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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 ?
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 .
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 .
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 .
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 )
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 .
Hi bobbym ,
I mean that the word " square " is replaced by " rhombus ",
others remain unchanged in the original post .
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 ?
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 bobbym ,
The use of function with min and max will save much labours , for it applies to every possible conditions but not just certain conditions .
I pay respect to your favour and hard works , but time and spirit should be spent on the right place .
Later I shall discuss some related topics in a new thread .
Hi Relentless ,
Thanks for your reply ! Though your formula is valid
for x=a=b=2/5 , but if you substitute x=a=b by 3/5 ,
you will get a not so correct result 9/20 instead of
7/15 .
Hi bobbym ,
Your second formula is too complicated , I can't even
check its validity for a = b = 2/5 .
Don't spend time on your formula anymore ! Just use
my formula for further investigation if you have checked its validity .
Hi bobbym ,
For example , let a = b = 1/10 , then your formula will get
P = ( 1/100 - 2/10 + 3/100 - 3/10 + 1 ) / ( - 27/10 )
= [( 1 +3 +100 - 20 - 30 ) / 100 ] / ( -27/10 )
= ( 54/100 ) *( - 10 / 27 )
= - 1/ 5
Another example , let a = b = 2/5 ,
my formula yields P = 28/135
while your formula yields P = 1/5 = 27/135 ;
which one is more correct ?
Hi bobbym ,
I think your formula may be correct for certain limited cases , but not always . Also the formula should be symmetric for a and b .
Moreover , if both a and b are small enough , the formula may yield a negative value !
Hi bobbym ,
I recognize that I can't derive a formula for 3 segments from your data .
The following is the problem involving 2 smaller segments and it's formula .
Problem :
Inside a segment E ( with length being 1 unit ) there are 2 smaller segments A and
B ( with respective lengths a and b units , where 0≦ a ≦ 1 and 0 ≦ b ≦ 1) which slide freely along E .
(1) Find the expectation of the length of common portion of A and B .
(2) If a point is chosen randomly on E , find the probability that the point lies
within the common portion of A and B .
Formula for (1) :
min ( a,b ) - [ min ( 1-a, 1-b )^3 - max ( 0, 1-a-b )^3 ] / 3(1-a)(1-b) unit
where min denotes the smallest value while max denotes the greatest value .
By dividing the expression by 1 unit , we get the answer of (2) readily .
( Welcome to check the validity of the formula by computer .)
Hi bobbym ,
Thank you very much for your hard working , but don't spend too much spirit by your way , it does not worth !
Please share some original data with me ! I can do it much more concisely since I already have a formula for 2 smaller segments .
I need time to analyse your formula , but what can I do will be limited unless I have enough original data in hand .
Hi bobbym ,
Yes , that's what I want . Can you give a help ?
In fact I am more interested to find a formula for problems involving 3 smaller segments with their individual lengths other than 1/2 with respect to e ( length of E ) .
Let P ( a, b, c ) denotes the probability ( estimated ) of a point chosen in E lies within A ,B , and C with their lengths being a, b, and c respectively ,( where a , b, and c may be values other than 1/2 , say from 1/10 to 9/10 ,for various a,b, and c , i.e. their values may be different .)
If I can obtain enough such data of P ( a, b, c) ,perhaps I can guess a general formula for it .
Hi bobbym ,
The 2 problems look like quite similar but in fact they are quite different . In that problem the probability of the last hit depends on the result of the previous throws , but in my problem the probability of the point lies within each segment is fixed .( Each segment has same length .) Moreover , in that problem n has to start from 3 but in mine n can be started from 1 .
However , I am amazed that the results of them will be exactly the same if one of them is shifted
by 2 positions .
Assume that the formula P(n)= 1/n+1 is true, then it can further be generalized .
Let P(n,r) denotes the probability that the point lies within exactly r smaller segments out of the n ,(where r varies from 0 to n ) then P(n,r) ≡ 1/n+1 ,i.e. no matter what is the value of r , P(n,r) always = 1/n+1 .
Since there are nCr items for each r , if their total P sum up to 1/n+1 ,then P of each item will be 1/[(n+1)(nCr )]. ( The total no. of various items will be nC0 + nC1 + nC2 + ........+nCn = 2 ^ n .)
E.g. ,let n = 3,
for r = 0, there is only 1 such item,( neither A nor B nor C ) and P(3,0) = 1/4;
for r = 1, P(3,1) also = 1/4 ,since there are 3C1 = 3 such items ,( either A or B or C ),thus each item will get
P = 1/4 * 1/3 = 1/12 ;
for r = 2, P(3,2) also = 1/4 , since there are 3C2 = 3 such items ,( either A with B ,or A with C , or B with C ),
thus each item will get P = 1/12 also ;
for r = 3, there is only 1 such item ( A with B with C ) and P (3,3) = 1/4 .
Hi ! bobbym
I think I am unable to prove the formula. I cannot even deduce P(2) to P(3),that's why I got a wrong answer for P(3).
However,somebody can solve P(3) by multiple integration and get a correct answer. Thus the formula may be proved by mathematical induction with assistance of multiple integration.
Much thanks to bobbym !
Previously I had never thought there may be such a relation between n and P. I will try to
prove that formula by mathematical induction .
Thanks again bobbym!
Now let n denotes the no. of smaller segments inside E( each with length being 1/2 of E). For n=1, surely P = 1/2 ; for n=2, we get P= 1/3 ; for n=3, you get P=1/4 ; for n=4, will you guess P= 1/5 and so on ?
Thank you very much bobbym ! I am looking to find why the 2nd answer is incorrect.
If there are only 2 smaller segments ( say A and B only) inside E,will you get P= 1/3 ?
Inside a segment E there are 3 segments A, B, and C which move freely along E, and all with lengths being 1/2 of E. If a point is chosen randomly on E, find the probability that the point lies within A, B, and C at the same time.
There are 2 possible answers to this problem:
(1)1/4
(2) about 1/5
I am not sure which one is correct,can the correct answer be verified by using computer?