geometric probability ---- squares

bobbym · 2016-02-22 01:40:32

Hi;

I do not agree with .17, I am still looking for a solution.

mr.wong · 2016-02-23 15:24:09

Related problem (II)
Inside a square E of 1*1 there are 2 rectangles A and B
both of 1* 1/2 and move freely inside E . If a point is chosen
randomly on E , find the probability that the point lies inside
A and B at the same time if
(i) A and B are mutually "parallel ".
(ii) A and B are mutually " perpendicular ".

bobbym · 2016-02-23 18:48:03

Hi;

Parallel to E also?

mr.wong · 2016-02-24 16:36:40

Hi bobbym ,

Once you put either A or B into E , you cannot move it to any
position non-parallel to E , otherwise you just can't put it into E .
Thus A or B must be parallel to E .
In fact for a rectangle with 1 unit * x unit which can be put into E
in a non-parallel position ( with 1 side parallel to a diagonal of E ),
the greatest value of x is about 0.447 .

bobbym · 2016-02-24 18:11:03

Okay, thanks for the info.

To take a step back the exact answer for the 3 triangle point problem is 13 / 120.

mr.wong · 2016-02-26 23:05:07

Hi bobbym ,

The value 13/120 is quite near 1/9 = 13/ 117 .

bobbym · 2016-03-10 17:54:47

Related problem (II)
Inside a square E of 1*1 there are 2 rectangles A and B
both of 1* 1/2 and move freely inside E . If a point is chosen
randomly on E , find the probability that the point lies inside
A and B at the same time if
(i) A and B are mutually "parallel ".
(ii) A and B are mutually " perpendicular ".

Can either side ( long or short ) of either smaller triangle be parallel? For instance could they be oriented like this

mr.wong · 2016-03-10 21:22:08

Hi bobbym ,

For the 2 smaller rectangles ( not triangles ) to be parallel ,
I mean that they are in the position like a " = " sign .
For they to be perpendicular , I mean that they are in the
position like a " τ " sign .
Thus in your diagram they should be mutually perpendicular .

Last edited by mr.wong (2016-03-10 21:25:25)

bobbym · 2016-03-11 05:08:42

For mutually perpendicular it is easy, I would say it is 1 / 4. This is easy to prove.

mr.wong · 2016-03-11 15:29:16

Hi bobbym ,

You are right ! For the case of perpendicular , the
answer is intuitively to be 1/4 .
For them to be parallel , the answer will be just the same
as the problem involving 2 segments .(i.e. P = 1/3 )

Last edited by mr.wong (2016-03-14 16:00:28)

bobbym · 2016-03-11 20:22:08

Hi;

Then we are done?

mr.wong · 2016-03-11 21:56:40

Hi bobbym ,

That's right !
Let us return to #1 . If the condition " keep parallel " is
not necessary , will the result be P = 1/4 * 1/4 = 1/16 ?

mr.wong · 2016-03-19 16:58:02

Related problem (III)

Inside a triangle E there is a smaller triangle X formed by joining
the mid-points of the 3 sides of E , and stays fixed in its position .
Another similar triangle A , with lengths of sides being 1/2 of that
of E and parallel to E with corresponding vertices facing the same
direction , can move freely but parallelly inside E .
If a point is chosen randomly on E , find the probability that the
point lies inside A and X at the same time .

Last edited by mr.wong (2016-03-19 17:00:26)

bobbym · 2016-03-26 19:09:36

Hi;

I am getting a probability of about 1 / 8.

mr.wong · 2016-03-27 21:01:35

Hi bobbym ,

You get P = about 1/8 , since P of the point lies within X = 1/4 .
Does this mean that P of the point lies within A = about 1/2 ?
I have not got the answer by myself yet . If I try to solve it with
solid geometry , I need to find the volume of a polyhedron consisting
of quadratic curve , and then its average height , but I have no confidence
to do so .
Let us return to the 3 triangles problem at # 6 , with reference of
your diagram at # 13 . If one of the smaller Δ , say B , is fixed
with its vertice corresponding the right angle , at coordinate
at either (i) (0,0) or (ii) (1/2, 0) or (iii) ( 0, 1/2) , while Δ A
is movable , will P all = 1/6 ? [ Where the coordinates of
the 3 vertex of the large Δ E are pre-set at ( 0,0) ,( 1, 0) and ( 0,1) ]

bobbym · 2016-03-27 21:10:19

Hi;

I am working on another method but unfortunately it does not yield the same answer of 1 / 8, so I will keep trying.

mr.wong · 2016-04-01 01:03:54

Hi bobbym ,

I don't know how to add a diagram as affix to my post !

bobbym · 2016-04-01 02:08:15

You must first upload it to http://imgur.com/

mr.wong · 2016-04-01 23:18:16

Hi bobbym ,

I mean a diagram generated with Geogebra .

bobbym · 2016-04-02 00:06:12

Use graphics view to clipboard from the File ->Export menu. Or, you can take a screenshot and save it as a jpeg.

mr.wong · 2016-04-02 22:56:19

Hi bobbym ,

I hope you can see the diagram !

Related problem (III) ( B )

Inside a triangle E there is a smaller triangle X formed by joining
the mid-points of the 3 sides of E , and stays fixed in its position .
Another similar triangle A , with lengths of sides being 1/2 of that
of E and parallel to E with corresponding vertices facing the same
direction , can move freely but parallelly inside E .
If a point is chosen randomly on X , find the probability that the
point also lies inside A .

Procedure to solve this problem with 3-dimensional geometry :
Let VA denotes the right - angle vertex of triangle A , with
co-ordinate lies within the triangle PBA . ( with co-ordinates (0,0) ,
(4,0) and (0,4) ) For various locations of VA , the area of overlapped
portion of triangles A and X can be found experimentally .
E.g. , when VA = (0,0) or (4,0) or (0,4) ( at point P , B and A resp ), the CA ( common area ) all = 0 ; if VA moves to the point S (2,2) ,
then common portion will be the square SC , with area 4 units .
Basically we should collect 6 co-ordinates of VA , besides the 4
points stated , 2 more co-ordinates (2,0) and (0,2) should be tackled .
After we have collected certain data , we try to illustrate the relation
between the CA and the various co-ordinates of VA with a 3-dimensional diagram : x-axis will be the line PB , y-axis will
show the line PA , while z-axis will show the values of CA .
If the curves formed by joining the various points of CA seem
to be linear ( straight lines ) , then it will be ok and we shall continue
to calculate the volume of the corresponding polyhedron , and then
the average height , representing the average value of CA . Divided this value by the area of triangle ABC , the probability required will be obtained .
Otherwise we have to collect more data , i.e. more values of CA corresponding to more locations of VA inside the triangle PBA . If
the curves obtained shown to be non-linear ( e.g. quadratic ) , to calculate the corresponding volume we need to use integration or even multiple integration .

bobbym · 2016-04-03 01:26:20

Hi;

I can see the diagram after enclosing it in img tags. Also, the intersection of X and A is usually not a square.

mr.wong · 2016-04-03 16:00:31

Hi bobbym ,

You are right , the intersection of X and A is usually not a square .
But we are only interested in their common area . Sometimes we have to
count the area part by part , thus a geometric diagram with co-ordinates
will be much helpful .
Perhaps in the 3-dimensional diagram being a polyhedron with x and y -axis
denoting various co-ordinates of VA , while z-axis denoting the common area ,
the corresponding values x , y and z may always be represented by the equation :
z = axx + bx + cyy + dy + exy + k
while the values of the coefficients a , b , c , d , e and k may be found if we have
collected enough data and solving the corresponding set of equations . In fact the
above equation represents the surface covering the polyhedron , using multiple
integration we may find the volume of the polyhedron directly without cutting it
to appropriate portions and find the corresponding volumes by solid geometry .

mr.wong · 2016-04-05 23:49:31

After collecting enough data and solving the corresponding
coefficients , we obtain the following equation :
z= 1/2 x - xx + 1/2 y -yy - xy
= 1/2 ( x-2xx + y - 2yy - 2xy )
( What will be the maximum value of z for
0 ≤ x≤1/2 and 0≤y≤1/2 ? ) ( in fact 1/2
means 1/2 of length of 8 units = 4 units as
in the diagram shown in # 46 .)

It's time for us to draw a 3-dimensional
diagram . But it is difficult for me to draw a
diagram consisting of non-linear curves using the
Geogebra , therefore I can only describe
the diagram in words .
The base of the polyhedron is a right -angle
triangle PQR with PQ = QR = 1/2 un.
For y=0 ,(i.e. the x-axis ) the above equation
becomes z=1/2 ( x-2xx) , being a quadratic
curve over QR . Similarly for x=0 ,
z=1/2 ( y-2yy) ,being a quadratic curve over
QP .
For x+y = 1/2 , the equation becomes
z=1/2(1/2 - 2xx-2yy-2xy)
= 1/4 -xx-yy-xy
also being a quadratic curve over PR .
( The above 3 curves all have maximum
values = 1/16 at the mid-points of PQ , QR
and PR .)
Can anyone please find the volume of the
polyhedron using integration for me ? Thanks
in advance !

bobbym · 2016-04-06 03:03:27

Hi;

If your equation looks like this,

it is easy to maximize z subject to the constraints 0 ≤ x≤1/2 and 0≤y≤1/2

z is at a maximum when x = 1 / 6 and y = 1 / 6 and the maximum value is 1 / 12

Math Is Fun Forum

#26 2016-02-22 01:40:32

Re: geometric probability ---- squares

#27 2016-02-23 15:24:09

Re: geometric probability ---- squares

#28 2016-02-23 18:48:03

Re: geometric probability ---- squares

#29 2016-02-24 16:36:40

Re: geometric probability ---- squares

#30 2016-02-24 18:11:03

Re: geometric probability ---- squares

#31 2016-02-26 23:05:07

Re: geometric probability ---- squares

#32 2016-03-10 17:54:47

Re: geometric probability ---- squares

#33 2016-03-10 21:22:08

Re: geometric probability ---- squares

#34 2016-03-11 05:08:42

Re: geometric probability ---- squares

#35 2016-03-11 15:29:16

Re: geometric probability ---- squares

#36 2016-03-11 20:22:08

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#37 2016-03-11 21:56:40

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#38 2016-03-19 16:58:02

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#39 2016-03-26 19:09:36

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#40 2016-03-27 21:01:35

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#41 2016-03-27 21:10:19

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#42 2016-04-01 01:03:54

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#43 2016-04-01 02:08:15

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#44 2016-04-01 23:18:16

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#45 2016-04-02 00:06:12

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#46 2016-04-02 22:56:19

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#47 2016-04-03 01:26:20

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#48 2016-04-03 16:00:31

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#49 2016-04-05 23:49:31

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#50 2016-04-06 03:03:27

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