This problem got me to thinking: what are there more of, acute triangles or obtuse triangles? Which naturally led me to wonder: what is the ratio of acute triangles to obtuse triangles?

Now, one way you could do the problem would be to make a "simplification" like say that the angles have to be whole numbers, and then count. But then you'd have to do a lot of counting, and you'd get all confused with the right triangles. So this would be a bad way to solve the problem.

The way I decided to solve the problem was to imagine that the first angle is plotted on the x-axis, the second angle is plotted on the y-axis, and the third angle is plotted on the z-axis. Then your x, y, and z values can all range between 0 and 180, but with the constraint that x+y+z=180. This is the equation of a plane that intersects the axes at (180,0,0), (0,180,0), and (0,0,180), respectively.

All the triangles are contained between this plane and the x-y, y-z, and x-z planes, not inclusive (because the planes contain the "triangles" with angles of 0 and 180, which are not really triangles at all). This region is a triangular pyramid, whose volume is 1/3*(180*180/2*180) because it's 1/3 times base (180 x 180 / 2) times height (180). This is better known as 1/6*(180

^{3}).

Then you can think about just acute triangles. Plot the angles as above, and notice that they are constrained both by x+y+z=180 and by x, y, z < 90. x, y, z < 90 is a cube with a side length of 90, with corner coordinates like (0,0,0), (90,90,90), (0,90,90), etc. Note, then, that the solid that includes all the acute triangles is the intersection of this cube with the pyramid described above. Note also that the cube is not contained within the pyramid, because one of its corners is (90,90,90) and the point (60,60,60) is on the surface of the pyramid, the plane x+y+z=180.

The key to knowing the volume of the acute triangles is, then, the intersection of these two solids -- or, more helpfully, the volume of the cube that is inside the pyramid. To find this, we will figure out how much of the cube is outside of the pyramid. Notice that three vertices of the cube -- (0,90,90), (90,0,90), and (90,90,0) -- are on the plane x+y+z=180. Thus, the part of the cube that is outside the pyramid is a small pyramid, similar to the one described above, except with edges of 90 instead of 180. So the volume of this is 1/3*(90*90/2*90), which is 1/6 the volume of the cube, so the volume of acute triangles is 5/6*(90

^{3}).

Thus, the ratio of acute triangles to all triangles is (5/6*90

^{3})/(1/6*180

^{3}), which is

(5/6*90

^{3})/(1/6*(2x90)

^{3}), which is

(5/6*90

^{3})/(1/6*2

^{3}x90

^{3}), and we can cancel out the 90

^{3}to get

(5/6)/(1/6*2

^{3}), and 2

^{3}is 8, so we get

(5/6)/(8/6), which is

5/8.

So, 5/8 of triangles are acute triangles, and so 3/8 of triangles are obtuse triangles. QED.

For example, if you asked the question: If you randomly chose three angles that added up to 180, and made a triangle with those angles, what is the probability that it would be an acute triangle? The answer would be 5/8. That sort of thing.

This may make you wonder: What about right triangles? There are no eighths left to be right triangles!

That is because a 90 degree angle is a very precise number. If you could pick any three angles that added up to 180 -- not just positive integers, not just rational numbers, but any positive angles -- the probability of your picking exactly 90 degrees for one of them would be exactly 0. Sorry, Pythagoras.

My question is, clearly there is some overcounting here. For example, the math above counts (40, 60, 80) and (80, 60, 40) -- six different points like this -- as different pairs of angles, but clearly they all describe the same triangle. So clearly the math above is wrong (you better have caught that). How to fix it?

## 3 comments:

Interesting analysys! I had never thought about that before. The "overcounting" you describe is not a problem. Since the probability of choosing a random triangle which has two or more angles equal to each other is zero, you are essentially counting every unique triangle six times when finding the volumes of the solids described above. To only count each triangle once, we could place the restriction that x <= y <= z, which geometrically speaking corresponds to... erm... well, I'm not sure I can do that in my head at the moment. But at any rate, that restriction defines an area of space, whose intersection with the aforementioned solids should still have the same ratio of acute triangles to total triangles.

BTW, this is a really interesting problem. Mind if I post it on my math blog (with prpoer attribution, of course)?

Brent -- are we treating triangles as chiral or non-chiral?

If they're chiral, then we have triple-counting rather than sextuple-counting.

As our former president once remarked, "It matters what the meaning of "is" is".

See the post after this for a resolution of the problems that my analysis has in this post.

Brent, feel free to post about it on the math less traveled.

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