How obtuse are triangles?

[bonanova]

What fraction of triangles are obtuse? Provide a proof.

[ThunderCloud]

Hmmm… I'd guess…

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Every triangle has three sides; let's call them a, b, and c, and say that 0 < a ≤ b ≤ c. In order to form a triangle, it must be the case that c < a + b, so then b² ≤ c² < (a + b)² = a² + b² + 2ab. The triangle is obtuse if c² > a² + b².

Since the scale of the triangle doesn't matter, let's assume that b = 1. So then, c² can fall anywhere between 1 and a² + 2a + 1, but only yields an obtuse triangle if it is larger than a² + 1.

The fraction of obtuse triangles is then ( (a² + 2a + 1) - (a² + 1) ) / ( (a² + 2a + 1) - 1 ) = 2a / (a² + 2a) = 2 / (a + 2), for a triangle whose shortest side is length a (and whose next shortest side is length 1). Integrating over a from 0 to 1, this gives 2( ln(3) - ln(2)) = ~81%.

 

[plasmid]

Two more answers to add to ThunderCloud's.

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"Proof" that "all" triangles are obtuse
Place the first point of the triangle and call it point A, and place the second point and call it B. Create a coordinate system around A and B such that A falls on (0, 0) and B falls on (1, 0). Then place the third point of the triangle and call it C. The triangle will be obtuse if the x-coordinate of C is either less than zero or greater than one, and can only be acute or right if C is within the range [0-1] (and not even for all cases within that range). Since there’s a finite range of x-values where the triangle might not be obtuse and an infinite range where it would be obtuse, the odds of the triangle being obtuse are essentially 1.

"Proof" that ~64% of triangles are obtuse
Take the longest edge of the triangle and define a coordinate system such that the two vertices on that edge are at (0, 0) and (1, 0), and since that’s the longest edge, the triangle will be obtuse if and only if the opposing angle is obtuse. Since the longest edge has length 1, the other vertex must lie within a circle of radius 1 around (0, 0) and within a circle of radius 1 around (1, 0). Resorting to an online calculator, that area is about 1.228. Also, based on trigonometry (or maybe hand waving) I assert that if and only if the other vertex falls inside a circle at (0.5, 0) with radius 0.5 then the triangle is obtuse. That circle with radius 0.5 has area 0.7854. So the odds that the third vertex is within that area, and therefore that the triangle is obtuse, is 0.7854 / 1.228 ~= 64%.

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Yeah, I admit my first answer was sort of trolling :3
But even bogus answers can serve as lessons about what sort of faulty logic to be on guard for.

 

 

[Pickett]

My brain isn't exactly firing on all cylinders today, but...

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So there are infinite number of triangles available, divided into 3 categories (or sets), each with an infinite number of elements

1. An infinite number of right triangles
2. An infinite number of acute triangles
3. An infinite number of obtuse triangles

As a quick example of that, imagine a right triangle with points at (0,0), (0,1), and (1, 0)...you can just move the point on the y-axis up and down however you want to stretch the triangle infinitely.

So then, it depends on how you define "what fraction".

• If you look at it from a statistical perspective and ask "if you draw 3 completely random, non-colinear points in the x-y plane, what are the odds of that triangle being obtuse?" I'd have to say it's 1/2 for the following reason:
  • While there are an infinite number of right triangles, it is statistically a zero percent chance of the 3 random points creating one. Which leaves 2 possible outcomes, each with equal probability.
• However, if you are asking for total number of obtuse triangles divided by total number of triangles, I'd say it's indeterminate.
  • It's similar to asking "what fraction of points are inside of a circle on a plane?"
  • Technically, since there are an infinite number of points inside the circle AND outside of the circle, you could map every single point on the inside of a circle to some point outside of the circle and vice versa...however, conceptually, it's obvious that there should be more points outside of the circle...but that defies the idea that there's an infinite number inside...

As I said, my brain is pretty scattered today, and I'm assuming that's not the answer you were looking for...

 

[bonanova]

  On 2/2/2018 at 2:39 PM, Pickett said:

My brain isn't exactly firing on all cylinders today, but...

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• However, if you are asking for total number of obtuse triangles divided by total number of triangles, I'd say it's indeterminate.
  • It's similar to asking "what fraction of points are inside of a circle on a plane?"
  • Technically, since there are an infinite number of points inside the circle AND outside of the circle, you could map every single point on the inside of a circle to some point outside of the circle and vice versa...however, conceptually, it's obvious that there should be more points outside of the circle...but that defies the idea that there's an infinite number inside...

 

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In some cases we can get around this point. We could for example divide the plane into increasingly small squares. If we picked a point at random we could say it lies in all squares with equal probability and then count squares. Since a finite circle includes a finite number of squares but excludes an (uncountably) infinite number, the fraction of squares inside the circle is no longer indeterminate -- it's zero. That would allow us to reach the reasonable conclusion say that the probability of hitting a finite circle embedded within an infinite dartboard is unambiguously zero. So in some cases where we're picking points at random we can start out picking very small areas, getting an answer, then take the limit of that answer as the areas go to zero. In cases where we're dealing with two infinite areas, however, this approach does not work. (Unless perhaps if the infinities are of different cardinalities.)

Using these "geometric probabilities" is something like saying that points have equal "density" everywhere. It's kind of a reasonable approach, but it's contradicted by the point that you make, namely that there is a surjection between the interior and exterior of a circle.

[bonanova]

This puzzle is an ancient one that doesn't have a definitive answer so far as I can find. Points given to ThunderCloud and plasmid for proofs of possible answers, but will leave the puzzle open for further comments.

I have a criticism of this puzzle, closely related to Pickett's comments, that I haven't seen raised elsewhere: There is no such thing as a random triangle in the plane.  How do you pick random points in the plane? We can impose a coordinate system that makes (0, 0) a reference point, and we can add (1, 0) to provides a scale factor and orient the axes, but that's it. What we can't do is pick three arbitrary points in the plane. The origin can be the first point, WOLOG, but the other two, if truly chosen at random, are both points at infinity. It's like asking the average value of the integers. A finite value would, by any measure be disproportionately "close" to the origin. I haven't found any reference to this objection in other discussions of this puzzle. The issue plagues any attempt at a solution.

In the analyses plasmid gave us, the two divergent answers are equally correct -- or equally incorrect. They assume one of the sides of the triangle has finite length. But any random line segment in the plane must have infinite length. And if so, then the analysis compares areas that are both of infinite extent. The analysis ThunderCloud gave us includes the premise that "scale doesn't matter" so let b=1, and then let a and c be anything. Same issue: if b is constrained, we can't let a and c be infinite. Or, if b=1, then it's not random over an infinite space, where things cannot be "scaled".

Other approaches that I found let the triangle, instead, be randomly chosen in a circle. This preserves angular randomness and permits comparison of side lengths, by eliminating the infinity problem. As a bonus, it gives a unique answer.

Follow-on puzzle: What fraction of triangles in a circle are obtuse?

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If we revisit the puzzle I recently posted that asked the fraction of random triangles in a circle that cover its center, there are at least three ways to convert that solution to an unambiguous solution of this puzzle.

 

[Cygnet]

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Plasmid's got this entirely right, but ThunderCloud and I were working on this separately and endeavored to find the actual fraction involved. I also made a picture, just for fun. 

The area of the shape that contains all triangles can be calculated using the formula for the arc (π/6), doubled, minus the equilateral triangle that is duplicated by the arc from the second circle (2π/6 - √3/4). The half circle that contains all obtuse triangles has an area of π/8, so the fraction of triangles that are obtuse is (π/8) / (2π/6 - √3/4). Simplifying this so it has no icky compound fractions, we get 3π/8π-6√3. This works itself out to ~0.639, just to confirm Plasmid's result.

Edit: I had this window open and didn't see bonanova had posted. Heh.

 

Edited February 7, 2018 by Cygnet

[bonanova]

Hi Cygnet, and welcome to the Den. And nice pic.

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Your and plasmid's result (.639) obtains if [0, 1] is taken to be the the longest of the sides. While p=.821 (similar to ThunderCloud's result) obtains if it is taken as the middle side. Finally, if it's the shortest side, so that the other two sides are unconstrained in length, plasmid's first result (p~1) obtains. Which makes putting all three points anywhere inside a circle a case that gives an unambiguous and pleasing result.

 

[ThunderCloud]

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I think plasmid's second analysis and Cygnet's answer will hold if we define random triangles by this process: Consider a straight stick of finite length. Break it into three pieces, choosing the two break points at random with uniform distribution (which we can define over a finite interval). So long as no stick fragment is larger than (or equal to) the sum of the other two, form them into a triangle. The probability that a triangle thus formed is obtuse should be as given by Cygnet and plasmid.

 

Edited February 9, 2018 by ThunderCloud