Would a comma have hurt anyone? Extra Credit

Having successfully hacked your way through the first keypad, the door opens to reveal a second door with yet another keypad that has six numerical inputs: “I,” “II,” “III,” “IV,” “V,” “VI,” “VII,” and “VIII.”

You were expecting this, which is why your accomplice had handed you a second scroll of paper. You unfurl this one as well, hoping they remembered to add spaces between the numbers.

No such luck. This paper reads: “IIIVIIIVIIIVIII.” That’s 15 characters in total. How many distinct combinations are possible for this second door?

Since there are three V characters, let's iterate on the various possibilities for these characters. Let's write the code word as $$C = R(w_1)R(x_1)R(w_2)R(x_2)R(w_3)R(x_3)R(w_4),$$ where $R(t)$ be the Roman numeral of the number $t,$ where we see that $w_i \in \{1, 2, 3\}$ and $x_i \in \{ 4, 5, 6, 7, 8 \}.$ In order for the our word $C$ to match the code provided, we would need to make sure that if $x_i = 4$ then $x_{i-1} \leq 7,$ for $i = 2, 3.$ Let $$\mathcal{X} = \{ (x_1, x_2, x_3) \in \{4, 5, 6, 7, 8 \}^3 \mid x_i = 4 \Rightarrow x_{i-1} \leq 7, i = 2, 3 \}$$ be the solution set for the values of the V characters.

Let's define the indicator function $$\delta(t) = \begin{cases} 1, &\text{if $t=1$}\\ 0, &\text{if $t=0$.}\end{cases}$$ Additionally let $N(t)$ be the number of possible combinations of numerals that in "I", "II", and "III" that can be found in $R(t).$ In particular, $N(0)=N(1)=1,$ $N(2)=2$ and $N(3)=4.$ For any $(x_1, x_2, x_3) \in \mathcal{X},$ we would also need to have the following conditions be met in order for our word $C$ to match the code provided: \begin{align*}w_1 = w_1(x_1) &= 3 - \delta(x_1-4) \\ w_2 = w_2(x_1,x_2) &= \min \{ 3, 8 - x_1 \} - \delta(x_2 - 4) \\ w_3 = w_3(x_2,x_3) &= \min \{ 3, 8 - x_2 \} - \delta (x_3-4) \\ w_4 = w_4(x_3) &= \min \{3, 8 - x_3 \}\end{align*} Then, the total number of distinct code combinations where the V characters are assigned to the numbers $(x_1,x_2,x_3)$ is given by $N(w_1(x_1))N(w_2(x_1,x_2))N(w_3(x_2,x_3))N(w_4(x_3)).$

Therefore, the total possible number of distinct combinations for the second door is $$\hat{N} = \sum_{(x_1,x_2,x_3) \in \mathcal{X}} N(w_1(x_1))N(w_2(x_1,x_2))N(w_3(x_2,x_3))N(w_4(x_3)) = 4000.$$

Would a comma have hurt anyone?

You are breaking into a vault that contains ancient Roman treasure. The vault is locked, and can be opened via a modern-day keypad. The keypad contains three numerical inputs, which are (of course) expressed using Roman numerals: “I,” “II,” and “III.”

It’s a good thing your accomplice was able to steal the numerical key code to the vault. Earlier in the day, they handed you this code on a scroll of paper. Once at the keypad, you remove the scroll from your pocket and unfurl it. It reads: “IIIIIIIIII.” That’s ten vertical marks, without any clear spacing between them.

With some quick mental arithmetic, you realize the combination to unlock the door could be anywhere from four digits long to 10 digits long. (Or is it IV digits to X digits?) How many distinct combinations are possible? If two combinations use the same numbers but in a different order, they are considered distinct.

Let $n_1$ be the number of "I" numerals, $n_2$ the number of "II" numerals and $n_3$ the number of "III" numerals. Since there are total of ten vertical marks, we must have $n_1 + 2n_2 + 3n_3 = 10.$ If we have a nonnegative solution to this Diophantine equation, then since any order of the different numbers is a distinct combination, then for any solution there are a total of $$\frac{(n_1+n_2+n_3)!}{n_1! n_2! n_3!}$$ possible combinations associated with the solution $(n_1, n_2, n_3).$

We can solve this Diophantine equation by iterating through the possible cases for $n_3 \in \{0,1,2,3\}.$ When $n_3 = 0,$ the remaining equation is $$n_1 + 2n_2 = 10,$$ which has nonnegative solutions $$n_1 = 10 - 2k, n_2 = k, k = 0, 1, 2, 3, 4, 5.$$ When $n_3 = 1,$ the remaining equation is $$n_1 + 2n_2 = 7,$$ which has nonnegative solutions $$n_1 = 7-2k, n_2 = k, k = 0, 1, 2, 3.$$ When $n_3 = 2,$ the remaining equation is $$n_1 + 2n_2 = 4,$$ which has nonnegative solutions $$n_1 = 4-2k, n_2 = k, k = 0, 1, 2.$$ Finally, when $n_3 = 3,$ the remaining equation is $n_1 + 2n_2 = 1,$ which has only a single nonnegative solution $n_1 = 1, n_2 = 0.$ So overall, the solution set is \begin{align*}\mathcal{N} &= \{ (0,2,2), (0,5,0), (1,0,3), (1,3,1), (2,1,2), (2,4,0), (3,2,1), \\ &\quad\quad (4,0,2), (4,3,0), (5,1,1), (6,2,0), (7,0,1), (8,1,0), (10,0,0) \}\end{align*} So, the total number of possible combinations is \begin{align*}N &= \sum_{(n_1,n_2,n_3) \in \mathcal{N}} \frac{(n_1+n_2+n_3)!}{n_1! n_2! n_3!} \\ &= \frac{4!}{0!2!2!} + \frac{5!}{0!5!0!} + \frac{4!}{1!0!3!} + \frac{5!}{1!3!1!} + \frac{5!}{2!1!2!} + \frac{6!}{3!2!1!} \\ & \quad\quad +\frac{6!}{4!0!2!} + \frac{7!}{4!3!0!} + \frac{7!}{5!1!1!} + \frac{8!}{6!2!0!} + \frac{8!}{7!0!1!} + \frac{9!}{8!1!0!} + \frac{10!}{10!0!0!} \\ &= 6+1+4+20+30+15+60+15+35+42+28+8+9+1 \\ &=274\end{align*}

Greedy Mowing Madness

You’re mowing a circular lawn with a radius of $1$ unit. You can mow in straight strips that are $1$ unit wide. The fewest number of passes you would need to mow the entire lawn is two, as shown below. In one pass (shown in blue) you can mow half the circle, and in the second pass (shown in red) you can mow the other half of the circle.

However, instead of minimizing the number of passes, you greedily choose how to orient each pass so it cuts as much of the unmowed grass as possible. A pass doesn’t have to go through the center of the circle and can be in any direction, but must be straight and cannot bend.With this “greedy” approach, how many passes will it take for you to mow the entire lawn?

First let's establish the first greedy mowing pass. Without loss of generality, we can set our circular lawn's center at the origin, and by rotating, we can assume that the first pass will be parallel to the $y$-axis. In this case, the center of the pass will be at the line $x = x_0$ for some $x_0 \in (-1,1)$ and the strip will cover the set $$\Omega_1(x_0) = \{ (x,y) \in B(0,1) \mid |x - x_0| \lt \frac{1}{2} \}.$$ The area of the circular lawn that this pass will mow is $$A_1(x_0) = \int_{\max \{ -1, x_0 - \frac{1}{2} \}}^{\min \{ 1, x_0 + \frac{1}{2} \}} 2 \sqrt{1-x^2} \,dx.$$ Differentiating under the integral sign, we get \begin{align*}A_1^\prime(x_0) &= 2\sqrt{ 1 - \min \{1, x+ \frac{1}{2} \}^2 } \frac{d}{dx} \min \{ 1, x + \frac{1}{2} \} - 2\sqrt{1 - \max \{ -1, x - \frac{1}{2} \}^2 } \frac{d}{dx} \max \{ -1, x - \frac{1}{2} \} \\ &= \begin{cases} 2\sqrt{1 - (x+\frac{1}{2})^2}, &\text{if $x \in (-1, -\frac{1}{2})$;} \\ 2\left( \sqrt{1 - (x+\frac{1}{2})^2} - \sqrt{1 - (x-\frac{1}{2})^2} \right), &\text{if $ x \in (-\frac{1}{2}, \frac{1}{2})$;}\\ -2 \sqrt{1 - (x-\frac{1}{2})^2}, &\text{if $x \in (\frac{1}{2}, 1).$}\end{cases}\end{align*} Solving we for $A^\prime_1(x_0) = 0$ we get the only critical point as $x^*_0 = 0,$ which yields $A_1(0) = \int_{-1/2}^{1/2} 2\sqrt{1-x^2}\,dx = \frac{\pi}{3} + \frac{\sqrt{3}}{2}.$ So let's assume that the first pass is $\Omega_1 = \{ (x, y) \in B(0,1) \mid |x| \lt \frac{1}{2} \}.$

Let's move on to the next greedy pass. Now obviously from the first pass we see that the optimal method is to have some path of the mower that goes through the origin, so without loss of generality, let's assume that the center of the second pass goes through the line $y = mx,$ for some $m \in (-\infty, \infty).$ The second pass is thus the area $$\Omega_2(m) = \{ (x,y) \in B(0,1) \mid | mx - y | \lt \frac{\sqrt{1 + m^2}}{2} \}.$$ The newly mowed area in pass two is the entirety of $A_1(0)$ minus the area of the overlap between the two passes $\Omega_1(0) \cap \Omega_2(m).$ For $m \in (-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}),$ we see that the overlap is the parallelogram formed by the lines $x = \pm \frac{1}{2},$ $y = mx + \frac{\sqrt{1+m^2}}{2},$ and $y=mx-\frac{\sqrt{1+m^2}}{2},$ that is $O(m) = \sqrt{1+m^2}.$ Therefore, the new area mowed by pass two is $$A_2 (m) = A_1(0) - O(m) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} - \sqrt{1+m^2}.$$ This is maximized when $m^* = 0,$ and when $$A_2^* = A_2(0) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} - 1.$$ We can see that if $|m| \gt \frac{\sqrt{3}}{3},$ the overlapping area of $\Omega_1(0) \cap \Omega_2(m)$ is larger than the area given by the parallelogram when $m = \pm \frac{\sqrt{3}}{3},$ so $A_2(m) \lt A_2( \sqrt{3}/3)$ for all $|m| \gt \frac{\sqrt{3}}{3},$ so we can claim that not only is $m^*=0$ the optimizer within the neighborhood $(-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3})$ but in fact for all $m \in \mathbb{R}.$

So after two passes we are left with the unmowed area $$U = \{ (x,y) \in B(0,1) \mid \max \{ |x|, |y| \} \gt \frac{1}{2} \}.$$ This has four sub-regions, one in each quadrant of the plane. It is easy enough to construct a mowing path that covers at least two of the four sub-regions, but you cannot cover all four sub-regions with an acceptable mowing path, for instance, take the path $$\Omega_3 = \{ (x,y) \in B(0,1) \mid |x + y| \lt \frac{\sqrt{2}}{2} \}.$$ Though there is in fact definitely a better approach that can cover strictly greater than two complete sub-regions, once at least two of the four sub-regions are covered, all remaining unmowed area is coverable within one more pass. Therefore, the greedy approach will take four passes to cover the lawn.

Zeno's Paradoxical 5K - Extra Credit

I still want to run a negative split, but upping my tempo in discrete steps is such a slog. Instead, my next plan is to continuously increase my pace in the following way:

At the beginning of the race, I’ll start with a 24-minute pace. Then, wherever I am on the race course, I’m always running 10 percent faster than I was when I was twice as far from the finish line. Also, my speed should always be increasing in a continuous and smooth fashion. Using this strategy, how long will it take me to complete the 5K? Once again, I’m hoping it’s faster than my steady 23-minute pace, even though I started out slower.

Let's again assume that the race start at $x=0$ and finish at $x=5,$ but we will need to understand some sort of history, since otherwise when I am one step into the race, I would need to know how fast I was going about 2 steps closer than $5$ kilometers before the beginning of the race. So here let's define $x_n = 5(1 - 2^{-n})$ for $n = \mathbf{-1}, 0, 1, 2, \dots$ be breakpoints where we will be interested in defining our speed. Let $v(x)$ denote the velocity you are running (in kmph), when you are at some $x \in (-5,5).$ Again, here we are given that $v(0) = 12.5$ and that at any point $x \in (0,5)$ we have $$v(x) = 1.1 v(2x-5).$$ Since $x_0 = 0,$ and $2x_n-5 = x_{n-1}$ for each $n = 0, 1, \dots,$ we can use the recursive formula to find the values of $$v_n = v(x_n) = 12.5 \cdot 1.1^n, \,\, n = -1, 0, 1, 2, \dots$$

Now let's tackle what smooth means here. The link provided in the prompt to Wolfram suggests that we are talking about some sort of $C^n(-5,5)$ space of functions on $(-5,5)$ that are continuous and have continuous derivatives up to order $n$ for some $n.$ We could of course overachieve and go with $C^\infty(-5,5),$ which we will start with here, but then see if we can relax this for some more Zeno's 5K fun. So without further adieu, let's start with the $C^\infty(-5,5)$ solution:

First, by design and analogy, we are likely to have a velocity function $v$ that blows up to $\infty$ as $x \to 5^-.$ Let's try something from the parametric family $$v_\alpha(x) = \frac{12.5}{(1-\frac{x}{5})^\alpha}$$ for $\alpha \gt 0,$ which satisfy $v_\alpha \in C^\infty(-5,5),$ $v_\alpha(0) = 12.5$ and $v_\alpha^\prime (x) \geq 0$ for all $x \in (-5,5).$ The only thing left to do is to decide which value of $\alpha$ satisfied the recursion formula. Note that $$v_\alpha(2x-5) = \frac{12.5}{\left(1 - \frac{2x-5}{5}\right)^\alpha} = \frac{v_\alpha(x)}{2^\alpha},$$ for all $x \in (0, 5)$ so in order to satisfy the recursion formula $v(x) = 1.1 v(2x-5)$ we must have $$\alpha = \frac{\ln 1.1}{\ln 2}.$$ So we have a $C^\infty (-5,5)$ solution for velocity at position $x \in (0,5),$ $$v^*(x) = 12.5 \left( 1 - \frac{x}{5} \right)^{-\frac{\ln 1.1}{\ln 2}},$$ which gives a completion time for the $C^\infty$ extra credit Zeno's 5K of \begin{align*}T = \int_0^5 \, \frac{dx}{v(x)}& = \int_0^5 \frac{ \left(1 - \frac{x}{5} \right)^{\frac{\ln 1.1}{\ln 2}} }{12.5} \,dx\\ &= \frac{5}{12.5} \int_0^1 u^{\frac{\ln 1.1}{\ln 2}} \,du \\ &= \frac{5}{12.5} \frac{1}{1 + \frac{\ln 1.1}{\ln 2}} = 0.351647262315\dots \,\,\,\text{hours},\end{align*} or $21.0988357389\dots$ minutes.

Obviously, $C^\infty$ qualifies as smooth, but what if we are willing to go with only $C^1(-5,5)$ solutions. Obviously, since $C^\infty \subsetneq C^1,$ we still have the solution $v^*$ from above, but what if we use our points $\mathcal{X} = \{x_{-1}, x_0, x_1, \dots \}$ as knots to define a $C^1$ quadratic spline. Can we still get a solution that is C^1 and satisfies the recursion formula? It turns out, $\dots$ yes!

Lets define the velocity function $u = u(x)$ by spline components $$u_n(x) = a_n + b_nx + c_nx^2$$ for $x_{n-1} \leq x \leq x_n,$ for $n = 0, 1, 2, \dots.$ In order to satisfy the recursion formula, since $2x-5 \in (x_{n-1},x_n)$ for all $x \in (x_n, x_{n+1}),$ we would need \begin{align*}u_{n+1}(x) &= a_{n+1} + b_{n+1} x + c_{n+1} x^2 \\ &= 1.1 u_{n}(2x-5)\\ &= 1.1 \left( a_n + b_n (2x-5) + c_n (4x^2 - 20x + 25) \right)\\ &= 1.1 (a_n - 5 b_n + 25 c_n) + 1.1 (2b_n - 20c_n) x + 4.4 c_n x^2,\end{align*} so we must have the recursion formulae \begin{align*} a_{n+1} &= 1.1 \left( a_n - 5b_n + 25 c_n \right) \\ b_{n+1} &= 2.2 \left( b_n - 10 c_n \right) \\ c_{n+1} &= 4.4 c_n \end{align*} for all $n = 0, 1, \dots.$

Using this recursion formula we can define each quadratic spline component in terms of the parameters $(a_0, b_0, c_0).$ We need to confirm that we can choose appropriate values of $(a_0, b_0, c_0)$ such that the resulting spline $u = u(x) \in C^1(-5,5),$ that $u^\prime(x) \geq 0$ for all $x \in (-5,5)$ and the $u(0) = 12.5.$ Since $x_0 = 0$ is on both the $u_0$ and $u_1$ components of the spline and we want both $u$ and $u^\prime$ continuous at $x_0$ we must have $$u_0(0) = a_0 = u_1(0) = a_1 = 1.1 (a_0 - 5b_0 + 25 c_0) = 12.5$$ and $$u^\prime_0(0) = b_0 = u^\prime_1(0) = b_1 = 2.2 (b_0 - 10c_0).$$ So we can group these conditions into the system of three equations and three unknowns \begin{align*}a_0 &= 12.5 \\ 0.1 a_0 - 5.5 b_0 + 27.5 c_0 &= 0 \\ 1.2 b_0 - 22 c_0 &= 0,\end{align*} which can be solved by $(a_0,b_0,c_0) = (12.5, \frac{5}{16}, \frac{3}{176}).$

From the choice of $(a_0,b_0,c_0)$ we have already established the base case $u_0(x_0) = u_1(x_0) = v_0 = 12.5.$ Suppose that for some $n$ we have $u_n(x_{n-1}) = u_{n-1}(x_{n-1}) = v_{n-1}.$ By construction, we have $u_n(x_n) = 1.1u_{n-1}(2x_n-5) = 1.1u_{n-1}(x_{n-1}) = v_n.$ Similarly, by construction, we have $u_{n+1}(x_n) = 1.1u_n(2x_n-5) = 1.1u_n(x_{n-1}) = v_n.$ Therefore, for each n = 0, 1, \dots, by induction we have $u_n(x_n) = u_{n+1}(x_n) = v_n,$ which proves that $u \in C(-5,5).$ Similarly, we have from the choice of $(a_0,b_0,c_0)$ already established that $u_0^\prime(x_0) = u_1^\prime(x_0).$ Assume that for some $n$ we have $u_n^\prime(x_n) = u_{n+1}^\prime(x_n).$ Since $u_{n+1}(x) = 1.1u_n(2x-5)$ we have $u^\prime_{n+1}(x) = 2.2u_n^\prime(2x-5),$ so in particular $$u^\prime_{n+1}(x_{n+1})=2.2u_n^\prime(2x_{n+1}-5) = 2.2u_n^\prime(x_n).$$ Similarly, we have $u^\prime_{n+2}(x) = 2.2u_{n+1}^\prime (2x-5),$ so in particular $$u^\prime_{n+2}(x_{n+1}) = 2.2u^\prime_{n+1}(2x_{n+1}-5) = 2.2u_{n+1}^\prime(x_n) = 2.2u_n^\prime(x_n) = u^\prime_{n+1}(x_{n+1}),$$ so we conclude that $u_n^\prime(x_n) = u_{n+1}^\prime (x_n),$ for all $n = 0, 1, 2, \dots,$ so $u \in C^1(-5,5).$ From the solution of $(a_0,b_0,c_0)$ we have $u_0^\prime(x) \geq 0$ for all $x \in (x_{-1}, x_0),$ and since $$u^\prime_{n}(x) = 2.2^n u_0(2^nx - 5(2^n-1)) \geq 0$$ for each $n = 1, 2, \dots,$ we have $u^\prime(x) \geq 0$ for all $x \in (-5,5).$ So the $C^1$ monotonic increasing quadratic spline $u$ is another $C^1$ solution to the extra credit Zeno 5K. This quadratic spline solution provides a slightly slower $C^1$ completion time of $$T = \int_0^5 \,\frac{dx}{u(x)} \approx 0.354718139\dots \,\,\text{hours}$$ or $21.28308834\dots$ minutes. Ultimately, it would appear that for any $n$ there is a $C^n$ solution comprised of a monotonically increasing $(n+1)$th order polynomial spline that satisfies the recursion formula of $v(x) = 1.1v(2x-5)$ for all $x \in (0,5).$

Zeno's Paradoxical 5K

I’ve been experimenting with different strategies in $5000$-meter races (known as “5K”s). If I run the distance at a steady pace, I’ll finish in precisely $23$ minutes. However, I tend to find a burst of energy as I near the finish line. Therefore, I’ve tried intentionally running what’s called a “negative split,” meaning I run the second half of the race faster than the first half—as opposed to a “positive split,” where I run a slower second half.

I want to take the concept of a negative split to the next level. My plan for an upcoming race—the “Zeno Paradox 5K”—is to start out with a $24$-minute pace (i.e., running at a speed such that if I ran the whole distance at that speed, I’d finish in $24$ minutes). Halfway through the race by distance (i.e., after $2500$ meters), I’ll increase my speed by $10$ percent. Three-quarters of the way through, I’ll increase by another $10$ percent. If you’re keeping track, that’s now $21$ percent faster than my speed at the start.

I continue in this fashion, upping my tempo by $10%$ every time I’m half the distance to the finish line from my previous change in pace. (Let’s put aside the fact that my speed will have surpassed the speed of light somewhere near the finish line.) Using this strategy, how long will it take me to complete the 5K? I’m really hoping it’s faster than my steady $23$-minute pace, even though I start out slower (at a $24$-minute pace).

Let the race start at $x=0$ and finish at $x=5.$ Let $x_n = 5(1 - 2^{-n})$ for $n = 0, 1, 2, \dots$ be the breakpoints where you are increasing your speed. Let $v(x)$ denote the velocity you are running (in kmph), when you are at some $x \in (0,5).$ Then $$v(x) = 12.5 \cdot 1.1^{\max \{ i : x_i \lt x \} }.$$ The time spent at the velocity $v_n = 12.5 \cdot 1.1^n$ is $$t_n = \frac{ x_{n+1}-x_n }{ v_n } = \frac{ 5 \cdot 2^{-(n+1)} }{ 12.5 \cdot 1.1^n } = \frac{2.2^{-n}}{5},$$ so the total time to complete the 5K in this fashion is $$T = \sum_{n=0}^\infty \frac{2.2^{-n}}{5} = \frac{1}{5} \frac{1}{1- \frac{1}{2.2}} = \frac{2.2}{5 \cdot 1.2} = \frac{11}{30}\,\,\text{ hours},$$ or $22$ minutes.

Overlapping bubble packing

Draw a unit circle (i.e., a circle with radius $1$). Then draw another unit circle whose center is not inside the first one. Then draw a third unit circle whose center is not inside either of the first two.

Keep doing this until you have drawn a total of seven circles. What is the minimum possible area of the region that’s inside at least one of the circles?

Let's work smarter rather than harder and first note that the if we want to pack a circle full of seven non-overlapping unit circles then the enclosing circle's radius would be $3$, with one circle concentric with the enclosing circle and the other $6$ circles arranged with their centers forming a regular hexagon, which even Wikipedia snarkily notes is "Trivially optimal". Of course, we're not here for non-overlapping circles, but let's use this same configuration. Situate a unit circle at the origin, $x_0=(0,0),$ and then $6$ unit circles centered at $x_i = ( \cos \frac{2\pi (i-1)}{6}, \sin \frac{2\pi (i-1)}{6} ),$ for $i = 1, 2, \dots, 6.$ See a Desmos illustration of the configuration below:

First we note that $\| x_i - x_j \|_2 = 1,$ for $i \ne j,$ so none of the $7$ centers of the unit circles are inside any other circle. Next we note that the bounding circle around these $7$ circles has radius $$R = \max \{ \|x\|_2 \mid \min_{i = 0, 1, \dots, 6} \{ \|x - x_i \|_2 \} \leq 1 \} = 2,$$ which is (perhaps again trivially, per Wikipedia) optimal. However, we do not want the minimal bounding radius, but rather the shaded area. Let's use inclusion / exclusion and note that there are six regions between the bounding circle of radius two and the shaded area, so once we know the area of one of these sections, say A^\prime, then the shaded area is $A = 4\pi - 6A^\prime.$ Using symmetry, we can denote the area of one of those sections as $$A^\prime = 2 \int_0^1 \left( \sqrt{4-x^2} - \left(\sqrt{1 - (x-\frac{1}{2})^2} + \frac{\sqrt{3}}{2}\right) \right) \,dx = A_1 - A_2 - \sqrt{3},$$ where $$A_1 = 2 \int_0^1 \sqrt{4 - x^2} \,dx$$ and $$A_2 = 2 \int_0^1 \sqrt{1 - (x-\frac{1}{2})^2} \,dx.$$ Using trig substitution $x = 2 \cos \theta,$ we see that \begin{align*}A_1 = 2\int_0^1 \sqrt{4-x^2} \,dx &= -8 \int_{\pi/2}^{\pi/3} \sin^2 \theta \,d\theta \\ &= 4 \int_{\pi/3}^{\pi/2} 1 - \cos 2\theta \,d\theta \\ &= \frac{2\pi}{3} + 2 \sin \pi - 2\sin \frac{2\pi}{3} = \frac{2\pi}{3} + \sqrt{3}.\end{align*} Similarly, using the trig substitution $x = \frac{1}{2} + \cos \theta,$ we see that \begin{align*}A_2 = 2 \int_0^1 \sqrt{1 - (x-\frac{1}{2})^2} \,dx &= -2 \int_{2\pi/3}^{\pi/3} \sin^2 \theta \,d\theta \\ &= \int_{\pi / 3}^{2\pi/3} 1 - \cos 2 \theta\, d\theta \\ &= \frac{\pi}{3} + \frac{1}{2}\sin \frac{2\pi}{3} - \frac{1}{2}\sin \frac{4\pi}{3} = \frac{\pi}{3} + \frac{\sqrt{3}}{2}.\end{align*}

Therefore, we have $$A^\prime = \left( \frac{2\pi}{3} + \sqrt{3} \right) - \left( \frac{\pi}{3} + \frac{\sqrt{3}}{2} \right) - \sqrt{3} = \frac{\pi}{3} - \frac{\sqrt{3}}{2},$$ so we have the minimal area of the region that's inside at least one of the circles as $$A = 4\pi - 6A^\prime = 2\pi + 3\sqrt{3} \approx 11.4793377299\dots$$