Overlaps of arcs

[Guest]

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 .


mr.wong

[Guest]

If the answer of this problem is got , then perhaps a
general formula of problems of this type can be obtained .

[Guest]

If the answer to this problem is got , then perhaps
the question the problem asks can be obtained, so we will all then know the question and maybe then can try and find the answer.

Is it hard to draw a picture....??????

[Guest]

I mean that if the answer of this problem is got , then perhaps
a general formula for problems involving n arcs may be found .

mr.wong

[Guest]

The problem is ...... What is the Question..????

Is there a problem explaining the Question ......??

[Guest]

Sorry ! I really do not understand your meaning .
However , can you solve the original problem ?

mr.wong

[Guest]

Can you explain the question in more detail OR draw a diagram to illustrate

[Guest]

It is still difficult for me to draw a diagram !
But the problem should not look too complicated
even without a diagram .
Imagine a circumference (named E with length being
1 unit ) of a circle , 2 arcs A and B can slide freely
and uniformly along E . A is 1/3 unit long while B is
1/2 unit long . Thus A and B will overlap with various
common portion . The problem is to find the expected
length ( the average length ) of common portion of A
and B . But there is a special condition that there is a
point X in E such that B cannot get through X ( blocked
by X ) , while A can get through it .

Hoped this will be clear enough .

mr.wong

[leesajohnson]

Please provide the diagram with the problem.

[Guest]

The answer is [tex]\tfrac{1}{6}[/tex]. The whole thing with X can be ignored if you think about it the right way, and if you continue thinking about it you can work out it's just the probability of a random point lying in A multiplied by the probability of it also lying in B.

Or alternatively you can write out double integrals and calculate it the long way, and you should still get the same answer:
Expected length [tex]= \int\limits_{0}^{\frac{1}{2}}\left(\int\limits_{b-\frac{1}{3}}^{b}\left(a-b+\frac{1}{3}\right) da + \int\limits_{b}^{b+\frac{1}{6}} \frac{1}{3}da +\int\limits_{b+\frac{1}{6}}^{b+\frac{1}{2}}\left(b+\frac{1}{2}-a\right) da \right) db / \int\limits_{0}^{\frac{1}{2}}\int\limits_{0}^{1} da db[/tex]
[tex]= 2\int\limits_{0}^{\frac{1}{2}}\left(

\left[\frac{a^2}{2}-ab+\frac{a}{3}\right]_{b-\frac{1}{3}}^{b}
+ \left[ \frac{a}{3}\right]_{b}^{b+\frac{1}{6}}
+ \left[ab+\frac{a}{2}-\frac{a^2}{2}\right]_{b+\frac{1}{6}}^{b+\frac{1}{2}}

\right) db[/tex]
[tex]= 2\int\limits_{0}^{\frac{1}{2}}\left(\frac{1}{18} + \frac{1}{18} + \frac{1}{18}\right) db[/tex]
[tex]= \frac{1}{6}[/tex]
where [tex]a[/tex] and [tex]b[/tex] represent the position of the arcs A and B.

Hope this helped,

R. Baber.

[Guest]

Thanks R. Baber much again !

I am quite astonished that the answer is so simple
and the point X can just be ignored .
But it should not be the same if there are more
than one arcs which can be blocked by X , while
the role of A is not so important .

Thanks also to leesajohnson for your reply !

mr.wong