Probably the best is to cut it radially (like a pizza). It can be tricky working out exactly where to make the cuts, but if you can pull it off, then all the pieces will be roughly identical (topping distribution notwithstanding).
But for the sake of arguing, lets say you want to divide the burger by making two parallel cuts: Where, then, do you make the cuts so that all three people get the same amount of burger?
NB/ This gets quite maths-heavy, so if you're not interested in that sort of thing, feel free to skip right to the end for the solution.
For simplicity, we're going to consider the burger as a circle, and make the cuts so that each chunk has the same area. The two cuts are going to be the same distance from, and parallel to, the central axis of the burger, so we only need to consider the position of one of the cuts.
Here's the set-up
- Aside: Radians
Radians are basically an alternative way of measuring angles. For maths and physics they're generally more useful than degrees.
They're relatively easy - there are 2pi radians in a full circle, so 2pi radians = 360 degrees
1 radian = 180/pi = 57.3 degrees
1 degree = pi/180 = 0.017 radians, etc.
* * *
Back to the circle; with angle x in radians, the area of the circular segment is
- Aside: Area of the Triangle
We start by splitting the triangle down the middle, so that we have two identical right angle triangles
The base, b = 2*(r*sin(x/2))
So the area of the triangle is (l*b)/2 = r^2 sin(x/2)cos(x/2)
Finally, use the identity sin(2x) = 2sin(x)cos(x)
* * *
So the area of the cut-off is
So first, we need to find x satisfying
The thing about this equation is it doesn't have an exact, analytical solution - to find the solution you have to use numerical methods. Well, I say you have to use numerical methods; these days you can just type the equation into WolframAlpha, and you'll get a solution like *snaps fingers*
Which is nice. I even have the WolframAlpha app on my phone. But when I thought up this question I was on holiday in Sherwood forest, where there was literally no mobile singal.
So that was out of the question. And since I'm not in the habit of carrying a scientific calculator around with me, I was stuck with the basic calculator on my phone. It looks like this:
Anyway, there are two ways of working this out with only a basic calculator. The first is 'easier', but only if you know some stuff, and the numbers happen to be nice (in this case, they kind of are). The second is harder, in that it requires more number crunching, but it'll work with any numbers, and can be more precise.
Again, feel free to skip to the solution if you're not interested in the gritty details.
First of all, here's a graph of the two sides of the equation
|hand-drawn with Skitch|
For the right hand side of the equation: 5pi/6 - 2pi/3 = 0.52
NB/ I'm using pi=3.1416 (rounded to 4 decimal places). If you prefer, you could use the approximation 22/7. The result should be roughly the same.
For the left hand side of the equation, we need to work out sin(5pi/6)
At A-level, we were expected to memorise sin() and cos() of angles 0, 30, 45, 60, 90, and 180 (degrees). We were also expected to know the formulas for sin() and cos() of sums of angles. For sin(), it works like
So our approximate value of x is 5pi/6 = 2.618
[Incidentally, the identity for sin(2x) is just a special case of the above, with a=b=x; i.e. sin(2x) = sin(x+x) = 2sin(x)cos(x)]
Now we just need to work out l/r = cos(x/2) = cos(5pi/12)
For this one, 5pi/12 rads = 75 degrees = 45 + 30, so we can use
But I'm just weird like that. Let's say you don't. How do you work it out?
- Aside: Square Roots
There are several ways of working out square roots with just basic operators. For two easy examples:
The first is 'Trial and Improvement' - pick a number, square it, does that give the right answer? If not, pick another number based on whether the last guess was too big or too small.
For example: sqrt(3)
1.5 -> 2.25 -> too small
1.7 -> 2.89 -> too small
1.8 -> 3.24 -> too big
1.75 -> 3.0625 -> too big
1.73 -> 2.9929 -> too small
1.74 -> 3.0276 -> too big
1.735 -> 3.010225 -> too big
1.7325 -> 3.00155625 -> too big
1.732 -> 2.999824 -> too small
The second method is the "Babylonian Method". It's more systematic, and can converge to the correct answer quicker than guessing. But it can be irritating if your calculator doesn't have a memory function.
It uses the recurrence relation
For example: sqrt(2)
x0 = 1.5 -> 2/1.5 = 1.33
x1 = (1.5 + 1.33)/2 = 1.4166.. -> 2/1.4167 = 1.41176..
x2 = (1.4167 + 1.41176..)/2 = 1.41421.. -> 2/1.41421 = 1.41421..
* * *
Whatever way you do it, you repeat the process until you get the degree of accuracy you're happy with.
You should get the answer around l/r = 0.259
We go back to the equation sin(x) = x - 2pi/3
We still have to find the solution numerically, we still don't have a calculator with a sin() function, and this time the numbers don't work out nicely.
So, the question is, how do we calculate sin(x)?
- Aside: Taylor Expansion
The Taylor Expansion of a function is a way of fitting a polynomial (sums of powers) to a more complicated function. It works like this
It's usually expanded around the origin (x0=0), since the equations work out neater. But you can do it around any point, x0=a. This is useful if the value you are trying to calculate is far from x=0. The closer x is to x0=a, the quicker the sum converges.
Even though the expansion is an infinite sum, it's usually sufficient to just take the first few terms, since each additional term makes a smaller and smaller contribution to the sum.
So the trick is working out how many terms you need to include to get some desired level of accuracy.
* * *
In this case, I'm going to use the Taylor Expansion of sin(x) around x0=pi, since the approximate value (2.6) is nearer to pi than 0.
Here's what the expansion looks like
So, how many terms do we need to include?
Here's what the graph looks like for different numbers of terms
|plotted with WolframAlpha|
So I would probably go to the third term (for 5dp), but only take the result to 3dp.
NB/ We shouldn't get too hung up on getting an extremely accurate value for x, since we're already getting rounding errors from the factors of pi in the expansion. Also, since we're calculating x to 5dp, we should use pi=3.14159
That means we want to solve
So, for finding the correct value (without WolframAlpha), we can use any root-finding method. For what it's worth, I used Trial and Improvement; the other methods are easier with a computer.
But, note that the function is decreasing
If you run through all that (I won't go into detail), it gives a value around x=2.605
Alternatively, you could expand around x0=5pi/6 (if you know/can work-out sin and cos of 150 deg without a calculator).
In this case you'd only need up to the term in x^2 (correct to ~4dp). Using this expansion would mean solving a quadratic equation, which is easy. But using this expansion can introduce more rounding errors from the factors of sqrt(3). It's a matter of preference, I guess. The answer should be about the same.
Finally, we need to calculate cos(x/2)
Again, we use the Taylor Expansion to calculate cos(). In this case, we're doing the expansion around x0=0; the expansion is
This gives a value around l/r = 0.265
So What is the Real Answer?
Once I got to somewhere where I could get at WolframAlpha, I checked the real numbers; here are the results:
acceptable. The approximation from Method Two (2.605) is correct to 3 decimal places, which is definitely acceptable.
And for the value of l/r
So, if the numbers happen to be convenient and you know some trigonometry, you're probably as well using Method One. If not, or if you just want more accuracy, then go for Method Two.
Applying the Results
The results are actually quite nice, in terms of practical application (dividing up a burger). The ratio of the radius (0.265) being close to one quarter, you find the cuts like this
So, now you know. Obviously, all this applies to dividing any circular thing evenly between three people. You could probably even adapt the methods for sharing between even more people.
And in theory, You could do all this with just pen and paper (no calculator). Though you probably wouldn't want to. I know I wouldn't..
[Wow, I really managed to stretch that one out.]