geometric probability ---- squares

mr.wong · 2016-04-06 21:59:18

Thanks bobbym !

Thus the polyhedron has an apex at the point
with co-ordinate (1/6 , 1/6 , 1/12) .
Waiting for a solution with integration , let us firstly
try to find an approximate answer of the problem with geometry .
Since the curve over QR = 1/2 x - xx ,
we can find the area under the curve
from 0 to 1/2 by using simple integration
and obtain 1/48 sq.un., thus the average
height ( representing the average common area of A and X )
= (1/48) / ( 1/2) = 1/24 .
Similarly the average height over PQ also = 1/24 .
We have omitted the portion of the polyhedron higher than 1/16
to the apex as its value is so small and will not affect much .
Divided 1/24 sq.un by the area of Δ ABC , i.e. 1/8 sq.un., we get
the probability being about 1/3 .

bobbym · 2016-04-08 03:10:13

Hi;

Which question is that answering?

mr.wong · 2016-04-08 15:09:09

Hi bobbym ,

That answer is for Problem (III B) at # 46 .

bobbym · 2016-04-09 06:55:03

Hi;

I will check to see if I can get the same answer but that will be later because I am being called away for now.

mr.wong · 2016-05-17 20:29:20

Hi bobbym ,

Finally I obtained the volume of the polyhedron to be 1/128 .
Thus its average height ( representing the average value of common
area ) = (1/128 ) / (1/8) = 1/16 . Therefore the probability required
= ( 1/16) / (1/8) = 1/2 . ( for Problem (III B) at # 46 )
Does this value coincide with your simulation ?

bobbym · 2016-05-17 20:58:20

Hi;

It does not coincide with what I had in my notes and I will need to rethink the whole problem.

My answer was 1 / 12 which came from a simulation and your maximized function in post #50.

Do you still want to use the diagram in post #46?

thickhead · 2016-05-18 00:12:36

I have got an idea which I am just initiating. For simplicity we can take the vertices as O(0,0) A(a,0) and B(0,b). let P be point (x,y) at which we want probability.Imagine the moving triangle with its vertices in their turn at P. say CDP where P corresponds to A of

we get C(x-a/2,y+b/2) D(x-a/2,y) Similarly for other positions E(x,y-b/2) F(x+a/2,y-b/2) G(x+a/2,y) H(x,y+b/2). the area CDEFGHC is the one where P is inside the moving triangle but only one condition, the hexagon must not step out of triangle ABC. this area divided by area of triangle ABC should give the probability distribution which when integrated over the triangle joining midpoints of triangle ABC gives the total probability. How to solve it I have not analyzed so far.

bobbym · 2016-05-18 08:15:28

Hi mr.wong;

I am getting an answer of 1 / 4.

mr.wong · 2016-05-18 17:06:54

Hi bobbym ,

I think the diagram in post #46 is still useful .
I am not sure that my answer of 1/2 is correct ,
I will keep on investigating .

mr.wong · 2016-05-18 17:08:09

Hi thickhead ,

Thanks for your reply ! But I don't quite understand
your statement . It would be much helpful if you
have a diagram or sketch with the post .
For simplicity just let a = b = 1/2 in this stage ,
and generalize the result if everything comes ok .
Integration is always useful in calculating areas and
volumes , but I am weak in calculus . Wish to receive your help in problem dealing with integration and other related fields .

bobbym · 2016-05-18 17:11:55

If I am interpreting the problem correctly the answer is 1 / 4 as stated above. This fits a simulation as well as an analytical solution.

thickhead · 2016-05-19 01:22:31

I was referring to your problem III I am getting the probability as 7/16 with the method I laid out before this. Did you come across this value sometimes in your analysis? I do not have the technique to give the diagram but I can just describe. I am taking the vertices of the outer triangle as O(0,0) A(2,0) and B(0,2) the midpoints are L(1,0)M(1,1) and N(0,1).just to have a glimpse of the hexagon I described before take a random pt P(x,y) at (0.8,0.8)
Then plot these pts C(-0.2,1.8) D(-0.2,0.8) E(0.8,-0.2) F(1.8,-0.2) G(1.8,0.8) and H(1.8,1.8) If the moving triangle is within this hexagon th point (x,y) is inside the triangle but part of the hexagon going out of triangle OAB is invalid. Only the common area of the hexagon and triangle OAB is valid. From the diagram we find 3 small triangular portions near the vertices are the left out portions. The fraction of sum of these areas over the area of triangle OAB when subtracted from 1 gives the probability that (x,y) lies inside the moving triangle. This expression has to be integrated over the area LMN gives the average probability.Don't consider (0.8,0.8) in any calculation. It is used just to draw a typical diagram.

Last edited by thickhead (2016-05-19 01:23:52)

thickhead · 2016-05-19 04:46:39

mr.wong wrote:

Inside a square E there are 2 smaller squares A and B ,both with length of sides being 1/2 of that of E. Both squares
can move freely inside E, but must keep "parallel " with E. If a point is chosen randomly on E , find the probability that the point lies inside A and B at the same time.

I am getting the average probability of a point to be inside one of the squares as 9/64 and inside both as 81/4096 using the method as given previously.

Now this is found to be incorrect.

Last edited by thickhead (2016-06-18 22:16:39)

mr.wong · 2016-05-20 15:08:58

Hi bobbym ,

You got an answer of 1/8 for Problem ( III) in # 38 , this is
equivalent to an answer of 1/2 for Problem (III)( B ) in
# 46 . I wonder why you give up this result ?
Though triangle X occupies 1/4 of the area of triangle E ,
but its position is at the centre of E , thus the probability
that it overlaps with triangle A should be greater .

thickhead · 2016-05-20 16:45:43

mr.wong wrote:

Inside a square E there are 2 smaller squares A and B ,both with length of sides being 1/2 of that of E. Both squares
can move freely inside E, but must keep "parallel " with E. If a point is chosen randomly on E , find the probability that the point lies inside A and B at the same time.

Hi wong,
I have been attracted by your problems recently.I propose to use a diagram which I may christen as "favourable mobility diagram" if it is not already existing. that is why I start with the simplest problem to check it and go to complicated problems later. irequest you and Bobbym to critically examine the same and correct me if and when necessary.
In this problem, consider a point P(x) in the left half of E. Way of approach is different for the right half but the results will be same. Consider the position of A when its right edge is flush with P. Consider another position of A with its left edge flush with P.Combine the two.This gives a diagram for the position of A when P is inside A.This I want to call "Favorable mobility diagram". Part of this is outside E; that part is invalid.Therefore the valid area (which is inside E also) is x+1/2 but out of this area 1/2*1 is self occupied area of A which is not available for mobility. Therefore only area x (i.e. x*1)is favorable area whereas total mobility is the area of E-area of A (again self occupied area) which comes out to be 1/2. therefore probability of P being inside A =x/(1/2)=2x.
Now let us compare it with analytcal approach. Consider position of A with its left edge flush with left wall of E. Now P is inside A. Now start moving A towards right. when its left edge clears p P starts falling outside A. So favorable distance is x. the triangle A however can move a maximum distance of 1/2.So the probability is x/(1/2)=2x. I hope this convinces the use of "favorable mobilty diagram"
Now coming to the solution of the problem,previously i had made the mistake of integrating this expression to get average probability for A and multiplied the result with itself for the combined probability of P being inside both A and B.this is incorrect. what one should do is at P itself multiply the 2 probabilities and integrate over the region.
Hence

Last edited by thickhead (2016-05-20 16:46:28)

bobbym · 2016-05-20 17:23:08

mr.wong. wrote:

You got an answer of 1/8 for Problem ( III) in # 38 , this is
equivalent to an answer of 1/2 for Problem (III)( B ) in
# 46 .

Problem (III)( B ): I meant the expected value of the overlapping area of the two triangles is 1 / 4. I am getting a probability of 1 / 2.

mr.wong: I have been solving all the problems based on the assumption the two smaller triangles and the two squares are fully inside the bigger triangle and square. No part of these triangles or squares is outside the larger triangle or square. Do you agree?

Hi thickhead;

I can not work on these anymore until mr.wong answers the above question.

thickhead · 2016-05-20 18:38:53

mr.wong wrote:

Thanks bobbym , you are right .
By formula , we have the probability of the point lies
within the axis of 1 side ( say horizontal side ) of the
common portion of A and B to be 1/3 . Similarly ,
for the vertical side , P also = 1/3 . Thus combinely
P of the point lies within both A and B will be 1/9 .
Will the answer be the same for other polygons , say
rhombus under similar conditions ?

Could you elaborate on the formula?

thickhead · 2016-05-20 19:32:41

mr.wong wrote:

Inside a square E there are 2 smaller squares A and B ,both with length of sides being 1/2 of that of E. Both squares
can move freely inside E, but must keep "parallel " with E. If a point is chosen randomly on E , find the probability that the point lies inside A and B at the same time.

For the purpose of clarity I take E to be of size 2*2 and A and B to be 1*1.
Consider a point P(x,y) 0<x<1,0,y<1; I bring the 4 vertices of A to point P one at a time to form my "favorable gross mobility diagram" I have added gross because self occupied area of a is to be subtracted from it. Now part of this diagram inside E is a rectangle of sides x+1 and y+1. Its area is (x+1)*(y+1). To get net favorable area we have to subtract area of A viz. 1*1. So the favorable net mobility area =(x+1)(y+1)-1=xy+x+y. To get the probability of P lying inside A we have to divide it total net mobility area=area of E-area of A=2*2-1*1=3.
So

We can check the value of p at selected points. At (0,0) ,p=0 and at (1,1) p=1;This is correct because at (0,0) just one corner of A coincides with it just for a fraction of moment and (1,1) will be inside or on A for all the time. the same value applies for B. Integrating we get,

Last edited by thickhead (2016-05-20 20:38:01)

mr.wong · 2016-05-20 20:06:57

Hi bobbym and thickhead ,

From the beginning , I have assumed that " the two smaller triangles and the two squares are fully inside the bigger triangle and square. No part of these triangles or squares is outside the larger triangle or square."
I hope this will help to clarify the problems .

Hi thickhead ,

I need time to read through your several posts .
I shall reply later if I understand your work .

bobbym · 2016-05-20 20:32:58

That is how I saw it too but as there is some disagreement in the 2 square in the bigger square problem maybe we should see if we can all agree on the answer...

thickhead · 2016-05-20 21:41:49

I have understood wong's condition that the 2 smaller squares are always within the bigger triangle. My "favourable mobility diagram" is virtual and any of its part lying outside the bigger square is to be discarded.

mr.wong · 2016-05-23 15:08:29

Hi thickhead ,

For the problem involving the squares , it seems you have
considered the area which has been covered by square A
during moving , but not the area being covered by A after
moving . This is not the aim of the original problem .
Even so the " gross mobility diagram " should looks like
a hexagon instead of a rectangle . ( if moving in a straight
line ) Thus the net area should be less than xy + x + y .
I don't know whether my comprehension of your statement
is right or not . Please correct me otherwise .
Nevertheless , I still persist that with the aim of the original
problem , the answer should be 1/9 .

thickhead · 2016-05-23 17:07:01

In the case of moving square the gross mobility diagram is also a square of double length and double height. It is not a hexagon. The idea is Take the moving square and slide all its borders over point P(x) so that P is just within the square. In the case of triangle it turned out to be a hexagon. The area of overlap of "gross mobility diagram" with the outer square( which I have taken as 2*2) comes out to be (1+x)*(1+y) From this I have subtracted 1*1 the area of the mobile square. This is the net area of mobility during which P remains inside/just on border of moving square A. the overall mobility of square a is 2*2-1*1=3. hence the probability p is calculated as (xy+x+y)/3. with 2 squares A and B probability would be p^2 and this is to be integrated over the area of south_west 1x1 square.Over the other squares the expression for p would be different and limits of integration would be different but will yield the same result.

thickhead · 2016-06-08 04:45:51

Last edited by thickhead (2016-06-08 16:20:56)

mr.wong · 2016-06-08 15:45:50

Thanks thickhead ,

I agree with your results . For n moving squares the
probability seems to be 1/ (n+1)^2 . ( but not yet proved )

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#51 2016-04-06 21:59:18

Re: geometric probability ---- squares

#52 2016-04-08 03:10:13

Re: geometric probability ---- squares

#53 2016-04-08 15:09:09

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#54 2016-04-09 06:55:03

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#55 2016-05-17 20:29:20

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#56 2016-05-17 20:58:20

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#57 2016-05-18 00:12:36

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#58 2016-05-18 08:15:28

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#59 2016-05-18 17:06:54

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#62 2016-05-19 01:22:31

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#63 2016-05-19 04:46:39

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#64 2016-05-20 15:08:58

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