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.

Fewest Firsts

As we just noted, in 2026, all seven days of the week appear as the first of the month at least once. But you know, I decided that I don’t like that at all. Instead, I want as few days of the week as possible to appear as the first of the month in a given year.

To accomplish this, I have been granted the authority to change the number of days in each of that year’s 12 months, provided that there are still 365 or 366 days in the year and each month has at least 28 days and at most 31 days.

What are the fewest days of the week that can appear as the first of the month in such a calendar year? (And for fun, rather than for credit: How many such calendars can you design with this property?)

Let $N^*$ be the minimal number of days of the week that can appear as the first of the month of any calendar year. First from the Classic problem we see that the furthest away that two consecutive months can start is 3 days of the week apart, so this means that the smallest that we could possibly hope for is to have a calendar with only three possible days of the week for firsts of the month. That is, $N^* \gt 2,$ or equivalently that $N^* \geq 3.$

Therefore, all we have to do is provide some calendar year where $N=3$ is attained. For instance, take the following setup where March, June, and September have 29 days and all other months have 31 days. This would give $3 \cdot 29 + 9 \cdot 31 = 87 + 279 = 366$ days in the year. Similarly, if January 1 occurs on $x \in \mathcal{D},$ then February 1 occurs on $x+3,$ March 1 occurs on $x+6,$ April 1 occurs on $x+7 \equiv x,$ May 1 occurs on $x+3,$ June 1 occurs on $x+6,$ July 1 occurs on $x+7\equiv x,$ August 1 occurs on $x+3,$ September 1 occurs on $x+6,$ October 1 occurs on $x+7 \equiv x,$ November 1 occurs on $x + 3$ and finally, December 1 occurs on $x+6.$ Putting these altogther gives a subset $\mathcal{N} = \{ x, x + 3, x + 6 \} \subsetneq \mathcal{D},$ with $N= |\mathcal{N}| = 3.$ Therefore, we have that $N^* \leq N = 3$ and hence the fewest days of the week that can appear as the first of the month is $N^*=3.$

Can every day be the first?

In 2026, every day of the week is the first day of the month at least once:

• Monday is June 1.
• Tuesday is September 1 and December 1.
• Wednesday is April 1 and July 1.
• Thursday is January 1 and October 1.
• Friday is May 1.
• Saturday is August 1.
• Sunday is February 1, March 1, and November 1.

Is 2026 special in this regard? If so, when is the next year when one of the days of the week is not represented among the firsts of the month? Otherwise, if 2026 is not special in this regard, then why not?

First let's note that if a month has $28$ days, looking at you February, then the next month will start on the same day of the week that it starts. If a month has $29$ days, again looking less frequently at you February, then then next month will start on the next day of the week. If a month has $30$ days then the next month will start two days after it, and finally for all of those $31$ day months, the following month will start three days of the week after it.

Let's further represent the days of the week as numerals, $\mathcal{D} = \mathbb{Z} / 7\mathbb{Z}.$ If January 1 occurs on some $x \in \mathcal{D}$ then February 1 occurs on $x+3.$ If the year is not a leap year then we have March 1 also on $x+3,$ followed by April 1 on $x+6,$ May 1 on $x+8 \equiv x+1,$ June 1 on $x+4$, July 1 on $x+6,$ August 1 on $x+9 \equiv x+2$, September 1 on $x+5,$ October 1 on $x+7 \equiv x,$ November 1 on $x+3$ and finally December 1 on $x+5.$ Gropuing these all together we see that we have all the days of the week covered $\{ x, x+1, x+2, x+3, x+4, x+5, x+6\} = \mathcal{D}.$

If on the other hand, this is a leap year, then we have January 1 and February 1 on $x$ and $x+3$ as before, but then March 1 on $x+4,$ April 1 on $x+7 \equiv x,$ May 1 on $x+2,$ June 1 on $x+5,$ July 1 on $x+7 \equiv x$, August 1 on $x+3,$ September 1 on $x+6,$ October 1 on $x+8 \equiv x+1,$ November 1 on $x+4,$ and finally December 1 on $x+6.$ Again, grouping these all togehter, we see that we have all the days of the week covered $\{ x, x+1, x+2, x+3, x+4, x+5, x+6 \} = \mathcal{D}.$

Therefore, there is nothing special about 2026. Each and every year under the Gregorian system has this property where each day of the week is represented among the firsts of the month.