Word Construct

Welcome to 2021! This belated greeting was the result of my laying-back since last December, in part due to lack of interesting topics, and in part because I was writing elsewhere. That "elsewhere" is a new puzzles section in which I have swapped out my creative writing genes with math cells as I tinker around ways to solve various interesting problems around the web, most notably the FiveThirtyEight Riddler. I have since my last blog post decided to separate my musings and solutions (or at least, partial results) to those interesting puzzles in two separate sections on my website. The blog is for blogging only, as I decreed.

Well, for my first post of 2021, I have decided to merge the two sides of the human brain together and bring my solution to an interesting Riddler puzzle once again on my blog. For one, I needed to chop off my writer's block in order to get going for the rest of the year, and secondly, the condition was ripe for it with a 3D geometry question coming out of the Riddler. My lack of graphic design experience meant in order for me to share my solutions, I can either go the route of publishing it in my puzzles section, in which I would need to draw out painstakingly figure after figure the steps to solve such a geometry problem (3D no less!), or, hone my "word construct" skills, and try to use words to tell my readers how I solved the problem. I chose the latter in part because I was too lazy to draw anything, and secondly, yes, a picture is worth a thousand words, but typing is much easier.

"If you have young children (or if you’re still a child at heart), you probably have small blocks somewhere in your home.

I recently found four cubic blocks in a peculiar arrangement. Three of them were flat on the ground, with their corners touching and enclosing an equilateral triangle. Meanwhile, the fourth cube was above the other three, filling in the gap between them in a surprisingly snug manner. Here’s a photo I took of this arrangement:

If you too have blocks at home (I mean, of course you do), see if you can make the same arrangement.

Now, if each of the four cubes has side length 1, then how far above the ground is the bottommost corner of the cube on top?"

When I first approached the puzzle, naturally I wanted to replicate the setup to gain some intuition into this 3D geometry problem. Unfortunately, blocks aren't a common item just happen to lie around in my humble abode, and it's not that I am not a child at heart, but rather I don't have young children at home as I proudly proclaim to be living alone. I do have a couple of Rubik's cubes, though, but I figured I needed four. I searched around for applicable models when lo and behold--I found a rather morbid, but serviceable, alternative:

The tiny blocks sitting on the carpet, almost unable to plop down due to its own lack of weight, are some of the disease cubes found as tokens in the board game "Pandemic", which I believe is quite popular now given the times. I had to arrange the three bottom cubes on the carpet rather than on a table, so that the added friction can prevent the cubes from sliding away when I try to insert the fourth cube into that equilateral triangular "hole" in the middle. The downside, of course, was that the carpet surface was not even, and so the cubes could not sit level and hence concluded my attempt to replicate the setup.

That leaves only one viable path forward--my attempt to visualize the whole thing. As I closed my eyes and imagined the positioning of the top cube, I paid special attention to the angle in which the top cube entered the hole. Thinking hard about the setup led me to one realization--the hole would be protruded by only one corner of the cube. I know that sounds quite pedantic, but it is a key observation because it narrows down the shape of the volume that's inside the hole to a pyramid. Then it dawned on me: since the three bottom cubes sit level to the surface (unlike that in my replication attempt), the hole would "cut" the base of the pyramid to be parallel to the surface! In addition, since the corner of the cube is formed by three faces, the base of the pyramid, cut from the hole, would be a triangle!

Like an experienced detective, I gathered all the "clues" and tried to work out the specifics of this pyramid, which would be the key to solving the puzzle. So far, I know the shape of the protruding piece into the hole is a triangular-shaped pyramid, also known as a "tetrahedron". Further, the base of the pyramid is parallel to the surface, which means that its height is perpendicular to the ground. This means that if we let that height be \(h\), and we know the hole is formed by the confluence of three unit cubes, then the distance between the tip of the pyramid, i.e. the bottommost corner of the cube, would simply be \(\boxed{1-h}\).

Now we need to find \(h\). Let's further zone in on the specifics of our tetrahedron. For the base, since it is formed by the triangular hole, the base would be an equilateral triangle with side length of 1. We also need to know the distance between a vertex of the base to the tip of the pyramid, which I will denote as \(s\). This value is NOT 1, as I almost falsely assumed. In order to figure out the value, I used a trick called "similar pyramids". I imagined that instead of a side-length-1 hole, we had the smallest possible triangular hole that could fit the top cube. What would its side length be? The answer would \(\sqrt{2}\), or the length of the face diagonal of the cube. In that case, assume the ground is low enough, the piece protruding into the hole would again be a tetrahedron with the base parallel to the ground, but furthermore, it would have the same shape as the tetrahedron in question. The hole would effectively "cut" that cube in half. The big pyramid would have base length of \(\sqrt{2}\) and the value of \(s\) equal to 1, i.e. the side length of the unit cube. Put everything together, the ratio of base length to \(s\) would be \(\sqrt{2} : 1\). Since that big pyramid is similar to our tetrahedron in question, it means our tetrahedron's base length to \(s\) would also be \(\sqrt{2} : 1\). We know that our pyramid's base length is 1, so with the ratio information, we can figure out the value of our \(s\):

\begin{align} \frac{\text{base length}}{s} &= \frac{\sqrt{2}}{1} \\ \frac{1}{s} &= \sqrt{2} \\ s &= \frac{1}{\sqrt{2}} \\ &= \frac{\sqrt{2}}{2} \end{align}

Now, since the base is an equilateral triangle, each of the slope faces would be congruent triangles with one of the sides sharing with that of the base. Furthermore, by symmetry, each of the slope triangles would also be isosceles. Let \(d\) be the length of the median of one of the slope triangles from the tip of the pyramid to the base. By definition, the median intersects the base at the halfway point of the base, such that the distance from one base vertex to the intersection point of the median is exactly \(\frac{\text{base length}}{s} = \frac{1}{2}\). In addition, the median of an isosceles triangle extended onto the unique side is also its altitude, which means the median is perpendicular to the base. Therefore, we can apply Pythagorean theorem to figure out the value of \(d\):

\begin{align} d &= \sqrt{s^2 - \left(\frac{1}{2}\right)^2} \\ &= \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 - \frac{1}{4}} \\ &= \sqrt{\frac{1}{2} - \frac{1}{4}} \\ &= \sqrt{\frac{1}{4}} \\ &= \boxed{\frac{1}{2}} \end{align}

The fact that \(d\) is exactly one half of the base length of the pyramid, meaning that the slope, half of the base, and the median forms a 45-45-90 triangle, is not surprising, because the cube was inserted in such a way that its center of axis along the body diagonal is perpendicular to the ground. Applying symmetry, the above construct holds.

We are almost done here. The last right-angle triangle to consider would be the one formed by \(d\), \(h\), and a line segment connecting the intersection point of the median and the base to the drop point of \(h\) inside the base of the pyramid, of which we will denote its length as \(r\). Once we've figured out what \(r\) is, we can once again use Pythagorean theorem to figure out \(h\) as \(h^2 + r^2 = d^2\).

Spoiler alert: the drop point is the centroid of the base triangle. This can be shown by symmetry, as the base is equilateral. This means that for the tetrahedron to form, the point right below the tip of the pyramid must be the base triangle's center of mass, which by definition is its centroid. The centroid is also the intersection point of the three medians of the triangle. In the case of an equilateral triangle, each median is also an altitude (or height) of the triangle. We can make use of a well-known property that the centroid bisects the median in two parts of ratio \(2:1\), as the vertex-to-centroid segment has twice the length as that from the centroid to the opposite side. Our \(r\) would be the length of the latter segment. We find the length of a median of the base equilateral by using the fact that the median is also the altitude and subsequently applying Pythagorean theorem on the 30-60-90 triangle formed by cutting the equilateral triangle in half. The short side has length \(\frac{1}{2}\), and so the long-side, being the length of the median, would have length \(\frac{\sqrt{3}}{2}\). The value of \(r\) would be a third of that length, meaning that \(r = \frac{1}{2\sqrt{3}}\).

Putting everything together: \begin{align} h &= \sqrt{d^2 - r^2} \\ &= \sqrt{\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2\sqrt{3}}\right)^2} \\ &= \sqrt{\frac{1}{4} - \frac{1}{12}} \\ &= \boxed{\sqrt{\frac{1}{6}}} \end{align}

And so the answer to our original puzzle would be: \(1 - h = \boxed{1 - \sqrt{\frac{1}{6}}}\) which is approximately 0.592 unit above ground.

The Volume

We can further figure out the volume of the pyramid inside the hole, by leveraging our value of \(h\) and the area of the base. The volume of the pyramid is:

\begin{equation} \text{Volume of tetrahedron} = \frac{1}{3} \times \text{base area} \times \text{height} \end{equation}

Since our base is an equilateral triangle, we can use its side length to evaluate the area as:

\begin{align} &\text{Area of equilateral triangle}\\ &= \frac{\sqrt{3}}{4} \times \text{side length}^2 \\ &= \frac{\sqrt{3}}{4} \times 1^2 \\ &= \frac{\sqrt{3}}{4} \end{align}

\begin{align} &\text{Volume of the protruding piece}\\ &= \frac{1}{3} \times \text{base area} \times \text{height} \\ &= \frac{1}{3} \times \frac{\sqrt{3}}{4} \times \sqrt{\frac{1}{6}} \\ &= \frac{1}{12\sqrt{2}} \\ &= \boxed{\frac{\sqrt{2}}{24}} \\ &\approx 0.0589 \quad \text{unit cubed} \end{align}