Ernie and the Geometric Ginger Cookies

Question

Ernie loves cooking and it's always worth the wait as he follows his curiously precise recipe to make Ginger Crunch Cookies. So I was happy to drop around when he invited me to help test his latest invention - a pneumatic cookie-cutter.

"I've made two of them," said Ernie. "They use pneumatic actuators to transfer all the dough inside the cut area onto the cooking tray. This one cuts squares and the other one cuts equilateral triangles."

"The triangular cutter seems a bit bigger", I commented.

"Only in linear dimensions", Ernie responded. "The triangle has exactly the same $10,000\text{ mm}^2$ area as the square".

Ernie already had the dough rolled out so it covered most of the surface of the large rectangular cooking bench and was about to cut the cookies, when he noticed that the readouts of the twin thermometers in his 'cooking furnace' disagreed by more than 0.6°K.

"Can't have that", he muttered as he reached for his screwdriver to make some necessary adjustments. "Can you cut them out for me? The instructions are at the end of the recipe."

I turned to the last page of his cooking notes and was more than a little nervous when I read the following instructions:

Cutting

Step A. Using the square cutter:
1) Position the cutter somewhere over the table
2) Rotate the cutter to an angle $\theta$ relative to one edge of the table
3) Make a cut and transfer the square piece of dough inside the cut region to the tray.
4) Repeat step 3, with the cutter at the same angle $\theta$, at all positions on the table displaced $md$ horizontally and $nd$ vertically (where $m$ and $n$ are integers $\{...-2,-1,0,1,2...\}$ and $d$ is a chosen displacement), and transfer the square piece of dough inside each cut square region to the tray.

Step B. Using the triangular cutter:
1) Position the cutter at a chosen position over the table
2) Rotate the cutter to an angle $\phi$ relative to one edge of the table
3) Make a cut and transfer all the pieces of dough inside the cut region to the tray.
4) Repeat step 3, with the cutter at the same angle $\phi$, at all positions on the table displaced $md$ horizontally and $nd$ vertically (where $m$ and $n$ are integers $\{...-2,-1,0,1,2...\}$ and $d$ is the same displacement as used for the square cutter), and transfer all the pieces of dough inside each cut triangular region to the tray.

Note. Choose positions and angles of the square and triangular cutters so you can use the largest value of $d$ possible, such that after following steps A and B, there is no dough left on the table (except maybe for a little bit round the edges of the table)

"So you want a set of perfectly square cookies - that don't overlap one-another?", I asked. "Correct", Ernie replied. "Plus, the odd-shaped bits that the triangular cutter will make in Step B...?". "Correct...", Ernie agreed again. "...and of course, make sure that you use the biggest value of $d$ possible! Because that's the only way to cut perfect cookies".

I tried sketching a few test patterns to find a suitable tiling on a spare copy of one of Ernie's recipes, but couldn't sort out the optimal value for $d$. So while still repairing the oven, Ernie drew a quick diagram on the dusty floor, and annotated it with angles accurate to the nearest second of arc, and distances to the nearest micron. I cut the cookies at the required spacing and sure enough, after the last triangular cut was made all the dough was gone. The cookies made it into the oven in time and were a great success.

Now I have a bit of a problem. I borrowed Ernie's cookie cutters and plan to make him some perfect ginger cookies for his birthday, but can't remember Ernie's diagram or distance measurements for efficient cuts. Can you help me? Accuracy to the nearest $\frac1{10}^{th}$ of a mm and $\frac1{10}^{th}$ of a degree (and maybe a simple diagram) should be enough.

Answer 1

OK, well first, it would be useful to calculate the dimensions of the cutters. The area of each cookie, square or triangular, is $10000\text{ mm}^2$, or $100\text{ cm}^2$. That's easy enough for the square; the cutter is $10\text{ cm}$ ($100\text{ mm}$) on a side. The triangle's a bit trickier; an equilateral triangle is basically two adjacent 30-60-90 triangles, and the unit measurements of such a triangle (assuming a hypotenuse of $1$) are $\frac12$ for the short side, and $\frac{\sqrt3}2$ for the long leg. So, the area of one 30-60-90 triangle with hypotenuse length $a$ is $\frac a2 \cdot \frac{\sqrt3a}2 \cdot \frac12$, or $\frac{a^2\sqrt3}8$. Two of those make an equilateral triangle, so the area of that is $\frac{a^2\sqrt3}4$. We need the value of $a$ that makes the area $10000$; substituting and rearranging variables gives us $\sqrt{\frac{40000}{\sqrt3}}$, or $151.9 \text{ mm}$ per side ($15.19\text{ cm}$).

So, we need an arrangement of squares, rotated from square with the table by some arbitrary angle $\theta$ and spaced by $m$ and $n$ multiples of $d$ in each direction such that they don't overlap, then an arrangement of triangles which can overlap, such that using the same $d$ but different $m$, $n$ and angle $\phi$, all dough in the area being cut is used.

The simple answer is $d = 10 \text{ mm}, \theta = 0$, and all $m,n = 1$. Additional measurements for the triangular blade are immaterial as this solution produces only the square cookies; there will be nothing left for the triangle blade to cut. This is a working answer but probably not the largest possible $d$ value. So, we must space out the squares to allow the triangular blade something to cut, but not so much that dough is left behind.

The ideal solution, IMO, is

an arrangement of rotated squares, such that the edges of adjacent squares touch along a portion of the full length of each side, and the space between them is the largest possible square small enough to fit within the triangle cutter, so the triangular cutter essentially doesn't make triangles, it simply picks up the smaller squares. $d$ is then the displacement that allows a rotation of the squares by the angle $\theta$ to create this leftover space, for the triangle cutters, also rotated to $\phi = \theta$ to come in behind and collect.

So, let's start by finding the dimensions of

the largest square that could fit into an equilateral triangle $15.2\text{ cm}$ on a side. By doing so, we essentially split the triangle into four regions; the square, an equilateral triangle above, and two 30-60-90 triangles to either side.

What we're looking for, given the length of the side of the triangle $a$, is a side length $b$ for a square, such that

the length of each side of the smaller equilateral triangle produced by inscribing the square will also be $b$.

Now, if the full triangle has height $h$ and side length $a$, then the height of the smaller triangle will be $h-b$ (as the square will have height $b$ and width $b$). The proportion of height to base of the smaller and larger triangles will be the same because they are geometrically similar (same angles, different dimensions), therefore $\frac{h-b}b = \frac{h}a$. We know $\frac{h}a = \frac2{\sqrt3}$, and so $h = \frac{a\sqrt3}2$, therefore substituting and solving for $b$, $b = \frac{a\sqrt3}{2 + \sqrt3}$. We know the triangle is $15.2\text{ cm}$ on each side, therefore with a little calculator work, the smaller squares will be

$7.054\text{ cm}$ on a side,

and therefore, adjacent squares' edges will touch along

$10 - 7.054 = 2.94\text{ cm}$

of their length.

So, what angle of rotation $\theta$ of the squares from being orthogonal to the cutting sheet gives us this amount of overlap? And what spacing between the centers of the squares $d$ will ensure the squares don't overlap? That's a bit more complex. We know that at $\theta = 0$, the overlap is $10\text{ cm}$, and the $d$ that causes the edges to line up perfectly is also $10\text{ mm}$. Now, squares are made of two 45-45-90 triangles, and the hypotenuse of such a triangle given unit side lengths is $\sqrt2 \approx 1.414$. So, a rotation $\theta=45°$, and $D=10\text{ cm}\sqrt2$ would cause the squares' edges to intersect at a single point.

Now, if I had something I could use to illustrate this next point, I would, but I don't, so you have to draw it out yourself and confirm it. If you put two squares together so that some but not all of an edge is touching, then draw a line between those two squares' bottom corners, that line would be parallel to and the same length as the line between their centers, because those two lines and lines connecting center to corner would create a parallelogram. In addition, that line forms the hypotenuse of a triangle with the bottom of one square and the side of the adjacent one, and we know the lengths of those sides

($10\text{ cm}$ and $7.054\text{ cm}$).

By the Pythagorean Theorem, $a^2 + b^2 = c^2$, therefore

$d = \sqrt{10^2 + 7.054^2} = \sqrt{149.763} = 12.2378\text{ cm}$.

In addition, we know two sides of a triangle, and we want the angle $\theta$ made between the longer leg and the hypotenuse. Remember "some old hippie caught another hippie tripping on acid"?, tan = opp/adj, so

$\tan\theta = \frac{7.054}{10}$, thus $\theta = 35.199°$

So there you have it:

$d = 12.24\text{ cm}$ or $122.4\text{ mm}$. Both $\theta$ and $\phi$ are $35.199°$ assuming that zero for both shapes makes at least one edge of the shape parallel to the top and bottom of the cutting sheet, and the grid for both square and triangle cutters is a unit grid of squares of length $d$, so all $m$ and $n$ are 1.

Any bigger, and

the square-shaped areas in between the square cutouts would be too big for the triangle cutter to remove completely, leaving small 30-60-90 triangles laying around.

I could, with a little more time, figure out the relative initial position of the triangle given the center of the square, but that was not part of the puzzle.

I have added a diagram of the tiling below - Penguino