30 Jul

Problematic Postcards

If you’ve come here as a result of a puzzling postcard you may have come across, welcome to Studio Infinity! We hope you’ll enjoy looking at some of the other content below as well, but here are the three posts corresponding to the problems you can find on those postcards, each of which links to a solution.

Insubordinate Integral
Smallish Sequence
Troubling Triangle

And, as I’ve mentioned here before, I invite you all to submit a problem or solution to Math Horizons Playground.

30 Jul

Insubordinate Integral

How does the value of the following improper integral compare to 1? I.e., is it smaller, larger, or exactly equal to 1?

(This problem was proposed to Math Horizons Playground by Mehtaab Sawney of Commack High School. And for all of you $\pi$-ists out there, $\tau$ is of course just the radian measure of a full circle, i.e., $\tau=2\pi$.)

Stumped? You can peek at the solution using the password “strapyb”.

30 Jul

Troubling Triangle

What is the area of the pink shaded triangle (as a fraction of ABC)?

For definiteness, the unlabeled points do in fact trisect each of the sides of the triangle.

Stumped? You can peek at the answer using the password “threedian”.

30 Jul

Smallish Sequence

What are the next few terms in this sequence of smallish numbers?

1 1 1 3 1 3 1 1 3 1 1 3 1 1 3 1 1 1 3 2 1 1 3 2 1 3 1 1 3 1 1 1 3 2 1 1 3 2 1 1 3 ? ? ?

(And, of course, what’s the rule generating the sequence?)

Stumped? You can peek at the solution using the password “1p1a2s”.

24 Jul

What’s My Radius?

I recently purchased a large number of styrofoam balls as supplies for an upcoming build (about which I will post later). The plans for that build required the diameter of the styrofoam balls, to pretty high accuracy. Although the balls were nominally 4 1/2 inches in diameter, I had noticed in a craft shop that their products also had diameters listed in whole numbers of millimeters that were close to the inch ratings, but not precisely equal. Not only could I not find the millimeter diameters listed for this particular size, I needed to know which was closer to reality, the U.S. or the metric measurement. (I had a hunch that given the world-wide nature of manufacturing and the fact that only the U.S. does not use metric, the metric measurements were more likely to be accurate.)

Unfortunately, my calipers did not have long enough jaws to clamp down onto the styrofoam balls, as you can see above. So how could I accurately measure the diameter of the sphere? You can’t exactly stick a ruler through the ball, and even if you could, how would you locate two diametrically opposed points? There’s a nice trick with a circle, in which you pick any two points, and then draw a line through one of them perpendicular to the line between them. Where that line intersects the circle is diametrically opposite the other point. (This works because the hypotenuse of any right triangle inscribed in a circle is a diameter.) But is there a three-dimensional version of this?

Fortunately, I was not the first person to face this conundrum; you can find this exact question on Math StackExchange. So first I tried the “accepted” answer: find two planes at exactly right angles with each other (I used a bookshelf, checked for true with a square), shove the sphere into the corner where the planes meet, and measure the distance between the plane and the point of contact to get the radius. Clever, huh?

       

But in practice, it’s pretty difficult to see exactly where the sphere touches the shelf. As you can see in the photo above, it looks consistent with a 57mm radius but it also looked consistent with a 57.5mm or even a 58mm radius. I wanted to double check with a more accurate method.

Fortunately, there was another answer on that page, involving measuring multiple distances on the surface of the sphere (easy to do with calipers), followed by a rather lengthy and involved calculation. But a little experimentation and figuring arrived at the following method, which is pretty quick and simple, so I thought it deserved its own post.

First,

Materials
sphere to measure
compass
calipers
pins

pick a convenient, round distance that appears to be in the neighborhood of a third to a quarter of the way around the sphere; for my styrofoam pieces, I chose 90mm. Using calipers (or a ruler), set the compass to exactly this distance. Choose an arbitrary point on the sphere, and draw a circle with the compass. (You can just see the circle faintly in the picture below.) Now choose another arbitrary point on this circle. We want to find the two points on the circle at that same chosen distance away from this second point. So put the point of the compass on this second point, and draw two small arcs intersecting the circle, one on either side of the chosen point. Carefully measure the straight-line distance between these two points of intersection (not the distance on the surface of the sphere, i.e., use calipers, not a bendy ruler or measuring tape). I stuck a pin into each of the two points as an aid to positioning the jaws of the caliper precisely at those points, as you can see at left. Call that measured distance F; in my case, I got 101.2mm. Call the original chosen distance (the one you drew the circle and arcs with) A. Then the radius of the sphere is ½A√((4A²-F²)/(3A²-F²)). With my numbers, A = 90 and F = 101.2, this came out numerically to 56.495… Hence, I conclude that the actual diameter of the styrofoam balls is 113mm, just slightly below 4 1/2 inches (numerically it comes out to 4.4488 in).

To see exactly why this works and how the formula is derived, check out the StackExchange page. A very pleasant case of the math actually working out to solve a practical problem!

14 Apr

Call for Puzzles

Hey, I have recently become problem editor for the undergraduate Math Horizons magazine of the Mathematical Association of America. So I’d love if you have problems/mathematical puzzles to submit to the column. The official submission blurb follows, and of course you will be credited in the Magazine.

The Playground features problems for students at the undergraduate and (challenging) high school levels. Problems or solutions (including more elegant or extended solutions to Carousel problems) should be submitted to MHproblems@maa.org or MHsolutions@maa.org, respectively. Paper submissions may be sent to Glen Whitney, ICERM, 121 South Main Street, Box E, 11th Floor, Providence, RI 02903 . Please include your name, email address, and school or institutional affiliation, and indicate if you are a student. If a problem has multiple parts, solutions for individual parts will be accepted. Unless otherwise stated, problems have been solved by their proposers.

11 Apr

Side of Spherical Square

This post contains the details of the claim made in “More Spherical Construction” that you can determine the side length of a spherical square from the ratio between the lengths of its diagonals. We’ll do this on a sphere of radius one; everything scales by a factor of the radius for a general sphere.

The advantage of working on a sphere of radius one is that then the side length is nothing other than the measure of the central angle (in radians) between two adjacent vertices of the square. Similarly, the diagonal is the central angle between opposite vertices of the square.

The easiest way to construct a square on the unit sphere is to place its center at the intersection of the sphere and the x-axis, and then choose a (positive) angle $\theta \leq \tau/4$ (where $\tau = 2\pi$ is the radian measure of a full circle) and place two vertices at plus or minus $\theta$ (from the square’s center) in the longitudinal direction and two vertices at plus or minus $\theta$ in the latitudinal direction. In spherical coordinates, the center is at $(1,0,\tau/4)$ and the four vertices are at $(1,-\theta,\tau/4), (1,\theta,\tau/4), (1,0,\tau/4-\theta), (1,0,\tau/4+\theta)$. For this square, the diagonal is obviously $2\theta$; it remains only to compute the side length of the square.

Using the great-circle distance formula on the two points $(1,\theta,\tau/4)$ and $(1,0,\tau/4-\theta)$ yields a side length of \[\arccos(\cos(\tau/4)\cos(\tau/4-\theta) + \sin(\tau/4)\sin(\tau/4-\theta)\cos(\theta)\;).\]

Since $\cos(\tau/4)$ is 0 and $\sin(\tau/4)$ is 1, this expression equals \[\arccos(\sin(\tau/4-\theta)\cos(\theta)).\]

And further, $\sin(\tau/4-\theta) = \cos(\theta)$, so the side length is just \[\arccos(\cos^2 \theta).\]

Therefore, the ratio of the diagonal to the side of the square is $2\theta/\arccos(\cos^2 \theta)$. Graphing this on the allowed interval $(0,\tau/4]$ for $\theta$ immediately demonstrates the claims made in the referring post: no value of the ratio is repeated for different values of $\theta$, so the ratio determines the diagonal (and side length) of the square; the maximum value is 2; and any value greater than $\sqrt2$ is achievable.

If you’d rather not rely on graphing software to extract these facts, but rather demonstrate them just from the formulas, here’s how you can proceed. The value 2 at $\theta=\tau/4$ comes from straight substitution. The value at 0 by substitution is the indeterminate 0/0, so we obtain the value by L’Hôpital’s rule: the derivative of the numerator is 2, but the derivative of the denominator is \[\frac{2\cos \theta}{\sqrt{1+\cos^2\theta}},\] so the diagonal-side ratio has the limiting value $\sqrt2$ at $\theta =0$. Finally, we need to show that the ratio is monotone increasing on the interval of interest. By the quotient rule, this is equivalent to showing that on this interval \[\arccos(\cos^2\theta) > \frac{2\theta\cos\theta}{\sqrt{1+\cos^2\theta}}.\] Both sides are 0 at $\theta=0$, so we can take derivatives one more time; the derivative of the left-hand side is of course $2\cos\theta(1+\cos^2\theta)^{-1/2}$ again, which on this interval is always larger than the derivative of the right-hand side, namely $2\cos\theta(1+\cos^2\theta)^{-1/2} – 2\theta\sin\theta(1+\cos^2\theta)^{-3/2}$. So the left-hand side is also always larger than the right hand side, i.e., the derivative of the diagonal-side ratio is positive on this interval.

10 Apr

Gengzhi Goblets

One might think that having produced prototypes of the Gengzhi Goblets, our work is just about done to produce sufficient quantity (roughly 300 of each) to serve as G4G13 giveaways. The question comes down to materials and expense. If the Gengzhi Goblets are actually to be used as measuring cups, then they need to be made from a food-safe material. There are not many options in the online 3-D printing world for food-safe materials. Other than bulky, heavy ceramic materials, I found just the Nylon PA12 from Sculpteo, and ordered a set of cups from them. The total came to over $200 for the set, clearly making 3D printing of 300 sets prohibitive.

For larger-quantity production, the industry standard appears to be injection molding, and there’s no difficulty in obtaining food-safe plastics from that process. So that’s the avenue I pursued in this case. The most economical supplier I found was Firedrake, although their lead times seem to be a bit longer than some other manufacturers. Ideally, you should be finalizing your designs at least three months before you need the items.

However, the next difficulty turned out to be that injection-mold designers don’t want an STL file, typically made up of hundreds or thousands of tiny triangular facets. They prefer STEP files, which consist of many fewer, more structured geometric elements: rectangular solids, cylinders, Bezier curves, and the like. And it’s an entirely different collection of programs for creating and manipulating STEP files. I did some modeling in FreeCAD (free and open-source), but ultimately had difficulty with intersections of two curved surfaces (e.g., in the attachment of the handle to the cup body) in that program, and so completed the cups in OnShape, (online, closed-source, but free for non-commercial use).

Many of the basic steps are similar in flavor with these “computer-aided design” programs. I could still import the STL from SageMath, but now I fit a Bezier curve to one ridge of the cup. Then I could make multiple copies of the ridge by rotating around an axis down the center of the cup, and then sweep out a surface from one ridge to the next. That created a membrane, which I had to thicken to a wall as before. Then I created a handle as a rectangular solid and used Boolean operations to unify it with the cup. To beef up the attachment to the cup I also combined in a half of an elliptical cone at the junction. Finally I could emboss the letters and puncture the hole in the handle (by Boolean-subtracting a cylinder). The final models look something like this:

Note, you can see that in such a design you typically include small bevels at most of the edges, and you also very slightly angle (by a few degrees) all of the “vertical” surfaces so that the piece will unmold more easily.

When your design checks out, you just send it off to the injection molding firm, and they do their magic from there. Here’s the mold that Firedrake produced,

and the completed Gengzhi Goblets have just arrived!