team is on offense, they now have a choice. They can still opt for a strategy that results in 7 points, 3 points, or 0 points, each with a 1-in-3 chance. Alternatively, they can opt for a more aggressive strategy that results in 7 points or 0 points, each with a 1-in-2 chance.

Your team remains on offense first. Assuming both teams play to maximize their own chances of Super Bowl victory, now what is your team’s probability of winning?

So let's introduce $\tilde{p}_7,$ which is the probability of a TD under the aggressive offensive strategy, and $\tilde{p}_0 = 1 - \tilde{p}_7,$ which is the probability of not scoring under the aggressive offensive strategy. Let's first confirm that the probability of winning conditional on getting to sudden death in the more complicated model, $\tilde{p}_{SD},$ remains the same as before ( that is $\tilde{p}_{SD} = p_{SD})$), if and only if $p_3 + p_7 \geq \tilde{p}_7.$ If the parameters are such that rather $\tilde{p}_7 \gt p_3 + p_7,$ then we have $\tilde{p}_{SD} = \frac{1}{1+\tilde{p}_0},$ that is, $$\tilde{p}_{SD} = \frac{1}{1 + \min \{ p_0, \tilde{p}_0 \} }.$$

So let's now go through the various cases. If you score a TD in your first drive, then your opponent will win with probability $\tilde{p}_7 ( 1 - \tilde{p}_{SD} )$ and lose with probability $\tilde{p}_0 + \tilde{p}_7\tilde{p}_{SD}$ when he pursues the aggressive strategy. If he goes with the standard strategy, he will win with probability $p_7 (1 - \tilde{p}_{SD})$ and lose with probability $p_0 + p_3 + p_7 \tilde{p}_{SD}.$ So since your opponent will chose his own maximal winning probability that is equivalent to your minimal winning probability, so your probability of winning if you score a TD in your first drive if $$\tilde{\Pi}_7 = \min \{ \tilde{p}_0 + \tilde{p}_7 \tilde{p}_{SD}, p_0 + p_3 + p_7 \tilde{p}_{SD} \} = \frac{1 + \min\{\tilde{p}_0, 1-p_7\} \min \{p_0, \tilde{p}_0\}}{1 + \min \{ p_0, \tilde{p}_0 \}}.$$

If you score a FG in your first drive, then your opponent will have a $\tilde{p}_7$ chance of winning and $\tilde{p}_0$ chance of losing in the aggressive strategy. On the other hand, if using the standard strategy, your opponent will have a $p_7 + p_3 ( 1 - \tilde{p}_{SD} )$ probability of winning and a $p_0 + p_3 \tilde{p}_{SD}$ chance of losing. Thus, if you score a FG in your first drive, your probability of winning is $$\tilde{\Pi}_3 = \min \{\tilde{p}_0, p_0 + p_3\tilde{p}_{SD}\} = \min \left\{ \tilde{p}_0, \frac{1 - p_7 + p_0 \min \{p_0, \tilde{p}_0\}}{1 + \min \{p_0, \tilde{p}_0\}} \right\}.$$

If you do not score in your first drive, then your opponent will have a $p_3 + p_7 + p_0 (1 - \tilde{p}_{SD})$ probability of winning in a $p_0 \tilde{p}_{SD}$ probability of losing in the standard offense strategy. On the other hand, your opponent will have a $\tilde{p}_7 + \tilde{p}_0 (1 - \tilde{p}_{SD}) = 1 - \tilde{p}_0 \tilde{p}_{SD}$ probability of winning and $\tilde{p}_0 \tilde{p}_{SD}$ probability of losing in the aggressive strategy. So, if you don't score in your first drive your win probability is $$\tilde{\Pi}_0 = \min \{ p_0 \tilde{p}_{SD}, \tilde{p}_0 \tilde{p}_{SD} \} = \min \{p_0, \tilde{p}_0\} \tilde{p}_{SD} = \frac{\min \{p_0, \tilde{p}_0\}}{1 + \min \{ p_0, \tilde{p}_0 \} }.$$

So the total probability of winning is $$\tilde{\Pi} = \tilde{\Pi}(p_0, p_3, p_7, \tilde{p}_0, \tilde{p}_7) = \max \{ \tilde{p}_7 \tilde{\Pi}_7 + \tilde{p}_0 \tilde{\Pi}_7, p_7 \tilde{\Pi}_7 + p_3 \tilde{\Pi}_3 + p_0 \tilde{\Pi}_0 \}.$$ Given our parameters where $p_0 = p_3 = p_7 = \frac{1}{3}$ and $\tilde{p}_0 = \tilde{p}_7 = \frac{1}{2},$ the probability of winning when you go first in the updated model is \begin{align*} \tilde{\Pi}_0 &= \frac{ \min\{ \frac{1}{3}, \frac{1}{2} \} }{ 1 + \min \{ \frac{1}{3}, \frac{1}{2} \} } &= \frac{1}{4} \\ \tilde{\Pi}_3 &= \min \left\{ \frac{1}{2}, \frac{ 1- \frac{1}{3} + \frac{1}{3} \min \{ \frac{1}{3}, \frac{1}{2} \} }{ 1 + \min \{ \frac{1}{3}, \frac{1}{2} \} } \right\} &= \frac{1}{2} \\ \tilde{\Pi}_7 &= \frac{ 1 + \frac{1}{2} \min \{ \frac{1}{3}, \frac{1}{2} \} }{1 + \min \{ \frac{1}{3}, \frac{1}{2} \} } &= \frac{7}{8}\end{align*} In this case, the probability $$\tilde{\Pi} = \max \{ \frac{1}{2} \frac{7}{8} + \frac{1}{2} \frac{1}{4}, \frac{1}{3} \frac{7}{8} + \frac{1}{3} \frac{1}{2} + \frac{1}{3} \frac{1}{4} \} = \max \{ \frac{9}{16}, \frac{13}{24} \} = \frac{9}{16}.$$

Altogether this was not the dunking on the Niners and their decision that I hoped it would be, but luckily, whether the decision was correct or not, the Niners lost. That, in the end, is all that really matters!!!