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Thanks thickhead ,
The answer 5/288 seems to be incorrect .
In fact the aim of finding the volumes of those
polyhedrons is to provide a method to solve the
problem involving 2 moving triangles but I failed .
Thus all the effort were wasted and I don't think
I will spend time dealing with the polyhedron any
more .
Thanks thickhead,
Due to my weakness in trigonometry , I solve the
problem in that clumsy method instead of your concise
way with trigonometry thus error will easily occur !
Hi thickhead ,
I don't mean to modify your formula to mine .
I mean perhaps I can generalize my formula for P(2)
to P(3) in my form , and I already have some ideas .
Solid geometry seems cannot be applied any more .
I will try by following the pattern of P(2) .
Hi thickhead ,
In fact I developed the formula more than 20 years
ago ! I haven't publish it to the public until last year
in a Chinese math . forum . I got the formula through
solid geometry and had not expected that the related
problem can also be solved with integration and derive
a formula which looks quite different with mine ! I hope
my formula can be modified and generalized to one involving 3 variables .
Hi thickhead ,
Does this mean ( 4 + √17)/ 768 can be simplified to
1/96 ?
Hi thickhead ,
It is quite astonished that for problem involving
2 segments , both formula which looks quite different
are in fact identical . ( In # 44, the term (1-c) should be
removed from my formula ! )
For my own convenience I have re-written your formula
for 2 variables in the following .( Where
m denotes min [min (a,1-a),min (b,1-b) ] and
n denotes max [min (a,1-a),min (b,1-b)] , thus there
is no need to consider the order of m and n . )
P(2) = [mn(1-n)- m^3 /3 ] / (1-a)(1-b)
= min ( a, 1-a, b, 1-b )* max [min (a,1-a),min (b,1-b)]
* { 1- max [min (a,1-a),min (b,1-b)]}
- min ( a, 1-a, b, 1-b ) ^3 / 3 / (1-a) (1-b)
Hi thickhead ,
E forms the complete circumference of the circle .
The answer is just simply be 1/3 * 1/2 = 1/6 un ,
the point X plays no role in this problem and can
be neglected .
The case for an arc which can be blocked by a
point X in the circumference is just equivalent to the case for a small segment moving freely inside a large segment but must keep entirely inside it .
While an arc not affected by the existence of X is
equivalent to a small segment moving in a large
segment and permitted to exceed both ends of it .
( of course still keep touching with it )
For cases involving only one arc which can be blocked by X ,
the existence of X can be neglected ,
but it will be different if there are more than 1 such
arcs . For example , if there is an additional arc C
,say also with length 1/2 unit and moving inside E , which can also can be blocked by X . Then the expected length of the overlapping portion of A , B
and C will be 1/3 * 1/3 = 1/9 un . ( by formula )
In this case the existence of X cannot be neglected .
Otherwise if arc C can get through X during moving
, then the answer will be 1/2 * 1/2 * 1/3 = 1/12 un ,
in this case the point X can also be neglected .
Hi thickhead ,
It seems you have developed another valid formula
for 2 moving segments which looks quite different
with mine , though it appears a bit too complicated .
Moreover , your formula seems not symmetric for
a and b . More data should be necessary for proving
its validity . ( I suggest you take reference to my formula .)
However , after your formula has been confirmed
I hope you can modify it to one applicable for 3 segments ( variables ) .
Hi thickhead ,
Should it be 28/135 for a=b=2/5 as well as
7/15 for 3/5 ? ( # 35 )
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?
Hi thickhead ,
Previously you have asked about the formula I used to find the
probability involving squares . You can find it in the thread
"geometric probability ---segments " # 22 .
Hi thickhead ,
For square LQMN I mean the square taken from
triangle P (0,2) ,Q (0,0) R (2,0) , thus its coordinates
will be L ( 0,1) , Q (0,0) ,M ( 1,0) and N (1,1) . So a diagram is always helpful !
Should L be (0,2) in your problem ?
Hi bobbym ,
When one end of B reaches the point X , then it
will stop there . ( also for X ) If B moves again then it can only move backwards . Whenever one end reaches X it will stop again . Thus X will never get inside of B .
Hi thickhead ,
You got the same answer of 41/420 with me .
(where the term 2/20 should be 2/10 ) But sometimes
it may cause confusion since P stands for 1/4 ( thus
( 1-P ) should be 3/4 ) while in the expression
P-2P^2 + P^3 , P^2 stands for P for 2 moving
triangles ( instead of 1/4 * 1/4 =1/16 ) and P^3 stands for P for 3 moving triangles .( instead of 1/4* 1/4 * 1/4 =1/64 ) .
But I wonder why it seems you can always get a correct answer by your way ! ( including P for triangle A or triangle X )
Hi bobbym ,
It means that there is a fixed point X lying in E which blocks B from
getting through it , thus B has to return its direction of moving , while A
is not affected by the existence of X , A can just get through it in moving .
Inside the circumference E with length 1 unit of a circle ,
there are 2 arcs A with length 1/3 unit and B with length
1/2 unit . Both A and B can move freely along E , but there
is a point X at E such that A can get through X while B
cannot . Find the expected length of the overlapping portion
of the 2 arcs .
Related problem (II) :
If a point is chosen randomly in square LQMN , find the probability
that the point lies inside at least 2 moving triangles .
( answer : 32/105 )
Hi bobbym ,
It was in # 74 .
Hi bobbym ,
I wonder why thickhead can get the formula for n squares directly
with double integration while proved by mathematical induction is not
necessary .
Thanks thickhead ,
Elegant and laborious work !
Since the common portion of 2 similar and parallel triangles
is still a similar and parallel triangle ( This is not true for squares ) ,
originally I had expected that the answer should be the square of certain
average ratio of the corresponding sides of overlapped triangle with E .
In fact it is much more complicated .
Hi bobbym ,
If the theory of that demonstration is too hard for me ,
then I shall skip it .
In the following we shall try to deduce some related results from
what we have got .
With reference to the original post in this thread :
Let P ( A | E ) denotes P of the point chosen in E which also lies in A .
and P ( ∼A | E ) denotes P of the point chosen in E which lies outside A
being 1- 1/4 = 3/4 .
(1) Let P ( A ∩ B | E ) denotes P of the point lies inside both A and B [ the intersection of A and B ] , with value being 1/10 ( according to thickhead 's result ) , thus P ( ∼A ∩ B | E ) = P ( B | E ) - P ( A ∩ B | E ) = 1/4 - 1/10 = 3/20 . being P of the point lies outside A but inside B .
( Notice that it should not be calculated as P ( ∼A | E ) * P ( B | E ) =
3/4 * 1/4 = 3/16 .)
(2) Since P ( A ∩ B ∩ C | E ) = 1/21 ( thickhead's result ) , thus
P ( ∼A ∩ B ∩ C | E ) = P ( B ∩ C | E ) - P ( A ∩ B ∩ C | E )
= 1/10 - 1/21 = 11/210 being P of the point lies outside A but inside
both B and C .
(3) P (∼ A ∩∼ B ∩ C | E ) = P ( C | E ) - P ( A ∩∼ B ∩ C | E ) -
P ( A ∩ B ∩ C | E ) - P ( ~ A ∩ B ∩ C | E ) = 1/4 - 11/210 - 1/21 - 11/210
= 41/ 420 ( may be illustrated more clearly with a Venn Diagram )
Related problem (I) :
Find the probability of a point chosen in E randomly which lies within
one and only one moving triangle . ( answer : 41/ 140 )
Hi thickhead ,
Will you expect apart from isosceles right angle triangles ,
the results will remain the same for any other triangles , say
equilateral triangles ?
Hi thickhead ,
Can you confirm your answer ( 1/10 ) of E involving 2 moving triangles
by getting the result directly with multiple integration other than indirectly deduced from results of the 4 smaller triangles of E ?
Hi bobbym ,
The answer of 13/120 was given by Andrew S at the 4th answer .
But I can't trace how he got this value .
In fact I prefer to accept thickhead 's result , i.e. 1/10 .
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 .The answer to this is 13 / 120, this is close to 1 / 10 which is what I got as an estimate of that probability. The guys at the SE would be glad to show you why it is 13 / 120 for the exact answer.
Hi bobbym ,
Would you please give me a link of that answer if convenient ?