So in a fit of desperate boredom in a meeting, my friend at work starts sketching triangles. I ask him what’s up with the triangles after the meeting, and he says he was trying to figure out whether you can put an equilateral triangle on a piece of graph paper such that all three points of the triangle are on spots where the vertical and horizontal lines meet. Basically, thinking of an xy coordinate system, where the x and y coordinates of all three points are integers.
One can quickly demonstrate that you can’t do this if one of the lines of the triangle is parallel to one of the axes of the coordinate system, because the height of the triangle is (square root of three)* (1/2 the length of a side). Thus, for any integer length of a side, you’ll never get an integer height. So having the bottom of the triangle line up with the x-axis of the graph paper you’re doodling on doesn’t work. Too bad, because that would be the easy way to do it.
I suspect that it’s similarly impossible to come up with three integer-value points that form an equilateral triangle in any case, whether one of the triangle legs is parallel to an axis or not. But since we’re not mathematicians, we couldn’t come up with a layman’s version of a proof that you can’t. This is maybe for the best since it allows the limited entertainment value of trying to find one that works by trial-and-error doodling on graph paper.
Nevertheless, can some mathematician verify for me that this is impossible, even with an infinitely large piece of graph paper? Assuming it is indeed impossible, can anyone explain why it’s not possible in terms that a couple guys who dropped out of math after Calculus 2 might understand?
If s is the length of any side of the equilateral triangle, then its area is s[sup]2[/sup]sqrt(3)/4. Since the vertices have integer coordinates, this area is irrational. However, Pick’s theorem implies the area is rational. Contradiction, hence impossible.
Alrighty, I’ll give this a try. Let’s see if I can scrape these cobwebs off…
Let’s focus on this.
Because we are talking about an equilateral triangle, let’s get a generic formula for the height of such a triangle. If we draw a line down the center of an equilateral triangle (splitting it into 2 equal, non-equilateral triangles), we can denote it as the height… let’s call it y.
Since we divided the triangle into two, we will just work with one of the halves. So, the hypotenuse of this “new” triangle is now x, where x is the length of one of the sides of the original equilateral triangle. The base of the new triangle is now x/2, since we divided the original triangle in half. The height of the new triangle, is y (from above).
So, let’s get a definition of y, in terms of x. This can be done using the Pythagorean Theorem, and basic algebra. My calculations brought me to the equation y = (sqrt(3)/2) * x (hopefully I didn’t make any algebraic mistakes).
We know that sqrt(3) is an irrational number, so (sqrt(3) / 2) is also an irrational number (a number that cannot be represented as a fraction).
Looking back at where we started:
let’s replace y = (sqrt(3)/2) * x with f(x) = (sqrt(3)/2) * x. We can see that f(1) will not yield an integer. Same with f(2). So we need to prove f(x) != integer, for all x, such that x is an integer.
Actually, let’s just look at the equation some more, instead (if you want to do a formal proof, you should be able to do these next steps using Mathematical Induction). We know that sqrt(3)/2 is an irrational number, and we know that sqrt(3)/2 * x is also an irrational number. Since it is an irrational number, then it can’t really have an inverse, can it? The only way to satisfy the above equation (f(x) = (sqrt(3)/2) * x) is essentially by finding the inverse of the right hand side of the equation. Since that cannot be done (yielding an integer as a result), there you have it.
(Hope that made sense)
LilShieste
On preview, it looks like Cabbage has provided a way too.
Without loss of generality, let one of the triangle’s vertices be fixed at (0,0). Then, let the second vertex be at (x1,y1). There are then two choices for placing the third and last vertex, (x2,y2), to make an equilateral triangle. I’ll pick the one that’s closest as you revolve “counterclockwise” around the origin from (x1,y1), but the argument I’m making would apply analogously to the other choice too.
The coordinates x2 and y2 are now determined, and can be expressed in terms of x1 and y1: (x2,y2) = 1/2 (x1,y1) + (sqrt(3)/2) (-y1, x1). Or:
Given that you’ve required x1 and y1 to be integers, it immediately follows from the above expressions that neither x2 nor y2 can be rational, and therefore certainly not integers. (An irrational multiplied or divided by a rational is irrational; also, an irrational plus or minus a rational is irrational.)
I’m probably missing something obvious, but does this theorem really help? Although the triangle’s coordinates are required to be integers, the side-length can certainly be irrational, and therefore the area as well.
With an infinitely large piece of paper and an infinitely large triangle, I think you can do it. In the language of Pick’s Theorem, (I+B/2-1)/s^2 = sqrt(3)/4, where I, B and s^2 are integers. By drawing larger and larger triangles, I can make (I+B/2-1)/s^2 as close to sqrt(3)/4 as you like.
For what it’s worth, this can be generalized somewhat.
The area of a regular n-gon with side s has area (s[sup]2[/sup]n/4) / tan(pi/n) (assuming I didn’t screw it up). I can’t think of a simple proof offhand, but I’m virtually certain that this is rational if and only if n=4, and, again by Pick’s theorem, we see that the only regular n gon with integer coordinates is a square.
Generalizing a bit more, I happen to know (but don’t know a proof) that if we extend beyond the plane to an arbitrary number of dimensions, the only regular polygons with integer coordinates are the triangle, square, and hexagon.
This is actually pretty easy to see. Label the points of the polygon P[sub]1[/sub],…,P[sub]n[/sub]; let P[sub]1[/sub]=(0,…,0) be at the origin, where P[sub]i[/sub] denotes the position vector from the origin to P[sub]i[/sub]. (So |P[sub]2[/sub]|=|P[sub]n[/sub]|=s the side length.) Now the integer-coordinate vectors in the plane of the polygon (including the P[sub]i[/sub]) form a lattice, so v = P[sub]3[/sub] - (P[sub]2[/sub] + P[sub]n[/sub]) is in the lattice as well. But v is parallel to P[sub]2[/sub] and has length 2 s cos (2 pi/n), so unless cos(2 pi/n) is rational the lattice has a dense one-dimensional sublattice (generated by P[sub]2[/sub] and v), not possible. But cos(2 pi/n) is only rational for n=2,3,4,6.
He probably would have been happy to find: (0,4) - (11,15) - (15,0). Not quite equilateral, but the distance difference is probably below the tolerances of your handy ruler: the sides are 15.52, 15.52, and 15.55
I’m not disputing the theory given before, just pointing out that in the real world, you can specify a triangle that meets the integer coordinate criteria to any given level of tolerance.
