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wlcm · 07-13-2003, 09:05 PM

Here is my go at it.

For the triangle to contain the circle, the shortest distance to traverse the square from on point, through the a second point to the third must be longer than 1/2 the perimeter of the circle.

This is one of those problems where it seems easier to find the probabilty of the converse and doing the P = 1 - !P way of finding things. So i will first try to find the probabilty of the triangle not containing the center.

So the probability that the second and third point (not necessarily the second and third point placed, since you don't know if they were placed one at a time or all at the same time, but the second and third point chosen for the traversal) lays a distance at most 1/2 the perimeter away from the first point is 1/2 * 1/2 = 1/4 But we say that the first point could have been any of the 3 points, so its 3 * 1/4 = 3/4. This should be the probability that the triangle did not contain the center since calculations were based on the distances being less than 1/2 the perimeter, so the probability that the triangle does contain the center is 1 - 3/4 = 1/4

Or we could say that the probablity of the first and second points don't really matter and that the problem only lies in the third point. so then we would need to say that in order for the triangle to not contain the center, the last point has to be at most 1/2 the perimeter away from one of the points in the direction of the other.
So the probabilty of that would be to add up the probabilty that the 3rd point is at most 1/2 the perimeter away from the 1st point in the direction of the 2nd is 1/2 and the same applies to the probabilty that the 3rd point is at most 1/2 the perimeter away from the 2nd point in the direction of the 1st point. But the areas of those two probabilties overlap and you don't quite know how much: ie P(A n B). So i'd say just take the average value of the distance between the 1st and 2nd points as the value of P(A n B), which is 1/4 the perimeter--which incidentally becomes a probabilty of 1/4.
So the probabilty of the triange not containing the center would be 1/2 + 1/2 - 1/4 = 3/4. and therefore the probabilty of the triangle containing the center would be 1 - 3/4 = 1/4. The probability of the center niether in nor outside of the triangle--that is on the edge itself--is 0 since placement of points are PDFs and the probabilty of placing the points in the exact places needed to make the center on the edge of the triangle is 0.

So after all that confusing mess.. i contend that the answer is 1/4.

Sorry i don't have any pictures to make it easier to follow. All distances are measured along the perimeter itself, or else this wouldn't make sense.

07-13-2003, 09:05 PM	#17 (permalink)
wlcm Tilted Location: Austin, TX	Here is my go at it. For the triangle to contain the circle, the shortest distance to traverse the square from on point, through the a second point to the third must be longer than 1/2 the perimeter of the circle. This is one of those problems where it seems easier to find the probabilty of the converse and doing the P = 1 - !P way of finding things. So i will first try to find the probabilty of the triangle not containing the center. So the probability that the second and third point (not necessarily the second and third point placed, since you don't know if they were placed one at a time or all at the same time, but the second and third point chosen for the traversal) lays a distance at most 1/2 the perimeter away from the first point is 1/2 * 1/2 = 1/4 But we say that the first point could have been any of the 3 points, so its 3 * 1/4 = 3/4. This should be the probability that the triangle did not contain the center since calculations were based on the distances being less than 1/2 the perimeter, so the probability that the triangle does contain the center is 1 - 3/4 = 1/4 Or we could say that the probablity of the first and second points don't really matter and that the problem only lies in the third point. so then we would need to say that in order for the triangle to not contain the center, the last point has to be at most 1/2 the perimeter away from one of the points in the direction of the other. So the probabilty of that would be to add up the probabilty that the 3rd point is at most 1/2 the perimeter away from the 1st point in the direction of the 2nd is 1/2 and the same applies to the probabilty that the 3rd point is at most 1/2 the perimeter away from the 2nd point in the direction of the 1st point. But the areas of those two probabilties overlap and you don't quite know how much: ie P(A n B). So i'd say just take the average value of the distance between the 1st and 2nd points as the value of P(A n B), which is 1/4 the perimeter--which incidentally becomes a probabilty of 1/4. So the probabilty of the triange not containing the center would be 1/2 + 1/2 - 1/4 = 3/4. and therefore the probabilty of the triangle containing the center would be 1 - 3/4 = 1/4. The probability of the center niether in nor outside of the triangle--that is on the edge itself--is 0 since placement of points are PDFs and the probabilty of placing the points in the exact places needed to make the center on the edge of the triangle is 0. So after all that confusing mess.. i contend that the answer is 1/4. Sorry i don't have any pictures to make it easier to follow. All distances are measured along the perimeter itself, or else this wouldn't make sense.