Math Is Fun Forum

juantheron

Member

Registered: 2011-10-19

Posts: 312

3 point in a plane

three non collinear points are marked in a plane. find the probability of getting an obtuse angled triangle by joining them.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: 3 point in a plane

Hi;

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: 3 point in a plane

thanks bobbym

would you like to explain your answer

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: 3 point in a plane

I can only copy the answer that was given to me when I was trying to solve this problem. I can not explain it better. It is what I have in my notes and of course I will give due credit to Doctor Anthony for his solution.

These problems can give a variety of answers depending on how you
define the way the random triangle is to be formed. You are probably
aware of the classic example of the random chord drawn in a circle,
and the probability that this chord is longer than the length of side
of the inscribed equilateral triangle. You can get answers of 1/4,
1/3 and 1/2 depending on the rule you adopt for drawing the 'random'
chord. Similarly, in the problem you quote above, a number of
different answers can be expected depending on the rule you adopt for
drawing the random triangle.

A good example of the rule could be as follows:

A circle can always be drawn through three points, so we could say
that the three points are to be chosen at random on the circumference
of a circle.

We select the first point anywhere on the circumference and then draw
a diameter and tangent to the circle at that point. For the other two
points we take the angle between the tangent and the chord to one of
the other points as x, and similarly the angle between the tangent and
the chord to the third point as y. In both cases measure x and y
anticlockwise from the tangent to the respective chords. Then both
x and y are uniformly distributed between 0 and pi.

The triangle will be obtuse if x and y are both less than pi/2
(probability of this (1/2)(1/2) = 1/4), or both greater than pi/2,
again with probability 1/4.  So altogether this probability is 
1/4 + 1/4 = 1/2.

We also get an obtuse angled triangle if x-y > pi/2, this gives
y < x-pi/2, and also if y-x > pi/2  i.e.,  y > x+pi/2

To find the probability of this we set up a two-dimensional sample
space, with the usual x and y axes, and with x varying uniformly from
0 to pi, y varying uniformly from 0 to pi.  This gives a square of
area pi^2.

The regions cut off by the line y > x+pi/2  and the line y < x-pi/2 
if put together will form a square of side pi/2, and an area pi^2/4.

So the probability of this producing the obtuse angle is 
(pi^2/4)/pi^2 = 1/4.

Now if we call first situation (probability = 1/2) event A and the
second situation (with probability 1/4) event B then:

P(A or B) = P(A) + P(B) - P(A and B)

           = 1/2 + 1/4 - P(A and B)

However, P(A and B) is zero, since if both x and y < pi/2, or both
x and y > pi/2, (event A), it is impossible for x-y or y-x to be
greater than pi/2, (event B).

We conclude that  P(A or B) = 1/2 + 1/4 = 3/4

So a triangle drawn at random has a probability of 3/4 of being
obtuse. This result depended on our definition of a 'random' triangle. 
Other definitions will produce a different result.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.