# Geometry Probability Question

Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.

Hi everyone I found this interesting question; help is appreciated! :)

We put 15 points on a circle O equally spaced. We then select two points A and B randomly from the 15 points. Find the probability that the perpendicular bisectors of OA and OB intersect inside Circle O.

My Progress:

The fact that these are perpendicular bisectors makes me want to think of triangles. So we have a triangle OAB and we want perpendicular bisectors of OA and OB to intersect inside the circle. The intersection of perpendicular bisectors is I believe the circumcenter. So we want the circumcenter of triangle OAB to inside circle O. After this point I have no idea what to do.

Thanks

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What do "$OA$", "$OB$" and "triangle $OAB$" mean when $O$ is a circle? –  joriki Sep 28 '12 at 20:57
I guess $O$ is the center? –  leonbloy Sep 28 '12 at 20:59
yes O is the center –  Mit Tripathi Sep 28 '12 at 21:03
Hint: Assumming $O$ is the center, and the points are on the circle (not inside it!), see that the event of interest (intersection inside circle) depends on the angle between $OA$ and $OB$. –  leonbloy Sep 28 '12 at 21:04
I'm voting to close this question as off-topic because it was asked while it was a question in an on-going contest. –  Joel Reyes Noche Mar 17 at 4:13

HINT: Show that the perpendicular bisectors of $OA$ and $OB$ meet on the circle precisely when $\angle AOB=\frac23\pi$. If $M$ is the midpoint of $OA$, and $P$ is the point where the bisectors meet, look at $\triangle OMP$ when $P$ lies on the circle: it’s a very familiar triangle
Fix a point $E$ on the circle as a bases for measuring angles. If $\angle EOA=\alpha$, then the perpendicular bisector of $OA$ intersects the circle at points at angles $\alpha-\frac\pi3$ and $\alpha+\frac\pi3$. Similar for $\angle EOB=\beta$. Therefore, the bisectorrs of $OA$ and $OB$ intersect inside the circle iff $\alpha-\frac{2\pi}3<\beta<\alpha+\frac{2\pi}3$. With very many points on the circle to select random points from, the probability would therefore be approximately $\frac 23$ (the share of $2\cdot\frac{2\pi}3$ in $2\pi$). However, here we know that $\beta$ differs from $\alpha$ by a multiple of $\frac{2\pi}{15}$, hence we need $\beta=\alpha+k\cdot\frac{2\pi}{15}$ with $-5<k<5$, i.e. only 9 out of 15 will do. This results in a probability of $\frac35$.
(It depends a little though on how we treat the case $A=B$: Do we say that the identical lines intersect? Or do we even impose the restriction that $A\ne B$ in the first place? In the latter case, we'd have a probability of $\frac8{14}=\frac47$ instead)
@MitTripathi: No, if $A$ and $B$ are not necessarily distinct, the answer is $\frac9{15}=\frac35$. It is clear anyway that the answer is a multiple of $\frac1{15}$. Only if we are to choose $B$ only among the remainig 14 points the answer must be replaced with $\frac8{14}=\frac47$. –  Hagen von Eitzen Sep 28 '12 at 21:28