dimbulb

On the perimeter of a square, generate 3 random points. (place 3 dots on the perimeter of a sqare randomly). These 3 points form the apex(tip) of a triangle.

What is the probability that the center of the square is bounded by this triangle?

Fake Alias

This is impossible to solve unless you give us more information about where to place the points.

dimbulb

the points are randomly generated.....
so if each point can appear anywhere on the perimeter of the square.....

what is the probability that the center is bounded by the triangle. One way of thinking about this is that if you have 1000 such squares and triangles, in how many of these sets would you have the center bounded by the triangle??

This puzzle is solvable. I couldnt solve it, but i approximated an answer through a simulation. My professor found a relatively simple proof for this though.

keep plugging away.

hrdwareguy

What if the three points are all on the same side of the square?

__________________
Gun Control is hitting what you aim at

Aim for the TFP, Donate Today

dimbulb

that's still a valid 'triangle'. So in this case, the center is obviously not in the triangle.

even if all 3 dots are in the same place, its a valid triangle....

Fake Alias

Im still confused abuot what kind of answer your looking for

spectre

It's late, so I may be wrong, but according to that, it seems like it would be impossible not to create a triangle.

__________________
"Fuck these chains
No goddamn slave
I will be different" ~ Machine Head

cheerios

true, but he's asking not if they make a triangle, but if that triangle contains the point in the very center of the circle.

dimbulb

SQUARE!!!

spectre

Ah, okay. Like I said, it was late. Thanks for clearing that up.

__________________
"Fuck these chains
No goddamn slave
I will be different" ~ Machine Head

Kadath

Uh, you say that the three points form the apex of the triangle. Do you mean the three points form the triangle? An apex(point) is one point, right?
Assuming I'm thinking about this correctly...
It's something more along the lines of degrees from the center of the square. Thinking about the four quandrants of the square formed by perpendicular lines bisecting the square, the points have to fall in certain areas inside those quadrants. Starting upper right at one and numbering them clockwise, we've got two cases I see right off the bat. If the first two points are in adjacent quadrants, the third must be...shit. I'm going to go think about this. God damn you for making me do math on a Saturday!

__________________
it's quiet in here

cheerios

that's what I said!

cheerios

ask if you can't read this.



btw: answer came to 1/16.

druptight

only problem i see with cheerio's answer, is that even if you do the first part, and put the third line on the side of the center, you can still make it fail to encompass the center. What i'm saying is that you can have B, but have it not encompass the center. still pondering.

Kadath

Woo, I was on the right track. cheerios' notes look like mine, only more developed.
Agreed with druptight. That's why I had to think some more.
To clarify, what about the centerpoint of each side? Which segment does that fall on?
I think we'll have to bring angles into this bastard a little bit, much as it pains me.

__________________
it's quiet in here

MacGnG

ok lets work together

http://tfptfp.grindhard.com/math/math.html

if you have a square divide in to quadrants. Label the sides a,b,c,d and each quadrant I,II,III,IV.

there are 8 parts of this square where points can go, A1,A2,B2,B3,C3,C4,D4,D1.

three points in 8 possible parts = 512 (8^3) possible combinations [ie; (a1, d4, c3) would make a right triangle].

since it would take a while to go thru every point and test it, we simply have to figure out where it does work because that would be less than where it doesn't.




upload into the "math" folder, go here for directions
http://www.tfproject.org/tfp/showthr...566#post376566

wlcm

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.

cheerios

wlcm: there is no circle. your proof seems to have confused me.
here's swing #2 from me, dimbulb shot down my first try.

in order for a triangle w/ points on the edges of a square to encompass the center of said square 2 things must be true.
#1: 2 points must be on opposide sides of the square.
#2: the third point must be within a certain area opposite the center. this area is defined by projecting a line between the first 2 points and the center. the distance between those 2 points is the area in which a third point will create a triangle that encompasses the center. lets call the distance between one point and a line through the center parallel to a side of the square a, and the similar distance with the other point b. so. we're back to the fact that the probility that the first and 2nd points will be on opposites faces is 1/4
P(#1) = .25
now, the P(#2) is true is a ratio between the defined length, and the complete length of the perimeter of the square. there are 2 possiblities. either the first 2 points are both on the same side of the center dividing lines, like in the first picture below. in this case, the area is defined by 2*l - a - b assuming that l is the length of a side.
if the points are on opposide sides of the center line, then the lenght is defined by 2*l - a + b. alternatively, one of the values can be considered negative. now, considering the probablities that each case will happen (1/2 for each),
P(#2) =[ 1/2 * ( 2*l -a -b) + 1/2 * (2*l -a +b)] 4*l

so, the answer will be the intersection of those 2 probablities.

P(1 n 2) = 1/4 * (above) = 1/16* (l-a)/l

That's stab #2. here's the piccies.

wlcm

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.

MacGnG

ok im just more confused, but i made a good picture. http://tfptfp.grindhard.com/math/math.html

schmuck

it's been 15 days since the last post on this thread. doesn't anybody care what the answer is to this pressing question?

here's my guess: one in six

cheerios

schmuck: I gave up in favor of more pressing issues in life... maybe I'll look at it again soon.

dimbulb

1/4 is the answer. Use angles. Fix the first point arbitrarily. look at the average distance between 2 random points.

Its hard to explain without a diagram, and i don't have web hosting.

ok. fix point A. consider half of the square, starting from point A to the reflection of point A through the center, O, which we shall call a.

randomly generate point B. let b be the reflection of B through the origin. B obviously lies between A and a.
Let AB be the shorter line segment connection A and B.

now, in order for the triangle to cover the center, O, point C has to fall between a and b. that is, on the line segment ab.

note that ab has the same length as AB.

wlcm

Well i'm glad to know the real answer to this finally. Been wondering if 1/4 was really the right answer or not. Thanks for posting, dimbulb.

Dimbulb, i don't think you finished that explaination but i think it started out about the same as the last one cheerios started on.

I believe that finding the average distance between two random points was the key to getting an answer indeed.

dimbulb

sorry.. was REALLY late at night... lol... and someone was waiting for me in bed....

so, if ab is the same length as AB,

then, what is the average value of AB/perimeter of the square?

its 1/4 . =)