What are the odds for a random triangle to have only acute angles?

Hmmmm....a rather strange puzzle...but I'll give _my_ version :-)

Since all triangles can in inscribed in a circle, points can be picked
on the perimeter of a circle in such a way as to produce an acute
triangle. This keeps this actual dimensions of the triangle from being
important since at any scale, there is a smaller similar triangle.
What about when two angles are infinitesimally close to 90 or some
situation where two sides are nearly parallel?  No matter, it can be
scaled down into a finite circle with one side of infinitesimal length.

  First, pick a point A on the perimeter of a circle O; it doesn't matter
where (as Nike says, "Just do it"). (p = 1)
  Next, pick another point B at random on perimeter of circle O such that
A, O, and B are not co-linear and B is not on A (p < 1)
  Next, pick a third and last point C at random on circle O such that C
lies in the opposite half of O from B.  The circle is cut in half by the
line AO.  A, O, and C also cannot be collinear and C cannot be on A.
(p < 1/2)
  Thus, the probability of all three independent events occuring is
infinitesimally smaller than 1/2.

For this method: p < 1/2

A simple simulation with a simple BASIC program agrees with this answer.

[ Editor: C on the other side of line AO from B is not enough for an
acute triangle; the triangle is acute exactly when the centre O lies
inside it.  For example A, B, C at 0, 10 and 200 degrees satisfy the
condition, but the angle at A, half the 190-degree arc BC not containing
A, is 95 degrees. ]

------------

A second attack:

Pick an angle A at random relative to the x-axis.  (We can work
symmetrically above and below the axis).  Now, pick an angle B relative
to a line with angle A relative to the x-axis. (This one was scary at
first)

Start by discreetizing the number of possible angles A can have to N
possibilities.

All N are equiprobable.
After N is picked, the probability that another line the crosses the line
defined by A and the x-axis will produce two acute angles (if A is acute)
is A / 180.
Since we will iterate though N/2 i's, A can be defined as 180i/N.
Substituting this into A / 180, the probability that the next two angles
will be acute (after A is chosen) is i / N.

Summing the all of the possible ways to get an acute triangle for an
even N:

   i=N/2 (N is even)
p = Sum (1 / N) * (i / N) = (N^2 + 2N) / 8N^2 = (N + 2) / 8N
    i=1
As N -> +inf, p -> 1/8

This value also agrees with a small BASIC simulation.

The moral of this story is that it is your METHOD of picking triangles
that determines the probability of picking an acute triangle.
Obviously, there are an infinite number of acute triangles as well an an
infinite number of non-acute triangles, thus, without a system for
randomly picking them (?!?), I don't think it's possible for one to
meaningfully consider the what the probability is of a random triangle
being acute.

-- Dennis A. Sevakis <spinman@ix.netcom.com> (signed "Mike")

[ Editor: from the rec.puzzles thread "acute angles of a random
triangle", October 1996. ]

conroym@euclid.Colorado.EDU (matthew conroy) writes:
>See the Journal of Recreational Mathematics, Vol. 12(2), 1979-80
>pages 88-94, Henry Kierstead, _An Acute Analysis of An Obtuse
>Problem From A New Angle_.  Therein, you'll find proofs for
>75%, 63.94%, 57.08%, 82.1%, and 100% obtuse.
>
>He also quotes V.G. Hobbes: "Whatever solution may be your
>preference we can at least agree on one thing: that triangles
>with obtuse angles are more common than those without.".

With all due respect to V.G. Hobbes (who I don't know), I cannot agree
with his statement.  Offering proofs of five contradictory statements
doesn't mean that the truth lies somewhere between the five cases, it
means that all bets are off.

If there is one thing that we should be able to agree on, it's that
there is no uniform probability distribution over an unbounded space.
Any result that starts out with the presumption that such a
distribution does exist will lead to a nonsensical result as a matter
of course.

-- Matthew Daly <daly@PPD.Kodak.COM>

In article <53e6s6$9fq@nntp5.u.washington.edu> marizac@u.washington.edu
(Mariza Cabral) writes:
>>Offering proofs of five contradictory
>>statements doesn't mean that the truth lies somewhere between
>>the five cases, it means that all bets are off.
>
>Or all but one is right.

My point is that if you can up with "proofs" for five contradictory
theories, then you have no specific reason to trust any of them.
Indeed, you should be extra certain that the statement is defined.  It
could be that four of those proofs are wrong, it could be that all five
are wrong and a sixth answer is correct, or it could be that the
question as stated is either imprecise or nonsensical.  But you
certainly can't make five data points from these "proofs".

>Indeed.  "Uniform probability in an unbounded space" is a sloppy way
>of saying uniform probability in a bounded space of increasing size.
>You can impose bounds on your space, and then stretch them out.
>The acute-triangle probability should converge to a constant as the
>space size is increased.

Not really.  R^2 is not a family of spaces that we can speak of it's
being of "increasing size".  It is a single space, and that single
space is unbounded.

If you would like to propose that the limit of uniform distributions
over a family of disks of increasing size approaches a uniform
distribution over R^2, you'll have to prove it.  But it is not unusual
in mathematics that every element of a sequence can share a property
that the limit of the sequence does not have.

For instance, consider the family of functions defined by:

f_n(x) = 0        if |x| > 1/n
f_n(x) = 1 - |nx| otherwise

for all real x and all positive integers n.  You can see for yourself
that each function is continuous.  But the limit of the family is the
function

g(0) = 1, g(x) = 0 otherwise

which is not continuous.

The same is true of the probability distributions over disks of
increasing size.  The common value of the distributions approaches 0
as the radius of the disk becomes large, but the zero function over R^2
is not a probability distribution, since the integral over the space is
not 1.  Even if the acute-triangle odds were CONSTANT over every finite
disk does not imply that it would have the same value over the
unbounded space, since the very question makes no sense over that
space.

[Note that I'm dealing with the question "What is the probability that
a random triangle is obtuse?", not "What is the probability that a
random triangle chosen from a disk of radius r is obtuse, as r becomes
large?"  I believe the second question has a solution, but it's
important to recognize that they're two seperate questions and that
the first is ill-defined.]

-- Matthew Daly <daly@PPD.Kodak.COM>