Tuesday, July 25, 2006

Actually, maybe I don't deserve rights

... where "rights" is a clever allusion to the clever pun I made in my last title. I actually do deserve rights.

An astute mathematician pointed out to me that there is a fatal error in my reasoning in my last post, so ridiculously terrible that I am thinking about deleting the whole post and pretending it never happened. But I'm honest, so I won't delete it; I'll leave it for historical interest. The key is this: Triangles actually have to be on the plane x+y+z=180. Inside the pyramid, there are points like (1,2,3) which clearly are not the angles of any triangle. So, welcome to the plane.

The plane intersects the first octant in an equilateral triangle with vertices at (180,0,0), (0,180,0), and (0,0,180). An equilateral triangle appears at the centroid (60,60,60), and isosceles triangles appear along the medians, which I am not going to bother expressing in parametric form. I think that right triangles appear along the midlines (connect the midpoints to make a small equilateral triangle).

Here is what I propose: Draw in all the medians. This creates six 30-60-90 triangles. I assert that each one carries exactly the same triangles, just with the angle coordinates in a different order. There are six triangles, which corresponds to the 3!=6 ways to rearrange three different numbers. There are three different hypotenuses and three different short legs, which correspond to the isosceles triangles. The 3 corresponds to the 3!/2! = 3 ways to rearrange three numbers when two of them are the same. And there is one center point, because there is only one way to "rearrange" 60-60-60.

Thus, we need only inspect one of these small 30-60-90 triangles. The coordinates of one of them, for example, are (60,60,60), (0,0,180), and (0,90,90). In my picture, this is the one on the top right.

Now look at the midline that cuts across each small triangle (from (45,45,90) to (0,90,90) in the example). The midline acts as an altitude drawn from the right angle to the hypotenuse. This cuts the 30-60-90 triangle into two smaller, similar triangles, one of which is 1/4 the area and one of which is 3/4 of the area of the original. The smaller one, whose vertices are (60,60,60), (45,45,90), and (0,90,90) in the example, contains all of the acute triangles. The larger triangle contains the obtuse triangles, and the right triangles are on the dividing line.

So, in conclusion, 1/4 of triangles are acute, 3/4 are obtuse, and none of them are right. QED.

6 comments:

Daniel said...

But what if I like my triangles to have their side lengths be random or the plane locations of their vertices be random? Of course, working with random angles is nice because you can put a uniform distribution on bounded intervals like [0, \pi). It's not so easy to do that on [0, \infty) although there some neat Bayesian statistical tricks where you pretend that you can.

Brent said...

Hrm, well, it seems I didn't notice that either. =) Well, at any rate, your new analysis seems correct, although I think it can be simplified a lot: starting with the equilateral triangle with vertices at (180,0,0) (0,180,0) and (0,0,180), draw in lines connecting the midpoints to form a smaller equilateral triangle. As you noted, these lines connecting the midpoints in fact represent all right triangles. But notice too that all the points *inside* the smaller equilateral triangle represent all acute triangles. Now we can simply note that this triangle has 1/4 the area of the big triangle.

It would be fun (and not too hard) to build a simple interactive Java applet or somesuch where you can drag around a point inside the "angle parameter space" (i.e. the big equilateral triangle) and see the resultant triangle updated in real-time. I might give it a whirl, but unfortunately I don't have much experience developing Java applets.

Diana said...

Brent, the smaller equilateral triangle with 1/4 the area is just six copies of the one that I was looking at, but mine contains no overcounting. I'm glad to see we're on the same page. It sounds like fun, the real-time triangle, but I have no experience in Java applets. Too bad.

Voj. Ses. said...

Hi,
came across this post via ephblog. I think that you guys are complicating the problem, and that there is an error in your reasoning.

In my opinion, you do not need 3 axes to represent all the triangles, but only 2 axes. The reason for this is that if you know 2 angles in a triangle, the third is NOT any random angle, but it is already determined (i.e., given x and y, z is 180-x-y).
Thus, all triangles can be represented by a triangluar surface in xy plane bounded by the coordinates (0,0), (0, 180), (180,0). Now, suppose we pick at random a point within that surface. The point we picked at random represents an acute triangle iff it is within (0,0), (0,90), (90,0), (90,90) square.

This square encompasses 50% of the area of the surface (0,0), (0, 180), (180,0). Thus, the probability of a random triangle being acute is .5, and so is the probability of it being obtuse. The probability of it being right is 0.

Voj. Ses.

Voj.Ses said...

Actually, no, I made a mistake. If we have my 2D representation, acute triangles are those within (90,0),(0,90), (90,90) triangle. Which is 25% of the whole area of (0,0), (180,0), (0,180). You were right. Well done.
Voj.Ses.

Jim Thompson said...

I like your problem and your honesty.

In my view, the proportion of acute triangles is given by the proportion of your triangular planar surface that intersects the cube defined by the planes x=90, y=90 and z=90. I think the line segments you describe are the result of this intersection.

It might be interesting to contemplate how the problem can be generalized.