From our friends at the Riddler:
Robert’s daughter has a set of Magna-Tiles, which, as their name implies, are tiles with magnets on the edges that can be used to build various polygons and polyhedra. Some of the tiles are identical isosceles triangles with one 30 degree angle and two 75 degree angles. If you were to arrange 12 of these tiles with their 30 degree angles in the center, they would lay flat and form a regular dodecagon. If you were to put fewer (between three and 11) of those tiles together in a similar way, they would form a pyramid whose base is a regular polygon. Robert has graciously provided a photo of the resulting pyramids when three and 11 tiles are used
If Robert wanted to maximize the volume contained within the resulting pyramid (presumably to store as much candy for his daughter as possible), how many tiles should he use?
Results
Let’s save the trig for a moment, and check out the results!
Instead of assuming that it lies flat with 12 tiles, let’s try out some other values.
Brace yourself - the trig is coming!
Approach
The base
Let’s start by looking at the base of the polyhedron.
The base is a regular polygon with \(n\) sides. We can divide that regular polygon into \(n\) triangles by connecting the vertices to the center. (We’re making wedges.).
The interior angle \(\theta\) for those \(n\) triangles is \(\theta = \frac{2\pi}{n}\).
And then we can divide each triangle again in half by dropping a perpendicular
from the center to the side. How long is that perpendicular, \(p\)? Well, we have
basically the same equation as before, just using a different number of sides
for the regular polygon.
$$
The tile
Now, let’s look at the tiles that form the pyramid’s walls. For a moment,
let’s imagine we had all the tiles and laid it out flat. Then the work we
just did for the base can tell us the height of each tile, \(h_t\)!
We’re using \(n\) to reference how many tiles we choose to use. So we’ll use \(f\) for the number of tiles that allows us to lay the shape out flat. (In our setup, \(f = 12\).)
\[ \begin{aligned} h_t & = (\tfrac{1}{2}a) \cot (\frac{\pi}{f}) \end{aligned} \]
The wedge
With this perpendicular distance \(p\), we can get the base area of a single wedge. \[ \begin{aligned} A_w & = \tfrac{1}{2} \cdot p \cdot a \\ & = \tfrac{1}{2} \cdot (\tfrac{1}{2}a) \cot (\frac{\pi}{n}) \cdot a \\ & = \tfrac{1}{4} a^2 \cot (\frac{\pi}{n}) \end{aligned} \]
Finally 3-D time! Let’s find how tall these wedges are! With the perpendicular distance along the base \(p\), and the tile height \(h_t\) we obtained earlier, we can express the wedge height \(h_w\):
\[ \begin{aligned} h_t^2 & = p^2 + h_w^2 \\ h_w^2 & = h_t^2 - p^2 \\ & = [ (\tfrac{1}{2}a) \cot (\frac{\pi}{f})]^2 - [(\tfrac{1}{2}a) \cot (\frac{\pi}{n})]^2 \\ h_w & = \sqrt{[ (\tfrac{1}{2}a) \cot (\frac{\pi}{f})]^2 - [(\tfrac{1}{2}a) \cot (\frac{\pi}{n})]^2} \\ & = (\tfrac{1}{2}a) \sqrt{ \cot^2 (\frac{\pi}{f}) - \cot^2 (\frac{\pi}{n})} \\ \end{aligned} \]
The volume
The volume of a single wedge \(V_w\) as a function of \(n\) is given by: \[ \begin{aligned} V_w & = \tfrac{1}{3} \cdot A_w \cdot h_w \\ & = \tfrac{1}{3} \cdot [\tfrac{1}{4} a^2 \cot (\frac{\pi}{n})] \cdot [(\tfrac{1}{2}a) \sqrt{ \cot^2 (\frac{\pi}{f}) - \cot^2 (\frac{\pi}{n})}] \\ & = \frac{1}{24} \cdot a^3 \cdot \cot(\frac{\pi}{n}) \sqrt{ \cot^2 (\frac{\pi}{f}) - \cot^2 (\frac{\pi}{n})} \end{aligned} \]
And let’s not forget that we’re trying to get the total volume, \(V\): \[ \begin{aligned} V & = n \cdot V_w \\ & = n \cdot [\frac{1}{24} \cdot a^3 \cdot \cot(\frac{\pi}{n}) \sqrt{ \cot^2 (\frac{\pi}{f}) - \cot^2 (\frac{\pi}{n})}] \end{aligned} \]
The function is a little hairy. There might be some secret trig identity or
limit to apply here, but let’s try just stuffing this into R’s optimize()
function and see what happens.
## [1] 0.8164883Something in the neighborhood of 81.6% of the number of tiles that lay flat seems to maximize the volume.
Code
Full code available as a RMarkdown file on Github
Calculate multiple scenarios
Summary table
# Summary for 12-tile
options(digits = 3)
summ %>%
filter(n_flat == 12) %>%
select(n, h_t = tile_height, base_perp = base_center_side_dist,
base_wedge_area, wedge_height, wedge_volume,
total_volume) ## # A tibble: 10 x 7
## n h_t base_perp base_wedge_area wedge_height wedge_volume total_volume
## <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 3 1.87 0.289 0.144 1.84 0.0887 0.266
## 2 4 1.87 0.5 0.25 1.80 0.150 0.599
## 3 5 1.87 0.688 0.344 1.73 0.199 0.995
## 4 6 1.87 0.866 0.433 1.65 0.239 1.43
## 5 7 1.87 1.04 0.519 1.55 0.268 1.88
## 6 8 1.87 1.21 0.604 1.42 0.286 2.29
## 7 9 1.87 1.37 0.687 1.26 0.289 2.60
## 8 10 1.87 1.54 0.769 1.06 0.271 2.71
## 9 11 1.87 1.70 0.851 0.763 0.217 2.38
## 10 12 1.87 1.87 0.933 0 0 0Plot main answer
summ %>%
filter(n_flat == 12) %>%
ggplot(aes(n, total_volume)) +
geom_point() +
theme_light() +
labs(
title = "Polyhedra volume for n-sided regular polygon base",
subtitle = "Assumes 12 tiles lay flat, side length = 1",
x = "n", y = bquote("Volume" ~ (units^3))) +
gghighlight(total_volume == max(total_volume), label_key = n,
unhighlighted_params = list(colour = "grey60")) +
scale_x_continuous(breaks = seq(0, 12, 2))Optimize for maximum volume
tot_volume <- function(q, n_flat = 10000, side = 1) {
n <- q * n_flat
tile_height <- (side / 2 ) / tan(pi / n_flat)
base_center_side_dist = (side / 2) / tan(pi / n)
base_wedge_area = (1/2) * base_center_side_dist * side
wedge_height = sqrt(tile_height ^2 - base_center_side_dist ^ 2)
wedge_volume = (1/3) * base_wedge_area * wedge_height
total_volume = n * wedge_volume
return(total_volume)
}
fr_flat_tiles <- optimize(tot_volume, lower = 0, upper = 1,
maximum = TRUE)$maximum
fr_flat_tiles