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How do I solve this geometry question involving probability?
58 viewed last edited 7 months ago
Anonymous
0

I saw this problem online and I am having trouble figuring how to answer this.


Here is what I have done so far: Let A, B and C be the randomly chosen points.


Let AA' be the diameter drawn from A; and BB' be the diameter drawn from B. Is the following claim correct?

The 3 points will not have a common semi-circle if and only if:

1) B and C lie on different sides of AA'; and

2) A and C lie on different sides of BB'

Prem Kumar
2

Here's another way to see the result and also to generalise to N points chosen randomly on the circle.


Pick one of these N points and draw the diameter through it. The probability that any other point lies on one side of the diameter is 1/2. So the probability that the remaining N-1 points all lie on one side of that diameter is  \frac{1}{2^{N-1}}.  


We have N points which are undistinguished, so the net probability that there is one point amongst them  to which the above condition applies = \frac{N}{2^{N-1}}


For N=2 this gives 1 as you might expect: two points chosen at random lie on a semicircle trivially.


For N=3 it yields \frac{3}{4} as the previous arguments did too.

Mahesh Godavarti
1

What you have is reasonable. The order of the points does not matter. So, what you are saying is: pick A, B and C in that order.


The probability that A and B, do not share a semicircle is zero (obviously).


So, we have to wait for C. For A, B and C to not share a semicircle, C has to be such that neither B nor C lie on the same side of the diameter AA' and neither A nor C lie on the same side of the diameter BB'.


This makes sense, however, we can't go anywhere from here.


I would do the following:


Let x be the smaller of the distance between A and B. Then 0 \leq x \leq \pi r .


Then the probability density function of x , f(x) = \frac{1}{\pi r} .


Then, the probability that C will not be in the same semicircle as A and B, given the distance between A and B is x , g(\text{not semi-C}|x)=\frac{x}{2\pi r}.


Therefore, the total probability density function that A, B and C will not share a semicircle is given by \frac{x}{2 \pi^2 r^2} . I.e. C has to fall in the unbolded portion of the circle whose length is also x .


Therefore, total probability that A, B and C do not lie on the same semicircle is given by \int_0^{\pi r} \frac{x}{2 \pi^2 r^2} dx = \frac{1}{4} .


I am sure there is a much more elegant argument given that the answer has turned out to be this simple.


Anonymous
1
If C and B have to be opposite sides of AA', the probability of this is 1/2. If A and C have to be opposite sides of BB', the probability of this is 1/2. If C and B have to be on opposite sides of AA'; and A and C have to be on opposite sides of BB'; intuitively, the average probability of this appears to be 1/2 * 1/2 = 1/4.
Mahesh Godavarti
0
Let's formalize it.
Mahesh Godavarti
1
I think this is the elegant answer. Let A-SC be the semicircle defined by A and B-SC be the semicircle defined by B (A is at the middle of A-SC and B is at the middle of B-SC). Then C should not lie in the semicircle defined by both A and B. The probability that C does not lie in the semicircle defined by A = 1/2. The probability that C does not lie in the semicircle defined by B = 1/2. Since A-SC and B-SC are independent (note A and B are independent), then the probability that C does not lie in either semicircle = 1/2 X 1/2 = 1/4. I think this is the answer.
Vivekanand Vellanki
1

Continuing with Mahesh's response, for a given x, the probability that A, B and C dont lie on a semi-circle is g(\text{not semi-C}|x)=\frac{x}{2\pi r}.


Given x, this determines accurately the probability that the 3rd point is not on the same semi-circle.


Since x is uniformly distributed from 0 to \pi r, we need to find the expected probability.


For every x, there is an x' such that x+x'\ =\ \pi r; and

g(not semi C | x) + g(not semi C | x') = 1/2.


Hence, the expected value of g(not semi C | x) = 1/4 over 0\le x\le\pi r