You probably could have just asked me to arrive at the right time ...

A standard analog clock includes an hour hand, a minute hand, and 60 minute markers, 12 of which are also hour markers.

At a certain time, both the hour hand and minute hand are both pointing directly at minute markers (either of which could also be an hour marker). The hour hand is 13 markers ahead (i.e., clockwise) of the minute hand. At what time does this occur?

Let's set up the problem. We can represent each hand by its phase on the clockface, that is the proportion of the entire rotation that the hand has already completed. Let $h(t)$ and $m(t)$ be the phase of the hour and minute hands, respectively, at time $t,$ measured in hours. Since the minute hand makes one full rotation every hour, we see that the minute hand's phase at time $t$ is always the fractional part of $t$, that is $m(t) = t - \lfloor t \rfloor.$ Similarly, since in one hour the hour hand only travels 1/12th of a rotation that we have $h(t) = t/12 - \lfloor t / 12 \rfloor.$

The difference in the phases can be transformed into the angle between the hour and minute hands, so we could look at $$\delta(t) = h(t) - m(t) = \left\lfloor t \right\rfloor - \left\lfloor \frac{t}{12} \right\rfloor - \frac{11t}{12}.$$ However, since the angle between two hands is always less than or equal to $180^\circ,$ or phase of $1/2,$ we instead should look at $$\Delta(t) = \delta(t) - \lfloor \delta(t) + 1/2 \rfloor = \begin{cases} -\frac{11}{12}t, & 0 \leq t \lt \frac{6}{11} \\ 1 - \frac{11}{12}t, & \frac{6}{11} \leq t \lt \frac{18}{11}; \\ \dots, & \\ k - \frac{11}{12}t, & \frac{6(2k-1)}{11} \leq t \lt \frac{6(2k+1)}{11}; \\ \dots, & \\ 11-\frac{11}{12}t, & \frac{138}{11} \leq t \lt 12. \end{cases}$$ We can see graphically that $\Delta$ is cyclical and that there are 11 full cycles within the 12 hour cycle of the clock face, that is, a period of $\tau = 12/11.$

Let's first try to find the solution to $$\Delta(t_0) = 1 - \frac{11t_0}{12} = \frac{13}{60},$$ with $6/11 \leq t_0 \lt 18/11.$ We see that we have $$t_0 = \frac{47}{55}.$$ Since $\Delta$ is periodic, we see that all of the solutions of the $\Delta = 13/60$ are of the form $$t_k = t_0 + k \tau = \frac{47}{55} + \frac{12k}{11} = \frac{47 + 60k}{55},$$ for $k = 0, 1, \dots, 10,$ as summarized in the table below.

$k$	$t_k$	hh:mm:ss
0	47/55	12:51:36
1	107/55	01:56:44
2	167/55	03:02:11
3	227/55	04:07:38
4	287/55	05:13:05
5	347/55	06:18:33
6	407/55	07:24:00
7	467/55	08:29:27
8	527/55	09:34:55
9	587/55	10:40:21
10	647/55	11:45:49

So we see that the appropriate time where the hour and minute hands are pointing at minute markers that are 13 markers apart is at 7:24.

Buzzing with excitement

The children's game called “Buzz” (also known as “Fizz buzz”) is where players take turns reciting whole numbers in order. However, in one particular variant of the game, anytime a number is a multiple of 7 or at least one of its digits is a 7, the player must say “buzz” instead of that number.

For example, here is how the first 20 turns of the game should proceed: 1, 2, 3, 4, 5, 6, buzz, 8, 9, 10, 11, 12, 13, buzz, 15, 16, buzz, 18, 19, 20.

How many times should “buzz” be said in the first 100 turns of the game (including those mentioned above in the first 20 turns)?

There are 14 multiples of 7 that are less than 100, since $100 / 7 = 14.2857\dots,$ so each of those numbers would be buzzes. Additionally, given that if we remove the possibility of there being a 7, then there are only 9 possible digits to choose from to make an up to two digit number with no 7s, that is a number from 0 to 99. Since we don't want 0 but do want 100, we see that there still are 81 possible numbers that do not contain any 7's, so that means that there would be 100 - 81 = 19 numbers that would buzz because they contain a 7. However, we have some overlaps, since 7, 70 and 77 are also multiples of seven, so from the inclusion-exclusion principle, the total number of buzzes in the first 100 numbers is $19 + 14 - 3 = 30$ buzzes.

As we just saw, in the first 20 turns of the game, 15 percent of the numbers were “buzzed.” But as the game proceeds, an increasing frequency of numbers get buzzed.

There is a certain minimum number $N$ such that, for the Nth turn in the game and for every turn thereafter, at least half the numbers up to that point have been buzzed. What is this value of $N$?

We can also encode a quick Python script to enumerate all buzzes in the first $N$ numbers:

def buzz(N):
    b = []
    for i in range(1,N+1):
        if i % 7 == 0:
            b.append(i)
        elif '7' in str(i):
            b.append(i)
    return b

By either a healthy amount of hunting and pecking, or some more systematic sounding of the phase space, we see that when $N = 708588$ that there are exactly 354294 buzzes, which is good for a ratio of 50% buzzes.

Extra amaze, amaze, amaze!!!

Next, Rocky wants to transport a solid crystal shaped like a regular dodecahedron to his human friend, Grace. As before, each edge has length 1.

This time, the long tunnel between the space ships can be any right prismatic shape, not necessarily a cylinder. Once again, Rocky needs the crystal to fit through the tunnel, and it’s okay if that fit is snug. What is the minimum possible cross-sectional area for the tunnel so that the crystal will fit through it?

A regular dodecahedron with unit side lengths centred at the origin is defined by the system of twelve linear inequalities \begin{align*} \pm x \pm \phi y &\leq \frac{1}{2} + \phi \\ \pm y \pm \phi z &\leq \frac{1}{2} + \phi \\ \pm \phi x \pm z & \leq \frac{1}{2} + \phi,\end{align*} where $\phi = \frac{\sqrt{5}+1}{2}$ is the golden ratio. Let's take a look at what happens when we intersect this dodecahedron by the plan $\phi x - z = 0,$ that is when $z = \phi x,$ that in a plane that goes through the origin and is parallel to the two faces $\phi x - z = \frac{1}{2} + \phi$ and $-\phi x + z = \frac{1}{2} + \phi.$ In this case, our system of linear inequalities simplifies to a system of 10 linear inequalities \begin{align*} \pm x \pm \phi y &\leq \frac{1}{2} + \phi,\\ \pm y \pm \phi^2 x &\leq \frac{1}{2} + \phi,\\ \pm 2 \phi x &\leq \frac{1}{2} + \phi.\end{align*} This system of 10 inequalities in $x$ and $y$ traces out a decagon in the $xy$-plane, with vertices as follows: \begin{align*} A &= \left( \frac{1+ \phi}{4}, \frac{\phi}{4} \right)\\ B &= \left( \frac{\phi}{4}, \frac{1+2\phi}{4} \right)\\ C &= \left( 0, \frac{1 + \phi}{2} \right)\\ D &= \left( -\frac{\phi}{4}, \frac{1+2\phi}{4} \right)\\ E &= \left( -\frac{1+\phi}{4}, \frac{\phi}{4} \right)\\ F &= \left( -\frac{1+\phi}{4}, -\frac{\phi}{4} \right)\\ G &= \left( -\frac{\phi}{4}, -\frac{1+2\phi}{4} \right)\\ H &= \left( 0, -\frac{1 + \phi}{2} \right)\\ I &= \left( \frac{\phi}{4}, -\frac{1+2\phi}{4} \right)\\ J &= \left( \frac{1 + \phi}{4}, -\frac{\phi}{4} \right) \end{align*}

We can divide this decagon into $4$ different trapezoids, that is, trapezoids ABIJ, BCHI, CDGH, EFGH, where the area of ABIJ and EFGH are equal to say, $A_1$ and the area of trapezoids BCHI and CDGH are equal, to say $A_2.$ We see that trapezoid ABIJ has a height of $1/4$ and bases of $\frac{1}{2} + \phi$ and $\phi / 2,$ so $$A_1 = \frac{1}{2} \frac{1}{4} \left( \frac{1}{2} + \phi + \frac{\phi}{2} \right) = \frac{1 + 3\phi}{16}.$$ Additionally, we see that BCHI has a height of $\phi / 4$ and bases of $1 + \phi$ and $\frac{1}{2} + \phi,$ so $$A_2 = \frac{1}{2} \frac{\phi}{4} \left( 1 + \phi + \frac{1}{2} + \phi \right) = \frac{\phi}{16} (3 + 4\phi) = \frac{4 + 7\phi}{16}.$$

Therefore the maximum cross sectional area is $$A = 2(A_1+A_2) = 2 \frac{5 + 10 \phi}{16} = \frac{5}{8} (2 \phi + 1) \approx 2.64754248594\dots.$$ There are other cross sections

Amaze, amaze, amaze!!!! (Classic)

An alien—more specifically, an Eridian—named Rocky, and needs to pass a solid xenonite crystal to his human friend named Ryland Grace, in a neighboring spaceship. The crystal is shaped like a regular tetrahedron, as shown below, and all its edges have length 1.

To safely transport the crystal, Rocky needs to create a long cylindrical tunnel between the spaceships, and then orient the tetrahedron so that it fits through the tunnel. It’s okay if the crystal fits snugly inside the tunnel—in this case, it can slide along without any friction. What is the minimum possible radius for Rocky's tunnel so that the crystal will fit through it?

Consider the regular unit tetrahedron, with one of its equilateral unit triangle faces lying in the $xy$-plane, in particular, let's say its vertices are at $(0,0,0),$ $(1,0,0),$ $(1/2, \sqrt{3}/2, 0)$ and $(1/2, \sqrt{3}/6, \sqrt{6}/3).$ If we take any plane that is parallel to the $xy$-plane, say the plane $z = h$, and intersect it with the tetrahedron, we will get an equilateral triangle with side length $s=1-h.$ This would be equivalent to orienting the tetrahedron so that one of its faces was always in the cross sectional plane of the cylindrical tunnel. In this case, the largest possible sidelength of the equilateral triangles was a side length of $s=1,$ which is circumscribed by a circle of radius $r_3 = \frac{\sqrt{3}}{3}.$ This is a pretty good start, but let's see whether any other orientations will give us a better answer.

If we take any plane that is perpendicular to the $xy$-plane, say the plane $x=h,$ and intersect it with the tetrahedron, then by diligent application of ratios, we will get an triangle with vertices \begin{align*} v_1&=(h,0,0), \\ v_2&=(h, \max\{h,1-h\} \sqrt{3},0), \text{ and }\\ v_3&=(h, \max\{h, 1-h\}\frac{\sqrt{3}}{3}, 2\max\{h, 1-h\}\frac{\sqrt{6}}{3}).\end{align*} Therefore we see that this triangle is isoceles, since $$d_{12} = \max\{h,1-h\} \sqrt{3}$$ and $$d_{13} = \sqrt{ h^2/3 + 24 h^2/9 } = \max\{h, 1-h\}\sqrt{3},$$ while the final length is given by $$d_{23} = \sqrt{4h^2/3 + 24h^2/9} = 2\max\{h, 1-h\}.$$ In this, case the largest possible sidelengths are given by the isoceles triangle with sidelengths $\sqrt{3}/2 - \sqrt{3}/2 - 1.$ Let's pretend like two of the side with length 1 are located at the points $(-1/2,0)$ and $(1/2,0)$, from which with some symmetry and Pythagorean allusions we get that the third vertex is at the point $(0, \sqrt{2}/2).$ The centroid of this triangle is at the point $(0, \sqrt{2}/6).$ The circumscribing circle has radius $$r_2 = \sqrt{ \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{2}}{6}\right)^2 } = \sqrt{ \frac{1}{4} + \frac{2}{36} } = \frac{\sqrt{11}}{6} \lt \frac{\sqrt{3}}{3}.$$

To recap, we have covered the cases where three vertices of the tetrahedron fit snugly against the edge of the cylindrical tunnel, in which case we need $r_3 = \frac{\sqrt{3}}{3}.$ We have also covered when two vertices of the tetrahedron fit snugly against the edge of the cylindrical tunnel. We can rule out a case where none of the vertices fit snugly against the edge of the cylindrical tunnel, since we could definitely reduce the radius until at least one of the vertices fit snugly against the cylindrical tunnel. So the only case that we have left to solve for is when only a single vertex fits snugly against the cylindrical tunnel.

Let's assume that the vertex at $(1/2, \sqrt{3}/6, \sqrt{6}/3)$ is the one that is riding snugly a long the cylindrical tunnel and that the at any moment, without loss of generality a cross section of the cylindrical tunnel through this vertex cuts out a triangle whose other vertices are at $(h,0,0)$ and $(k, \max\{k, 1-k\}\sqrt{3},0)$ for some $0\leq h, k \leq 1.$ In particular, this would mean that the centroid of the triangle is at $$C(h,k) = \left( \frac{2h+2k+1}{6}, \frac{\sqrt{3}(6\max\{k,1-k\}+1)}{18}, \frac{\sqrt{6}}{9} \right).$$ The distance to the centroid to the point $(1/2, \sqrt{3}/6, \sqrt{6}/3)$ is \begin{align*}d(h,k) &= \sqrt{ \left(\frac{1 - h -k}{3}\right)^2 + \left(\frac{\sqrt{3}(1-3\max\{k, 1-k\})}{9}\right)^2 + \left( \frac{2\sqrt{6}}{9} \right)^2 } \\ &= \sqrt{ \frac{(h+k)^2}{9} + \frac{1}{27} \left( 9 \max \{k, 1-k\}^2 - 6 \max \{k, 1-k\} + 1 \right) + \frac{8}{27} } \\ &= \sqrt{ \frac{(h+k)^2}{9} + \frac{1}{3} \max \{k, 1-k\}^2 - \frac{2}{9} \max \{k, 1-k\} + \frac{1}{3} }\end{align*} Let's assume that $k \leq \frac{1}{2}$ for the time being, then we get \begin{align*}d_1(h,k) &= \sqrt{ \frac{h^2 + 2hk + k^2}{9} + \frac{k^2}{3} - \frac{2k}{9} + \frac{1}{3} } \\ &= \sqrt{ \frac{4}{9} \left( k - \frac{1-h}{4} \right)^2 + \frac{h^2}{12} + \frac{h}{12} + \frac{11}{36} }\\ & \geq \frac{\sqrt{11}}{6},\end{align*} since $0 \leq h \leq 1$ and $(k - (1-h)/4)^2 \geq 0$ as well. On the other hand, if we assume that $k \geq \frac{1}{2},$ then we get \begin{align*}d_2(h,k) &= \sqrt{ \frac{h^2 + 2hk +k^2}{9} + \frac{(1-k)^2}{3} - \frac{2 (1-k)}{9} + \frac{1}{3} } \\ &= \sqrt{ \frac{4}{9} \left( k - \frac{2 - h}{4} \right)^2 + \frac{h^2}{12} + \frac{h}{9} + \frac{1}{3}} \\ &\geq \frac{\sqrt{3}}{3} \geq \frac{\sqrt{11}}{6},\end{align*} since $0 \leq h \leq 1$ and $(k - (2-h)/4)^2 \geq 0.$ So we see that even if we can get an orientation where only one vertex is flush with the cylindrical wall, we cannot get a circumscribed radius that is any smaller than $r_2 = \frac{\sqrt{11}}{6}.$ Therefore, we see that the smallest possible radius for Rocky's tunnel is $$\frac{\sqrt{11}}{6} \approx 0.552770798393\dots$$ units.

Classical Looping Loci

I have a loop of string whose total length is $10$. I place it around a unit disk and pull a point on the string away from the disk until the string is taut, as shown below.

I drag this point around the disk in all directions, always keeping the string taut, tracing out a loop. What is the area inside this resulting loop?

For sake of exposition, let's assume that the unit circle is centered at the origin, $O = (0,0),$ and that I am pulling the string taut along the $x$-axis until it reaches the point $P = (r,0)$. Let's first determine the value of $r.$

If the string is taut, then the string will form a straight line that is tangent to the circle at say some point $Q$. If we draw a radius from $O$ to $Q$, then $\Delta OQP$ is a right triangle, where $\angle OQP$ is the right angle. Since the hypotenuse has length $r$ and the radius $OQ$ has unit length, we see that $|PQ| = \sqrt{r^2-1}$ from some guy named Pythagoras. Additionally we see that $m \angle QOP = \tan^{-1} \sqrt{r^2-1}.$ Therefore, if we try to calculate the total length of the string, we get $$\ell (r) = 2\pi + 2 \sqrt{r^2-1} - 2 \tan^{-1} \sqrt{r^2-1},$$ which is an increasing function from $\ell: [1,\infty) \to [2\pi, \infty).$ Since $\ell$ is monotonic it can be inverted, so that if we know that the string has length $10$, then we get $$\hat{r} = \ell^{-1} (10) \approx 3.2753266928\dots.$$

Since the choice of stretching the string along the $x$-axis was arbitrary, we see that as we drag the point around the disk in all directions, the maximum extent from the origin will constantly be $\hat{r},$ so the locus traces out a circle with radius $\hat{r},$ which has area $$A = \pi \hat{r}^2 \approx 33.7022675393\dots.$$

Average Cantor Distance

It’s possible to pick a random point in the Cantor set in the following way: Start with the entire number line from 0 to 1. Then, every time you remove a middle third, you give yourself a 50 percent chance of being on the left remaining third and a 50 percent chance of being on the right remaining third. Then, when you remove the middle third of that segment, you again give yourself a 50 percent chance of being on the left vs. the right, and so on.

Suppose I independently pick two random points in the Cantor set. On average, how far apart can I expect them to be?

Let's define $\Delta$ as the average distance between two different randomly chosen points in the Cantor set, say $a$ and $b \in \mathcal{C}.$ Based on the self-similarity of the Cantor set $\mathcal{C},$ we see that $\mathcal{C} = \frac{1}{3} \mathcal{C} \cup \left( \frac{2}{3} + \frac{1}{3} \mathcal{C} \right).$ Then from the law of total expectation and the fact that the probability of being in either the left or right half of the Cantor set is independently $\frac{1}{2}$ for each of $a$ and $b$, we get \begin{align*} \Delta &= \mathbb{E} \left[ |a-b| \mid a, b \in \frac{1}{3} \mathcal{C} \right] \mathbb{P} \left\{ a, b \in \frac{1}{3} \mathcal{C} \right\} \\ &\quad + \mathbb{E} \left[ |a-b| \mid a, b \in \frac{2}{3} + \frac{1}{3} \mathcal{C} \right] \mathbb{P} \left\{ a, b \in \frac{2}{3} + \frac{1}{3} \mathcal{C} \right\} \\ & \quad\quad + \mathbb{E} \left[ |a-b| \mid a \in \frac{1}{3} \mathcal{C}, b \in \frac{2}{3} + \frac{1}{3} \mathcal{C} \right] \mathbb{P} \left\{ a \in \frac{1}{3} \mathcal{C}, b \in \frac{2}{3} + \frac{1}{3} \mathcal{C} \right\} \\ &\quad\quad\quad + \mathbb{E} \left[ |a-b| \mid a \in \frac{2}{3} + \frac{1}{3} \mathcal{C}, b \in \frac{1}{3} \mathcal{C} \right] \mathbb{P} \left\{ a \in \frac{2}{3} + \frac{1}{3} \mathcal{C}, b \in \frac{1}{3} \mathcal{C} \right\} \\ &= \frac{ \Delta_1 + \Delta_2 + \Delta_3 + \Delta_4}{4}.\end{align*}

Since they are simply scaled copies of the original problem, we see that $\Delta_1 = \Delta_2 = \frac{1}{3} \Delta.$ However, let's take a look at $\Delta_3$ and $\Delta_4.$ For any $a \in \frac{2}{3} + \frac{1}{3} \mathcal{C},$ there is some $a^\prime \in \mathcal{C}$ such that $a = \frac{2}{3} + \frac{1}{3} a^\prime.$ So if $b \in \frac{1}{3} \mathcal{C}$ then there is some $b^\prime \in \mathcal{C}$ with $b=\frac{1}{3} b^\prime$. Therefore, we then have $|a-b| = \frac{2}{3} + \frac{1}{3}(a^\prime - b^\prime) \geq \frac{1}{3} \geq 0.$ So taking the expectation over all values of $a^\prime, b^\prime \in \mathcal{C}$ we get $$\Delta_3 = \Delta_4 = \mathbb{E} \left[ |a-b| \mid a \in \frac{2}{3} + \frac{1}{3} \mathcal{C}, b \in \frac{1}{3} \mathcal{C} \right] = \frac{2}{3} + \frac{1}{3} \mathbb{E} \left[ a^\prime - b^\prime \right] = \frac{2}{3}.$$

Thus putting everything together we get $$\Delta = \frac{1}{4} \sum_{i=1}^4 \Delta_i = \frac{ \frac{2}{3} \Delta + \frac{4}{3} }{4} = \frac{1}{6} \Delta + \frac{1}{3},$$ which means that $\frac{5}{6} \Delta = \frac{1}{3}$ or finally that the average distance between two randomly chosen points in the Cantor set is $$\Delta = \frac{6}{5}\cdot \frac{1}{3} = \frac{2}{5}.$$

Team Centroid is slacking

Two teams of shovelers plan to remove all the snow from a parking lot that’s shaped like a regular hexagon. Team Vertex initially places each of its six shovelers at the six corners of the lot. Meanwhile, Team Centroid initially places all its shovelers at the very center of the lot.

Each team is responsible for shoveling the snow that is initially closer to someone on their own team than anyone on the other team. What fraction of the lot’s snow is Team Centroid responsible for shoveling?

We can draw a straight line from Team Centroid's central spot to each of the vertices. Drawing a perpendicular bisector of each of these lines we obtain a new smaller, rotated hexagon, shown in blue in the figure below. We note that the apothem of the smaller hexagon is one half of the circumradius of the larger hexagon. Since the area of a regular $n$-gon is given by $$A_n = na^2 \tan \frac{\pi}{n} = \frac{1}{2} nR^2 \sin \frac{2\pi}{n},$$ we see that the apothem and circumradius are related by $a = R \cos \frac{\pi}{n},$ so in this case we have $a = R \cos \frac{\pi}{6} = \frac{R \sqrt{3}}{2},$ or equivalently $R = \frac{2a}{\sqrt{3}}.$

Since we have that the apothem of the Team Centroid hexagon is one half of the circumradius of the Team Vertex hexagon, we have $$R_C = \frac{2}{\sqrt{3}} a_C = \frac{1}{\sqrt{3}} R_V.$$ Since areas scale with the square of the circumradius, we have that the area shoveled by Team Centroid is $\left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3}$ of the entire hexagonal parking lot.