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wlcm · 07-14-2003, 02:26 AM

Sorry, i seemed to have confused myself a bit. Let me clearify a bit: i seemed to have used the word circle when thinking of center or square. Since it is really late at night, try to bear with me if i make that mistake again.

#1: if you draw any line through the center of a square, you end up cutting the square in half.
#2: there exists a line through the center of the square that does not cut through the triangle if and only if the triangle does not enclose the center of the square.
#3: distances around the perimeter of the square from one point to another can be measured.
#4: Using the #1-3, Given a triangle that does not bound the center of the square, the sum of the two shortest distinct distances found by way of #3 has to be less than 1/2 the perimeter of the square.

#4 can be easily seen by drawing the line mentioned in #2 and seeing that if you follow the perimeter from where the line intersects the square to the other point where the line intersects the square, you must past through either all or none of the vertices of the triangle--if the triangle does not enclose the center. Either path will be 1/2 the perimeter of the square since the line cuts the square into halves.

With that said, i also say that the probability P(that any two points on the edge of a square has a distance, measured along the square's perimeter, less than 1/2 the perimeter of the square) = 1

probability that distance = exactly 1/2 is zero, since the integration of pdf at a singular point is 0.

So all that is to be found is if the third point makes everything work out.

We'll have A, B, and C represent the first, second, and third point respectively, and O represent the center of the square.

If i were to draw a line LA made by connecting A and O, i would cut the square into halves--one half containing B, and the other not. We would like C to be on the half that contains B. Probability of this happening is 1/2.

But we could think the other way and make line LB by drawing the line between O and B and wanting C to be on the half that contains A. Again the probability of that happening is 1/2.

Now i will say that the probability that the outcome that we want after drawing that line LA and LB will be P(A)=1/2 and P(B)=1/2 respectively.

So it comes down to saying that the probability that the triangle does not contain the center of the square is P(A U B) = P(A) + P(B) - P(A n B).

Now to find P(A n B). The "area" of overlap in the favorable regions varies with the distances between A and B. In order to do this, we need tobfind the expected value (average) of the distance between A and B.
Finding the expected value involves integrating, along its appropirate bounds, the uniform pdf that describes the point placements, but i won't go much into that.
It is easy enough to see that since the max distance is 1/2 the perimeter and the min distance is 0, and that the probability of placement is uniform, the average distance should be 1/4 the perimeter.

So if the expected value of the distance between points A and B is 1/4th the perimeter, the probability of point C landing in that area is 1/4. This makes it P(A n B) since this area is the overlap in the situations dealing with P(A) and P(B). P(A n B) = 1/4.

So the probability P(A U B) that the triangle does not contain the center of the square is:
P(A) + P(B) - P(A n B) = 1/2 + 1/2 - 1/4 = 3/4.

Since we are looking for the probability that the triangle contains the center of the square, P = 1 - !P, so 1 - 3/4 = 1/4.

I hope that clears up my explaination a bit. I'm thinking this is right, but not 100% sure. And of course there should be other ways to do this.

Cheerios i think you might need to refine your P(#2) a bit more, since in the second picture, it is B that turns negative, yet the actual length becomes 2L + a - b. So and i am not sure right now what it is, but something just feels like its missing from P(#2). Keep working on it and try to get a final answer in one number. I have a feeling that there is supposed to be a numerical answer for this problem.

07-14-2003, 02:26 AM	#19 (permalink)
wlcm Tilted Location: Austin, TX	Sorry, i seemed to have confused myself a bit. Let me clearify a bit: i seemed to have used the word circle when thinking of center or square. Since it is really late at night, try to bear with me if i make that mistake again. #1: if you draw any line through the center of a square, you end up cutting the square in half. #2: there exists a line through the center of the square that does not cut through the triangle if and only if the triangle does not enclose the center of the square. #3: distances around the perimeter of the square from one point to another can be measured. #4: Using the #1-3, Given a triangle that does not bound the center of the square, the sum of the two shortest distinct distances found by way of #3 has to be less than 1/2 the perimeter of the square. #4 can be easily seen by drawing the line mentioned in #2 and seeing that if you follow the perimeter from where the line intersects the square to the other point where the line intersects the square, you must past through either all or none of the vertices of the triangle--if the triangle does not enclose the center. Either path will be 1/2 the perimeter of the square since the line cuts the square into halves. With that said, i also say that the probability P(that any two points on the edge of a square has a distance, measured along the square's perimeter, less than 1/2 the perimeter of the square) = 1 probability that distance = exactly 1/2 is zero, since the integration of pdf at a singular point is 0. So all that is to be found is if the third point makes everything work out. We'll have A, B, and C represent the first, second, and third point respectively, and O represent the center of the square. If i were to draw a line LA made by connecting A and O, i would cut the square into halves--one half containing B, and the other not. We would like C to be on the half that contains B. Probability of this happening is 1/2. But we could think the other way and make line LB by drawing the line between O and B and wanting C to be on the half that contains A. Again the probability of that happening is 1/2. Now i will say that the probability that the outcome that we want after drawing that line LA and LB will be P(A)=1/2 and P(B)=1/2 respectively. So it comes down to saying that the probability that the triangle does not contain the center of the square is P(A U B) = P(A) + P(B) - P(A n B). Now to find P(A n B). The "area" of overlap in the favorable regions varies with the distances between A and B. In order to do this, we need tobfind the expected value (average) of the distance between A and B. Finding the expected value involves integrating, along its appropirate bounds, the uniform pdf that describes the point placements, but i won't go much into that. It is easy enough to see that since the max distance is 1/2 the perimeter and the min distance is 0, and that the probability of placement is uniform, the average distance should be 1/4 the perimeter. So if the expected value of the distance between points A and B is 1/4th the perimeter, the probability of point C landing in that area is 1/4. This makes it P(A n B) since this area is the overlap in the situations dealing with P(A) and P(B). P(A n B) = 1/4. So the probability P(A U B) that the triangle does not contain the center of the square is: P(A) + P(B) - P(A n B) = 1/2 + 1/2 - 1/4 = 3/4. Since we are looking for the probability that the triangle contains the center of the square, P = 1 - !P, so 1 - 3/4 = 1/4. I hope that clears up my explaination a bit. I'm thinking this is right, but not 100% sure. And of course there should be other ways to do this. Cheerios i think you might need to refine your P(#2) a bit more, since in the second picture, it is B that turns negative, yet the actual length becomes 2L + a - b. So and i am not sure right now what it is, but something just feels like its missing from P(#2). Keep working on it and try to get a final answer in one number. I have a feeling that there is supposed to be a numerical answer for this problem.