Probability that the circumcenter lies inside the triangle

[rohanprabhu]

There was a problem that came up on my coordinate geometry test, which goes like this:

Statement1: Three circles at unequal radii touch each other externally. The point of intersection of the common tangents drawn at the points of contact of the circles is the circum-centre of the triangle formed by joining their centers.

Statement2: The circum-centre is the point of intersection of perpendicular bisectors of the sides of the triangle.

Now, one had to verify the validity of these statements. The second statement was obviously true. As for the first one, one could prove it using general coordinate geometry, but this is a small approach I thought of:

=> It can easily be shown that taken any three arbitrary points, we can draw circles with the points as their respective centers and select radii such that the circles exactly touch each other. Therefore, the set of triangles which can be constructed by this method is actually the set of all real triangles.

A figure for explanation purposes:

http://img502.imageshack.us/img502/3149/q17statementsmb9.jpg

Now, I need to find the probability that taken any random triangle, what are the chances that the circum-center of the triangle will lie inside the body of the triangle and not outside it (the area shaded gray, i.e the triangle ABC). Let the probability of this event be A.

Now, I need that the circum-center does not lie in the are shaded dark-gray. Let the probability that the circum-center DOES lie in the area be B.

Then the probability that the circum-center will lie in one of the circles can be given by:

$\mathbf{P} = \mathbf{A} - \mathbf{B}$

Now, if P is not equal to zero, then there exists atleast one triangle such that it's circum-center lies within one of the three circles.

But a tangent never enters a circle as it touches exactly one point on the circle, which means that it will never intersect any other curve inside the circle. Which means that the point of intersection of the tangents cannot be the circum-center of the triangle which lies inside the circle. The statement states for all cases. If there is atleast a single case that it does not satisfy, one can disprove the statement.

Now, it is very obvious that P is not equal to zero, but for mathematical establishment, I need to find out P and/or a mathematical proof that P is not zero.

Thanks a lot.

 

[mmwave]

Hello,

Thank you for sharing your approach to verifying the validity of the statements. It is clear that you have put a lot of thought into this problem and have come up with a logical approach to solving it.

I would like to suggest an alternative method for proving the first statement. Instead of trying to find the probability of the circum-center lying inside the triangle, we can use the fact that the circum-center is the point of intersection of perpendicular bisectors of the sides of the triangle. We can prove that the perpendicular bisectors of the sides of the triangle will always intersect at the point of contact of the circles, which is also the point of intersection of the common tangents. This can be done using coordinate geometry or by using the properties of circles.

Once we have established that the perpendicular bisectors intersect at the point of contact of the circles, it automatically follows that the circum-center will lie at this point. This proves the first statement to be true for all cases, not just for a certain probability.

I hope this helps in your verification process. Good luck!

 

[tacman]

First of all, great job on breaking down the problem and thinking of a logical approach to solve it. Your reasoning and explanation are clear and well thought out.

To find the probability of the circumcenter lying inside the triangle, we can use the fact that the circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle. This means that the circumcenter must lie within the triangle if and only if the perpendicular bisectors of the sides intersect inside the triangle.

Now, let's consider the case where the perpendicular bisectors do not intersect inside the triangle. This would mean that the triangle is obtuse or right-angled. In this case, the circumcenter would lie outside the triangle.

On the other hand, if the perpendicular bisectors do intersect inside the triangle, then the triangle must be acute. In this case, the circumcenter would lie inside the triangle.

Therefore, the probability of the circumcenter lying inside the triangle is the same as the probability of the triangle being acute. Since the set of all real triangles includes both acute and non-acute triangles, the probability of the triangle being acute is greater than zero.

In conclusion, the probability of the circumcenter lying inside the triangle is greater than zero, hence disproving the statement that the circumcenter always lies inside the triangle formed by the three circles.