Please be obtuse!

[karthickgururaj]

Pick three points on a plane at random (assume uniform probability density across the plane). What is the probability that the triangle so formed is obtuse angled?

 

Source: general internet, with some slight changes.

 

I got a surprising answer to this one..

[karthickgururaj]

Thanks all for the interesting discussion. I understand the folly in my own solution (at the least!).

 

I'm summarizing some of the interesting posts for a reader who might come across this thread in future. Might be worthwhile to also check the initial set of solutions that attempted to solve the problem "directly" on the infinite plane (including my own). The solutions varied from 1, 0.75, 0.64.. Also, you can check a related discussion: 

 

Didn't discuss the approach below much further, but same result as bonanova's (closest to simulation result):

seems to me every triangle will have a greatest angle between 60 and 180 degrees and cant see how any of those would be more likely than another so three quarters of those will have a greatest angle greater than 90 degrees?

 

Bonanova's solution:

My take:

 

[spoiler='Quick and dirty answer for finite circle']Place any three points A B and C in a unit (or any finite radius) circle.

We prohibit ABC from all lying on a straight line.

This is OK, because such configurations constitute 0% of all cases.

 

Construct the points A' B' and C' so that the centers of the line segments AA', BB' and CC' lie on the origin.

Inspect the eight triangles ABC, A'BC, AB'C, ABC', A'B'C, A'BC', AB'C' and A'B'C'.

In every case exactly six of the eight triangles are obtuse.

 

The probability of an obtuse triangle in a finite circle is 0.75.

 

---------

 

What is the probability that a random triangle inside a circle covers its center?

Examine the same eight triangles!

 

 

If the OP has a solution, it is the same as for a circle of finite radius.

 

Simulation results when trying over a square and a circle. Not the same result as what was predicted above, but close:

I'm not sure I follow. If you pick any three points randomly (keeping it in a finite area to make things comprehensible) and call them X, Y, and Z, then one of the three sides XY, YZ, or ZX must be the longest, and you can relabel the points so that the longest side has points A and B and make the leftover point be C. The points were all uniformly chosen in whatever area was available; the limitation of where point C could lie purely stems from having labeled A and B as the termini of the longest edge. Put another way: point C could fall outside the red and blue areas, but if it did then you would just relabel the points so it's no longer labeled C.

 

I really do think the answer is probably 75% instead of 64%, but I'm not completely sure what the hole in the 64% argument is. I get the feeling that there's something subtle going on that makes the probability distribution of point C within the red and blue regions become non-uniform and denser in the blue region, but I can't tell exactly what it is.

 

Edit: Plus, Java is telling me that it's about 72.5%   It's seldom that I favor theory over experiment, but in this case I still think 75% is more likely correct.

package obtuse;
import java.util.*;
 
class Triangle {
  /* Contains x and y coordinates of the three points of the triangle in
   * variables x[0], y[0]; x[1], y[1]; x[2], y[2]
   */
  double[] x, y;
  
  public Triangle() {
    /* Picks the three points of the triangle randomly from the unit square
     */
    x = new double[3];
    y = new double[3];
    Random rand = new Random();
 
    for(int i=0; i<3; i++) {
      x[i] = rand.nextDouble();
      y[i] = rand.nextDouble();
    }
  }
  
  public boolean checkObtuse() {
    /*  This calculates the dot product of each pair of vectors from each
     *  vertex and sees if any of them are negative, which would indicate
     *  an obtuse angle.
     */
    boolean obtuse;
    
    obtuse = (((x[1]-x[0])*(x[2]-x[0]) + (y[1]-y[0])*(y[2]-y[0])) < 0) ||
             (((x[0]-x[1])*(x[2]-x[1]) + (y[0]-y[1])*(y[2]-y[1])) < 0) ||
             (((x[0]-x[2])*(x[1]-x[2]) + (y[0]-y[2])*(y[1]-y[2])) < 0);
    return obtuse;
  }
}
 
public class Obtuse{
  public static void main(String[] args){
    int iters = 100000;
    int obtuse = 0;
    Triangle myTriangle;
    
    for (int i=0; i<iters; i++){
      myTriangle = new Triangle();
      if(myTriangle.checkObtuse()) {
        obtuse++;
      }
    }
    System.out.println((double)obtuse/iters);
  }
}

 

 

Sure enough, when I modified the Java code to also calculate the probability of a triangle drawn from a unit circle to be obtuse, it wasn't ~72.5% like a unit square. It was ~72%. Gyah.
 

package obtuse;
import java.util.*;
 
class Triangle {
  /* Contains x and y coordinates of the three points of the triangle in
   * variables x[0], y[0]; x[1], y[1]; x[2], y[2]
   */
  double[] x, y;
  
  public Triangle(String shape) {
    /* Picks the three points of the triangle randomly from the unit square
     * or unit circle
     */
    x = new double[3];
    y = new double[3];
    Random rand = new Random();
    
    switch (shape) {
      case "UnitSquare":
        for(int i=0; i<3; i++) {
          x[i] = rand.nextDouble();
          y[i] = rand.nextDouble();
        }
        break;
      case "UnitCircle":
        for(int i=0; i<3; i++) {
          do {
            x[i] = 2 * rand.nextDouble() - 1;
            y[i] = 2 * rand.nextDouble() - 1;
          } while (Math.sqrt(x[i]*x[i] + y[i]*y[i]) > 1);
        }
        break;
    }
  }
  
  public boolean checkObtuse() {
    /*  This calculates the dot product of each pair of vectors from each
     *  vertex and sees if any of them are negative, which would indicate
     *  an obtuse angle.
     */
    boolean obtuse;
    
    obtuse = (((x[1]-x[0])*(x[2]-x[0]) + (y[1]-y[0])*(y[2]-y[0])) < 0) ||
             (((x[0]-x[1])*(x[2]-x[1]) + (y[0]-y[1])*(y[2]-y[1])) < 0) ||
             (((x[0]-x[2])*(x[1]-x[2]) + (y[0]-y[2])*(y[1]-y[2])) < 0);
    return obtuse;
  }
}
 
public class Obtuse{
  public static void main(String[] args){
    int iters = 100000;
    int obtuseSquare = 0;
    int obtuseCircle = 0;
    Triangle myTriangle;
    
    for (int i=0; i<iters; i++){
      myTriangle = new Triangle("UnitSquare");
      if(myTriangle.checkObtuse()) {
        obtuseSquare++;
      }
      myTriangle = new Triangle("UnitCircle");
      if(myTriangle.checkObtuse()) {
        obtuseCircle++;
      }
    }
    System.out.printf("%d unit square and %d unit circle", obtuseSquare, obtuseCircle);
    System.out.printf(" triangles out of %d trials were obtuse.\n", iters);
  }
}

 

 

Counter example to Bonanova's solution, which may be one of the reasons why the simulation result is not the same as predicted:

I'm not sure why the post gets messed up when I try to quote Bonanova's argument, but I can't quote it without wrecking its formatting.

Triangle ABC is roughly equilateral, and point C is very close to the origin.

ABC - acute
ABC' - acute
AB'C - obtuse
AB'C' - obtuse
A'BC - obtuse
A'BC' - obtuse
A'B'C - acute
A'B'C' - acute
Triangles in circle.jpg

 

One more counter example to Bonanova's solution:

here is an example of 2/6 ratio. Only ABC' and A'B'C triangles are obtuse. Others are acute.

obtuse.png

[bonanova]

A uniform probability density function across an infinite domain is zero everywhere. To see this, first choose an origin. Then the likelihood of three randomly chosen points lying within a circle of finite radius is zero. That's because the area ratio of any finite circle and the entire plane is zero. So, with probability one any randomly chosen point in the plane lies at infinity, where angles are not defined.

 

So the question can only be answered for a finite portion of the plane, for example by choosing points at random within a circle of finite radius.

 

Since that result does not depend on radius, one can then argue that the finite result holds on the plane and that the answer solves the OP. For finite circular domains, obtuse triangles are more plentiful than acute ones, but I'll leave the exact ratio for another to find.

[karthickgururaj]

A uniform probability density function across an infinite domain is zero everywhere. To see this, first choose an origin. Then the likelihood of three randomly chosen points lying within a circle of finite radius is zero. That's because the area ratio of any finite circle and the entire plane is zero. So, with probability one any randomly chosen point in the plane lies at infinity, where angles are not defined.

 

So the question can only be answered for a finite portion of the plane, for example by choosing points at random within a circle of finite radius.

 

Since that result does not depend on radius, one can then argue that the finite result holds on the plane and that the answer solves the OP. For finite circular domains, obtuse triangles are more plentiful than acute ones, but I'll leave the exact ratio for another to find.

 

I agree we need to assume the points to be contained in a finite circle, and then apply the limit as the radius approaches infinity. The question can be considered suitable re-worded.

 

[spoiler=Why circle, and not any other shape?.. ]

..because a circle is "fair" Fairness was implicit in the question. It is like asking: a point is chosen at random on x-axis, what is the probability that it is positive? "Intuitively", it looks to be 0.5. But we can get to that result only if we assume a line segment centered at origin, and then apply the limit as the segment length approaches infinity. A line segment centered at origin is "fair" to both positive and negative directions.

 

I solved the puzzle in a slightly different way (however..), without starting with a bounding circle. I'll post that solution later, not sure if it is entirely correct.

 

[bonanova]

I'll still hold off with a solution for a finite circular domain.

I'm very interested to see your infinite-domain solution as well.

[plasmid]

Both of these are considering the infinite plane case, not the finite area case.

Define point A to be at the origin.

Define point B to be the point that is farthest away from point A, and without loss of generality define its position to be (1, 0).

Define point C as the third point, which must lie a distance less than 1 from A because we defined B to be the point that's farthest from A.

 

Angle A-B-C will always be acute if C is within the unit circle around A and B is at the edge of that unit circle.

 

Angle B-A-C will be acute if C is on the same half of the unit circle as B; that is, if it has a positive x-coordinate.

 

Angle B-C-A will be acute if C is NOT within the circle centered at (0.5, 0) (halfway between A and B) with radius 0.5 so that A and B are at opposite ends of its diameter. Remember that if you consider two points at opposite ends of the diameter of a circle, then each point on the circumference of the circle will form a right triangle with those points on the diameter.

 

So we started off given that C is within the unit circle around A.

To form an acute triangle, C must have a positive x-coordinate, which excludes 50% of the initial area.

To form an acute triangle, C must also not lie in the circle with diameter from A to B, which excludes another 1/4 of the original area.

So the final probability that C will form an acute triangle is 1/4, and the probability that it will form an obtuse triangle is then 3/4.

 

Define point A to be at the origin.

Define point B to be the point closest to A, and without loss of generality define its position to be (1, 0).

Define point C as the third point, which must lie a distance greater than 1 from A. Define D as the distance from A to C.

 

Now for angles A-B-C and B-A-C to both be acute, point C must lie between the two lines that are defined as perpendicular to A-B with one line going through A and the other going through B. That is, C's x-coordinate must be between 0 and 1. But if C's x-coordinate has an infinite possible range (well, excluding the unit circle around A since it must be farther from A than B), then the probability that it's between 0 and 1 to form an acute triangle is zero.

Edited October 2, 2014 by plasmid

[koren]

Let's join the points A and B by line and draw perpendiculars to this line at points A and B. The space between perpendiculars is Q. The space beyond the perpendiculars is INFINITY - Q = INFINITY .

The probability of obtuse-angled triangle (point C is beyond Q) is equal to INFINITY/Q, where Q tends from 0 to INFINITY.
However, since Q is an integer number the final value of this probability is INFINITY.

Edited October 2, 2014 by koren

[plasmid]

Is there any reason why my argument above that the probability is 3/4 is incorrect?

For that matter, is there any reason why the following argument that the probability is about 64% is incorrect?

Are they ALL correct?!?

Define A and B as the two points that are farthest apart out of A, B, and C.

Also, without loss of generality, define your coordinate system such that A is at (0, 0) and B is at (1, 0).

 

Since the distance from A to C and the distance from B to C are each less than the distance from A to B, you can draw circles of radius 1 around points A and B, and point C must lie within the overlap of those circles. In the figure below, it would fall somewhere in either the red or blue areas. That total area is 2*pi/3 - sqrt(3)/2 = (4*pi - 3*sqrt(3)) / 6 (reference: http://math.stackexchange.com/questions/402858/area-of-intersection-between-two-circles)

 

If point C lies within the blue circle then angle A-C-B would be obtuse. Otherwise all three angles would be acute. The blue circle has radius 1/2, so its area is pi/4.

 

The probability that triangle ABC is obtuse is equal to the blue area divided by the blue plus red areas,

= [pi/4] / [(4*pi - 3*sqrt(3)) / 6]

= 3*pi / [8*pi - 6*sqrt(3)]

approximately 0.64, or 64%

 

[karthickgururaj]

 

Both of these are considering the infinite plane case, not the finite area case.

Define point A to be at the origin.

Define point B to be the point that is farthest away from point A, and without loss of generality define its position to be (1, 0).

Define point C as the third point, which must lie a distance less than 1 from A because we defined B to be the point that's farthest from A.

 

Angle A-B-C will always be acute if C is within the unit circle around A and B is at the edge of that unit circle.

 

Angle B-A-C will be acute if C is on the same half of the unit circle as B; that is, if it has a positive x-coordinate.

 

Angle B-C-A will be acute if C is NOT within the circle centered at (0.5, 0) (halfway between A and B) with radius 0.5 so that A and B are at opposite ends of its diameter. Remember that if you consider two points at opposite ends of the diameter of a circle, then each point on the circumference of the circle will form a right triangle with those points on the diameter.

 

So we started off given that C is within the unit circle around A.

To form an acute triangle, C must have a positive x-coordinate, which excludes 50% of the initial area.

To form an acute triangle, C must also not lie in the circle with diameter from A to B, which excludes another 1/4 of the original area.

So the final probability that C will form an acute triangle is 1/4, and the probability that it will form an obtuse triangle is then 3/4.

 

Define point A to be at the origin.

Define point B to be the point closest to A, and without loss of generality define its position to be (1, 0).

Define point C as the third point, which must lie a distance greater than 1 from A. Define D as the distance from A to C.

 

Now for angles A-B-C and B-A-C to both be acute, point C must lie between the two lines that are defined as perpendicular to A-B with one line going through A and the other going through B. That is, C's x-coordinate must be between 0 and 1. But if C's x-coordinate has an infinite possible range (well, excluding the unit circle around A since it must be farther from A than B), then the probability that it's between 0 and 1 to form an acute triangle is zero.

 

 

I think there is a subtle problem with the highlighted text above.

By saying that B is the farthest (or closest) point from A, we already assume the point C is chosen. It is like the choice of the third point is restricted because the second point must be farthest (or closest) to A.

[karthickgururaj]

Let's join the points A and B by line and draw perpendiculars to this line at points A and B. The space between perpendiculars is Q. The space beyond the perpendiculars is

INFINITY - Q = INFINITY . (since Q is also INFINITY, you can't be doing INFINITY-Q)

The probability of obtuse-angled triangle (point C is beyond Q) is equal to INFINITY/Q, where Q tends from 0 to INFINITY.

However, since Q is an integer number the final value of this probability is INFINITY.

I guess you meant to say the probability is "1"    My approach is very similar to above.. but there are couple of issues you might need to address..

[plasmid]

I think it would generally be allowable to say "pick three points", then have someone pick three points, and then say "calculate the distance between points 1&2, points 2&3, and points 1&3, and then define A and B as the two points that are farthest apart", and then proceed with the rest of the argument. But I'm not 100% sure.

[karthickgururaj]

Ok, here is my solution, similar to how koren approached it..

 

Pick the first point on the plane, call it O. Pick the second point, call it X. Join O and X, call that line x-axis. Draw a line perpendicular to x-axis, passing through O, call it y-axis. And finally, draw a line perpendicular to x-axis (and parallel to y-axis) passing through X, call it y'-axis. The y and y' axis divide the plane into three infinite parts. Now, for the triangle to be acute, the third point should be chosen in between y and y' (see note [1]). If a point is selected on any other part (far side of y or y'), the triangle is obtuse. So the question comes to: what is the probability of selecting a point in the infinite region between y and y'?

I will claim, that probability is 0. Or in other words, probability the triangle is acute is zero.

This can be proved through contradiction. Suppose the probability is not 0, but some number 'p'. Then there exists a finite integer N such that n > 1/p. On x-axis mark points X1, X2, ... Xn at equal intervals such that the distance between them equals OX. Now draw a line perpendicular to x-axis, passing through Xn. Consider the region bounded by such a line and y-axis. What is the probability an arbitrarily selected point falls in this region? It must be: (1/p)*n (see note [2]). But (1/p)*n is greater than 1, which can't be. So our initial assumption is wrong, implying the probability should be 0.

 

Notes:

[1] Some points in the region between y and y' will produce an obtuse triangle, but that doesn't change the end result.

[2] Yes, that is a bit informal and takes a "leap of faith". A more formal way would be to first solve it for a circle with radius R and then take the limit as R -> infinity, like bonanova suggested.

 

[koren]

 

Let's join the points A and B by line and draw perpendiculars to this line at points A and B. The space between perpendiculars is Q. The space beyond the perpendiculars is

INFINITY - Q = INFINITY . (since Q is also INFINITY, you can't be doing INFINITY-Q)

The probability of obtuse-angled triangle (point C is beyond Q) is equal to INFINITY/Q, where Q tends from 0 to INFINITY.

However, since Q is an integer number the final value of this probability is INFINITY.

I guess you meant to say the probability is "1"    My approach is very similar to above.. but there are couple of issues you might need to address..

 

"since Q is also INFINITY , you can't be doing INFINITY-Q" 

Q is  not an INFINITY but number  which varies from 0 to INFINITY having a fixed value in each test point. 

Of course I had in mind the probability = 1. Sorry.

[plasmid]

Suppose you were to ask someone to pick two points, defining A and B as the first and second points that they pick, and then pick a third point C. This is essentially what's being done in the arguments that the the probability of being obtuse is one.

 

After they've picked points A and B but before they've picked point C, you can define the distance A-B based on the coordinates of those points. You can then a circle of radius A-B around point A. When you place point C somewhere in the infinite plane, there will be a finite amount of space within that circle and an infinite amount of space outside that circle where C could fall. So the probability that C will lie inside that circle (and therefore the distance A-C will be less than A-B) is zero while the probability that C will lie outside that circle (and therefore the distance A-C will be greater than A-B) is one.

 

That appears to violate the OP's statement that the points are chosen randomly -- intuition says that if you pick three points randomly, then there should be a 50/50 chance that the second randomly chosen point is closer to the first point than the third randomly chosen point.

[karthickgururaj]

 

Suppose you were to ask someone to pick two points, defining A and B as the first and second points that they pick, and then pick a third point C. This is essentially what's being done in the arguments that the the probability of being obtuse is one.

 

After they've picked points A and B but before they've picked point C, you can define the distance A-B based on the coordinates of those points. You can then a circle of radius A-B around point A. When you place point C somewhere in the infinite plane, there will be a finite amount of space within that circle and an infinite amount of space outside that circle where C could fall. So the probability that C will lie inside that circle (and therefore the distance A-C will be less than A-B) is zero while the probability that C will lie outside that circle (and therefore the distance A-C will be greater than A-B) is one.

 

That appears to violate the OP's statement that the points are chosen randomly -- intuition says that if you pick three points randomly, then there should be a 50/50 chance that the second randomly chosen point is closer to the first point than the third randomly chosen point.

 

I agree that the a-priori probability that the second chosen point is closer to first is 0.5. But the posteriori probability is zero.

[karthickgururaj]

 

 

Let's join the points A and B by line and draw perpendiculars to this line at points A and B. The space between perpendiculars is Q. The space beyond the perpendiculars is

INFINITY - Q = INFINITY . (since Q is also INFINITY, you can't be doing INFINITY-Q)

The probability of obtuse-angled triangle (point C is beyond Q) is equal to INFINITY/Q, where Q tends from 0 to INFINITY.

However, since Q is an integer number the final value of this probability is INFINITY.

I guess you meant to say the probability is "1"    My approach is very similar to above.. but there are couple of issues you might need to address..

 

"since Q is also INFINITY , you can't be doing INFINITY-Q" 

Q is  not an INFINITY but number  which varies from 0 to INFINITY having a fixed value in each test point. 

Of course I had in mind the probability = 1. Sorry.

 

I didn't understand you then.. I thought Q was the area between the two perpendiculars, in which case Q is infinite..

[koren]

Q really is the area between the two perpendiculars and it is really infinite but only in DIRECTION PARALLEL TO PERPENDICULARS.  

This direction is not relevant  because the movement in this direction does not change the triangle to obtuse.

The probability we are looking for depends  actually on dynamics in the perpendicular direction and can be calculated by comparing  two line segments outside the perpendiculars which are infinity.with the line segment Q which is tending to infinity NUMBER.

So, the probability is INFINITY/NUMBER = 1.

 

This solution is correct for one point let's say A, which means it will be correct also for B and for C, which in turn means it is correct for the entire case. 

[plasmid]

Can you counter my arguments that the answer is 75% or 64%?

 

The total of the three angles of the triangle must sum to 180; that is, C-A-B + A-B-C + B-C-A = 180. Finding the probability that any of those angles are >90 degrees is the same as asking: if you have a stick and break it at two random places, what are the odds that one of the resulting three pieces will be larger than 1/2 the original length of the stick? The answer to that question is: there's a 1/4 chance that both breaks will fall to the right of center to make the left piece more than half of the original size, there's a mutually exclusive 1/4 chance that both breaks will fall to the left of center to make the right piece more than half of the original size, and of the remaining cases where we know that one break must be on the left and one break must be on the right so there's a 1/2 chance that the center piece will be greater than half of the original size (if the distance from the left end of the original stick to the break on the left side is less than the distance from the middle to the break on the right side). So the total probability that any of the three breaks will be >1/2 of the original stick is 3/4.

Edited October 4, 2014 by plasmid

[koren]

 

Can you counter my arguments that the answer is 75% or 64%?

 

OK

The total of the three angles of the triangle must sum to 180; that is, C-A-B + A-B-C + B-C-A = 180. Finding the probability that any of those angles are >90 degrees is the same as asking: if you have a stick and break it at two random places, what are the odds that one of the resulting three pieces will be larger than 1/2 the original length of the stick? The answer to that question is: there's a 1/4 chance that both breaks will fall to the right of center to make the left piece more than half of the original size, there's a mutually exclusive 1/4 chance that both breaks will fall to the left of center to make the right piece more than half of the original size,

 

= In my opinion you should not consider separately the left and the right sides, as the same combination on both sides is the same triangle. Accordingly, the issue should be the probability of two breaks on one side of the stick which is 0.5 - the probability that the second break falls on the first break side (whatever it is).=

 

and of the remaining cases where we know that one break must be on the left and one break must be on the right so there's a 1/2 chance that the center piece will be greater than half of the original size (if the distance from the left end of the original stick to the break on the left side is less than the distance from the middle to the break on the right side). So the total probability that any of the three breaks will be >1/2 of the original stick is 3/4

 

= I'm not quite sure about the last conclusion, but if it is true, we have 1. =  

.

 

 

[plasmid]

The first point in blue is fine -- you can simply say that there's a 1/2 chance that both breaks will fall on the same half of the stick if you prefer.

 

There's a 50% chance that both breaks will NOT fall on the same half of the stick (so neither the leftmost nor rightmost fragments are greater than half the original length).

Out of that 50% where the breaks are on different sides, 50% of those scenarios will have the middle piece be larger than half of the entire stick.

So 25% of the time all three pieces will be less than half the size of the original stick.

[karthickgururaj]

 

Can you counter my arguments that the answer is 75% or 64%?

 

The total of the three angles of the triangle must sum to 180; that is, C-A-B + A-B-C + B-C-A = 180. Finding the probability that any of those angles are >90 degrees is the same as asking: if you have a stick and break it at two random places, what are the odds that one of the resulting three pieces will be larger than 1/2 the original length of the stick? The answer to that question is: there's a 1/4 chance that both breaks will fall to the right of center to make the left piece more than half of the original size, there's a mutually exclusive 1/4 chance that both breaks will fall to the left of center to make the right piece more than half of the original size, and of the remaining cases where we know that one break must be on the left and one break must be on the right so there's a 1/2 chance that the center piece will be greater than half of the original size (if the distance from the left end of the original stick to the break on the left side is less than the distance from the middle to the break on the right side). So the total probability that any of the three breaks will be >1/2 of the original stick is 3/4.

 

Well, something must be wrong, since you got two different answers IMHO, as I have already stated, your initial assumption that A and B are the closest (or farthest) points seems to be wrong. I'll try to justify my point in a subsequent post, please give me some time.

 

And on the solution above, you are solving a different problem How can you be sure that the solution to both the problems are the same?

 

I'm going to attempt what bononova suggested, perhaps you can try that too. Solve first for three points in a circle of radius R, instead of an infinite plane.

[plasmid]

The point I'm trying to make is that, unless there's a clear reason why one of the 100% or 75% or 64% answers should be "correct" while the others have some sort of flaw, then the types of arguments on which all of these answers are based should be viewed with suspicion.

 

This seems to have a similar logical trap as the famous problem: "There are two envelopes filled with money and you are allowed to take one of them for yourself. They appear identical, but you know that one of them has twice as much money as the other. You take one of the envelopes and open it to find out how much is inside, and then you are given the opportunity to switch and take the other envelope instead. Should you switch?"

[karthickgururaj]

The point I'm trying to make is that, unless there's a clear reason why one of the 100% or 75% or 64% answers should be "correct" while the others have some sort of flaw, then the types of arguments on which all of these answers are based should be viewed with suspicion.

 

This seems to have a similar logical trap as the famous problem: "There are two envelopes filled with money and you are allowed to take one of them for yourself. They appear identical, but you know that one of them has twice as much money as the other. You take one of the envelopes and open it to find out how much is inside, and then you are given the opportunity to switch and take the other envelope instead. Should you switch?"

I agree completely. Which is why, to be sure, we need to solve it for a bounded circle.

[koren]

The first point in blue is fine -- you can simply say that there's a 1/2 chance that both breaks will fall on the same half of the stick if you prefer.

 

There's a 50% chance that both breaks will NOT fall on the same half of the stick (so neither the leftmost nor rightmost fragments are greater than half the original length).

Out of that 50% where the breaks are on different sides, 50% of those scenarios will have the middle piece be larger than half of the entire stick.

So 25% of the time all three pieces will be less than half the size of the original stick.

Look, I'm not sure in my solution too. Samething is wrong with the methodology.

If we will take two arbitrary points A, B and begin to sort through all possible options by moving the third point C, we obtain an infinite number of obtuse triangles relatively a limited number of acute-angled (my solution).

If we will take one arbitrary point and sort through all possibilities by moving two other points we will get completely different result.

The funny thing is that according to the task conditions we have to "throw" on the plane the three points at once.

(I always hated this probability stuff, now I know why )

[plainglazed]

seems to me every triangle will have a greatest angle between 60 and 180 degrees and cant see how any of those would be more likely than another so three quarters of those will have a greatest angle greater than 90 degrees?

[Barcallica]

My simple mind tells

Probability=1/3. Because out of all triangle there is, third of them are obtuse.