geometric probability ---- squares

mr.wong · 2016-01-24 15:49:54

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.

Last edited by mr.wong (2016-01-24 15:50:41)

bobbym · 2016-01-24 21:50:32

Hi;

mr.wong · 2016-01-26 15:23:26

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 ?

bobbym · 2016-01-26 15:59:39

Hi;

You mean with sides that are 1 / 2 of E?

mr.wong · 2016-01-27 20:51:49

Hi bobbym ,

I mean that the word " square " is replaced by " rhombus ",
others remain unchanged in the original post .

Last edited by mr.wong (2016-01-27 20:53:47)

mr.wong · 2016-01-31 20:37:43

Now let us consider triangles :
Inside a triangle E there are 2 smaller similar triangles
A and B , both with length of relative sides being 1/2 of
that of E . All the 3 triangles are parallel with vertices upwards .A and B 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 .

bobbym · 2016-02-03 04:50:05

Hi;

There are several ways to have vertices upward, which do you want? Are these equilateral triangles?

mr.wong · 2016-02-03 15:52:03

Hi bobbym ,

The triangles are not necessarily equilateral , just any 3 similar
triangles. Any 1 of the 3 vertices may do , just that the triangles are parallel .(i.e. for all the 3 corresponding sides )

bobbym · 2016-02-03 15:59:11

Then how do we know that the answer will not depend on the shape of the triangle chosen?

mr.wong · 2016-02-04 15:16:57

Hi bobbym ,

Since the 3 Δs are similar , the result should be
the square of the ratio of any 1 side of the
expectation of common portion of A and B with
the corresponding side of E .

bobbym · 2016-02-05 03:16:29

Hi;

I could be making some giant mistake but everything keeps pointing to 1 / 10 as the probability. Of course I had to assume that what works for one triangle ( in this case I used a right triangle with sides of 1 and 1) is gonna work for them all.

mr.wong · 2016-02-06 16:07:27

Hi bobbym ,

I think that whenever A , B and E are similar figures ( nomatter squares , rhombuses or Δs ) and in parallel position , and with the same proportion of corresponding
sides ( i.e. 1/2 ) , the probability required will remain the
same , i.e. 1/3 * 1/3 = 1/9 .

bobbym · 2016-02-07 03:47:08

I will recheck and if necessary post it to the mathematica gurus but simulations are not confirming your conjecture.

This is what I am simulating.

The red and blue triangles are 1 / 2 the length of the larger outer triangle and they can move anywhere inside the largest triangle. The bases stay parallel to the largest one and they are similar triangles.

mr.wong · 2016-02-07 15:51:11

Hi bobbym ,

I have to emphasize that not only the bases ,but in fact all the 3 sides of the Δs stay parallel . Otherwise if you turn 1 of the smaller Δ by 90 degree anti-clockwisely ( it seems possible ) then only 2 of the 3 sides remain parallel .

Last edited by mr.wong (2016-02-07 15:53:47)

mr.wong · 2016-02-10 16:50:53

Related problem ( I ) : To find the volume of a polyhedron

Let X denotes a polyhedron with 6 vertices PQRSTU ,
11 sides and 7 faces :
(1) base PQRS being a square with sides 1/2 unit .
(2) Δ PTS with PT = 1/12 unit ( in fact sq.unit)
where TP is perpendicular to the base .
(3) Δ PTQ being congruent to Δ PTS .
(4) Δ RUQ with RU = 1/8 unit ( or sq.unit )
where RU is perpendicvlar to the base .
(5) Δ RUS being congruent to Δ RUQ .
(6) Δ TQU .
(7) Δ TSU being congruent to Δ TQU .

How to find the volume of X ?

bobbym · 2016-02-10 18:43:02

Hi;

Yes, I have simulated the condition that all three sides are parallel. I got an answer of about 1 / 10. I am currently moving and there is much confusion because management is inept. I will be a bit slow in replying for a while.

mr.wong · 2016-02-11 21:06:53

Hi bobbym , wish you will have a smooth move !

Now let us continue to the polyhedron X :
If we cut X into 2 halfs at TU , we get 2 pyramids .
Let us choose the left one PRUTS , denoted by Y . We may
treat Y as a pyramid having 5 faces with PRUT ( being a trapezium ) as base , and S as vertice .
We may find that PR = √1/2 unit ; ST = √ 37/144 unit ;
SU = √ 17/64 unit and UT = √ 289/576 unit .
The area of base PRUT may also be found as
√1/2 * 5/48 sq.unit .
How to find the height of the pyramid Y so that we can
calculate the volume of Y ?

bobbym · 2016-02-12 05:05:45

I would like to go back to the sliding triangles first for several reasons:

1) Are answers did not coincide.

2) My simulation techniques were not fast enough.

These need to be addressed.

mr.wong · 2016-02-13 16:21:57

Hi bobbym ,

In fact the problem of the polyhedron is much related
to that of the 3 Δs , perhaps we can obtain the
same result through different ways .

In polyhedron Y ,since trapezium PRUT and Δ PRS are mutually
perpendicular , thus the height of the pyramid is in fact the
height of Δ SPR .
Let SW denotes the height meeting PR at W ,
in fact SW = PW = WR = 1/2 PR = 1/2 √1/2 unit .
Now we can calculate the volume of Y :
V = 1/3 * 1/2 √1/2 * √1/2 * 5/48 cu. unit
= 5/ 576 cu.unit
Thus V of X = 5/ 288 cu. unit

Thus the average height ( expected area ) on PQRS
= ( 5/288)/ ( 1/4 ) sq.unit
= 5/72 sq.unit

The above value denotes the expectation of the common
portion of the 2 smaller Δs in # 13 , thus the probability
required = (5/72)/(1/2) = 5/36 ( about 0.14 )

bobbym · 2016-02-13 19:52:00

I am unable to verify that experimentally at the moment. I am still getting 1 / 10.

mr.wong · 2016-02-15 16:06:53

Hi bobbym ,

I made some mistakes in finding the volume of
the polyhedron in the past 3 posts . Now I shall
start again from the beginning and use your diagram
in #13 as reference .
Let X denotes a polyhedron with 6 vertices PQRSTU and 7 faces .
(1) base PQRS being a square with sides 1/2 unit .
(2) Δ PTS with PT = 1/3 unit where TP is perpendicular to the base .
(3) Δ PTQ being congruent to Δ PTS .
(4) Δ RUQ with RU = 1/2 unit where RU is perpendicular to the base .
(5) Δ RUS being congruent to Δ RUQ .
(6) Δ TQU .
(7) Δ TSU being congruent to Δ TQU .

In fact PT denotes the expectation of the length
of the common portion of the bases of the 2
smaller Δs ( the red and blue ) when the distance
of their bases with the base of the large Δ are
fixed to be 0 during moving in parallel direction ,
while RU denotes the expectation of the length
of the common portion of the bases of the 2
smaller Δs when the distance of their bases
with the base of the large Δ are fixed to be 1/2
unit .
By formula we find that PT = 1/3 unit while RU is
fixed to be 1/2 unit .
In this case UT will be a straight line ( linear )
otherwise it may be a curve of 2nd order if it
represents certain areas etc .
The polyhedron represents the distribution of data
of various expectation of the common portion of
the bases of the 2 smaller Δs according to the
various distances of their bases with the large Δ .
If we divide the volume of X by its base , we
get the overall expectation .
To find the volume of X , it may be easier to cut
X into 2 halfs at TU and get a pyramid Y .
The base of Y is a trapezium PRUT with PT // RU ,
since the height of the trapezium PR = √1/2 unit ,
∴ the area of PRUT may be found to be
( 1/3 + 1/2 ) / 2 * √1/2 = √1/2 * 5/12 sq.unit .
( If UT is a curve instead of a straight line , then the
area of PRUT should be found by integration if
we know its equation .)
Since trapezium PRUT and Δ PRS are mutually
perpendicular , thus the height of the pyramid is
in fact the height of the Δ SPR .
Let SW denotes the height meeting PR at W ,
in fact SW = PW = WR = 1/2 PR = 1/2 √1/2 unit .
Now we can calculate the volume of Y :
V = 1/3 * 1/2 √1/2 * √1/2 * 5/12 cu. unit
= 5/ 144 cu.unit
∴ V of X = 5/ 72 cu. unit
Thus the average height on PQRS
= ( 5/72)/ ( 1/4 ) unit
= 5/18 unit
The above value denotes the expectation of
the base of common portion ( also a similar
and parallel Δ ) of the 2 smaller Δs in # 13 .
Thus the probability required
= (5/18)*(5/18) = 25/324 ( about 0.077 )

mr.wong · 2016-02-17 23:29:50

Perhaps we can find the probability in another way .
In the diagram of #13 let the coordinates of the large Δ
denoted by (0,0) ( 1,0) and ( 0,1) , and let the 2 smaller Δs
denoted by A and B . Their vertices at right angle will be
represented by VA and VB . It is clear that both VA
and VB lie within the Δ (denoted by D) with coordinates (0,0) (1/2,0) and (0,1/2) or intersections of the 3 lines with equations
(i) y=0 ; (ii) x=0 and (iii) x+y = 1/2 .
If a point is chosen randomly inside D , what will be the expected
values of its coordinates ?

mr.wong · 2016-02-20 20:37:52

Let us try to solve the original problem step by step from special case to general case .
Let (a1,a2) denotes the coordinate of VA and (b1,b2) denotes the coordinate of VB .
Suppose a2 = b2 , i.e A and B have same distance with the base of the large Δ .
The length of the base of the similar Δ formed by overlapping A and B
will be 1/2 - [ max ( a1,b1 ) - min ( a1,b1 )] i.e. 1/2 - |a1-b1| ( absolute value )
Since both a1 and b1 lie between 0 and 1/2 , thus |a1-b1| also lie between
0 and 1/2 . We can find the expected value of 1/2 - |a1-b1| by solid
geometry .
Let PQRST denotes a pyramid with base PQR being a Δ where PQ
= 1/2 unit ( length of |a1-b1| ) while PR = 1/2 unit ( range of a2 or b2 : also from
0 to 1/2 ) . RPTS being a square with length 1/2 unit , where PT denotes the value of
1/2 - |a1-b1| = 1/2 since |a1-b1| = 0 when both a1 = b1 = 0 .
Similarly RS = 1/2 when a1 = b1 = 0 with a2 and b2 = 1/2 .
To calculate the volume of the pyramid we treat the square as base with
Q as the vertex .
Since V = 1/3 * 1/2 * 1/2 * 1/2 = 1/24 cu.unit
Since the area of the original base Δ PQR = 1/2 * 1/2 * 1/2 = 1/8 sq.unit
Thus the average height ( expected value of 1/2 - |a1-b1| ) = 1/3 unit .
Thus the area of the similar Δ formed by overlapping A and B will be
1/3 * 1/3 * 1/2 sq.unit , dividing by the area of the large Δ being 1*1*1/2
sq.unit , we get the probability required = 1/9 .

Nextly we shall find what will be the result if a2 ≠ b2 .

bobbym · 2016-02-21 02:36:20

It might be 1 / 9 but I would like the theory and the simulation to agree.

mr.wong · 2016-02-22 00:31:55

Hi bobbym ,

I apologize that I had made mistakes again in the last post .
Under the condition of a2 = b2 , the probability required should
be 5/12 * 5/12 = 25/144 ( about 0.17)
I am not sure which answer provided by me is correct , or may
be all wrong .
It seems it is impossible to solve this problem by solid geometry ,
for there are too many variables thus out of the capacity of solid
geometry . Perhaps this problem may be solved by multiple integration
which I am not familiar . Thus I shall postpone to discuss this problem
for myself , but wish someone skilful in multiple integration may find a solution for it .
Later I shall discuss various related problems other than triangles .

Math Is Fun Forum

#1 2016-01-24 15:49:54

geometric probability ---- squares

#2 2016-01-24 21:50:32

Re: geometric probability ---- squares

#3 2016-01-26 15:23:26

Re: geometric probability ---- squares

#4 2016-01-26 15:59:39

Re: geometric probability ---- squares

#5 2016-01-27 20:51:49

Re: geometric probability ---- squares

#6 2016-01-31 20:37:43

Re: geometric probability ---- squares

#7 2016-02-03 04:50:05

Re: geometric probability ---- squares

#8 2016-02-03 15:52:03

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#9 2016-02-03 15:59:11

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#10 2016-02-04 15:16:57

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#11 2016-02-05 03:16:29

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#12 2016-02-06 16:07:27

Re: geometric probability ---- squares

#13 2016-02-07 03:47:08

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#14 2016-02-07 15:51:11

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#15 2016-02-10 16:50:53

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#16 2016-02-10 18:43:02

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#17 2016-02-11 21:06:53

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#18 2016-02-12 05:05:45

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#19 2016-02-13 16:21:57

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#20 2016-02-13 19:52:00

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#21 2016-02-15 16:06:53

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#23 2016-02-20 20:37:52

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