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.