Perpendicular bisectors in a circle

We know that each of the 31 points in S is equally spaced within a circle centered at O. From this, we can deduce that the angle between two neighboring points in S (Say, point X and Y) is $\angle XOY = \frac {360}{31}^ \circ$

Next, we assume that we have taken a random point A within S , leaving 30 points remaining in set S . Which point we take first does not matter, as the remaining points within set S will still contain the same angle relative to A 's position.

From these 30 points, half of them are on the one side side of OA , while the remaining half are on the other side. Since these halves are perfectly symmetrical, we only need to consider one of these halves, which we will call set S' from now on.

Now the problem has been reduced to finding out how many points within S' has a perpendicular bisector which intersects with OA 's perpendicular bisector within the circle O .

From the above, we know that for $i=1..15 , \angle AOS_i = i \times \frac{360}{31}^\circ$

Next, we ssume that OA 's perpendicular bisector intersects OA at point A' . OA is equal to circle O 's radius ( r ), and thus $OA' = \frac{1}{2} r$

Then assume we take a random point in the remaining set S' and call it point B . OB 's perpendicular bisector intersects OA 's perpendicular bisector at point C .

The points A' , O and C forms a right triangle, with $\angle A'OC = \frac{\angle AOB}{2}.$

For point C to still be within circle O , OC must be less than r . And thus, $OC \leq OA$ ,

$OC \leq 2OA'$ ,

$\cos (\angle A'OC) \geq \frac{1}{2}$ ,

$\angle A'OC \leq 60 ^\circ$

And since $\angle A'OC = \frac{1}{2} \angle AOB$ , we can deduce a single rule: for any given point A and B ,

$\angle AOB \leq 120.$

From the 15 points in set S' , only 10 of them fulfills this requirement. Thus $\frac {m}{n} = \frac {10}{15} = \frac {2}{3}$ , and $m+n = 5.$

Suppose C = midpoint of OA, D = midpoint of OB, and E = intersection of perpendicular bisectors of OA and OB. [ Note that OC = OD = $\frac r 2$ , where r = radius of circle and $\angle OCE = \angle ODE = 90^\circ$ . ]

We wish to compute the probability that $OE < r$ . If $\theta = \angle COE$ then $OE = \frac{OC}{\cos\theta} = \frac{r}{2\cos\theta}$ so the condition $OE < r$ holds if and only if $\cos\theta < \frac\pi 3$ , i.e. if and only if $\angle AOB < \frac{2\pi} 3$ .

Now it's easy to compute the probability. Without loss of generality, fix A. Then B can be at most $[\frac{2\pi} 3 / \frac{2\pi}{31}] = 10$ points away from it. Thus B can take 20 possibilities and the probability is 20/30 = 2/3, i.e. m=2, n=3.

We ask a related question: If A and B are two distinct points on circle O, and if the perpendicular bisectors of OA and OB intersect at a point also on circle O, what is the measure of angle AOB?

So let C be the midpoint of OA, let D be the midpoint of OB, and let the perpendicular bisectors of OA and OB meet at point E on circle O. Let r be the radius of circle O, and consider triangle OCE. Note that C is a right angle, OC has length r/2, and OE has length r. Therefore OCE is a 30-60-90 triangle, with angle COE being 60 degrees. Similarly, angle DOE is also 60 degrees. We conclude that angle AOB is 120 degrees.

It follows that, given any two points A and B on circle O, the perpendicular bisectors OA and OB intersect inside O if angle AOB is less than 120 degrees, they intersect on O if angle AOB equals 120 degrees, and they intersect outside O if angle AOB is greater than 120 degrees.

We are now ready to solve the original problem. Let A be one of the 31 equally spaced points on the circle, and then choose B at random from the 30 remaining points in set S. From the discussion above, we want the probability that angle AOB is less than 120 degrees. Note that if M and N are elements of S that are adjacent to each other on circle O, then angle MON is 360/31, or approximately 11.61 degrees. So, starting from point A, to meet the conditions of the problem, point B can be chosen to be any of the twenty points in S closest to A (ten points on each side of A), since angle AOB will be approximately 116.1 degrees for the choice of B that makes angle AOB as large as possible.

We conclude that there are 20 "good" choices for point B out of the 30 possible choices. So the probability that OA and OB intersect inside the circle is 20/30 or 2/3, and so the answer to the problem is 2+3=5.

To evaluate the probability, we first find the 'area' in which the required condition is true.

Take 2 points $A$ & $B$ on the circle of radius $R$ such that their perpendicular bisectors intersect at the circumference,at say $P$ . Join $OP$ . Then $OP = R$ . If $M$ & $N$ be midpoints of $OA$ & $OB$ , then $OM = \frac{R}{2}$ . So $\cos \angle POM= \frac{\frac{R}{2}}{R}= \frac {1}{2}$ , hence $\angle POM = 60$ . Likewise $\angle PON = 60$ . [By 'radius' in the following,I mean a line joining centre to a point on circumference]

Now $31$ points are equally spaced. So angle at centre bounded by any $2$ consecutive radii is $\theta = \frac{360}{31}$ .

Now, in a $120$ -sector,with $OA$ as one of it's boundary radii, if $n$ more radii fit in,joining such points to the centre,then $\theta$ -sectors in this larger 120 sector are GIF: $\frac{120}{\frac{360}{31}}$ in number i.e., $10$ . So corresponding more radii which satisfy the condition w.r.t. $OA = n + 1 - 1 = n = 10$ .

This is also true for the other side of $OA$ opposite to $OB$ . So we get that within the arc with midpoint $A$ subtending an angle $60*2 * 2 = 240$ at the centre, any radius satisfies the conditon with respect to $OA$ .

Such radii are $10*2=20$ .

So for each radius,we have $20$ more in a $240$ -sector ABOUT it satisfying the condition.

So for $31$ radii,total number of pairs is $\frac{31*20}{2}$ . [excluding the doubly counted sets]

So probability is $\frac{\frac{31*20}{2}}{\binom{31}{2}}$ . It is easy to compute it's $\frac{2}{3}$ .

Let $A, B$ be any two points on the circumference of a circle with radius $r$ and center $O$ .
Let $C, D$ be the midpoints of $AO$ and $BO$ , respectively.
Let the perpendiculars from there meet at $X$ .
Look at $\Delta COX$ (which is congruent to $\Delta DOX$ ). This is a right triangle with one leg = $|AO|$ = $r/2$ .
Let $\theta$ denote $\angle AOX$ . Then $|OX|$ = $|CX| / cos \theta$ .

$X$ is inside the circle iff $|OX| < r$ , which a little manipulation shows is equivalent to $cos θ > 1/2$ . In other words, $-π/3 < θ < π/3$ .

If $A$ and $B$ are any two of 31 equally spaced points on the circumference, $∠AOB = 2θ = 2πn/31$ for some $n$ running from -15 to +15.
$2πn/31 < 2π/3$ is equivalent to $n < 31/3$ , or $n ≤10$ . Similarly on the negative side: $n ≥ -10$ . Excluding $n=0$ leaves 20 values of $n$ that put the intersection $X$ inside the circle.
Hence the required probability is $\frac{20}{30} = \frac{2}{3}$ . Therefore $m+n=5$

Refer to the following diagram when reading my solution:

alt text

First, we will find the maximum angle between two points such that the perpendicular bisectors intersect inside the circle. Consider a circle with center $A$ (sorry for the overlapping labels) and a point $B$ . Now, consider a point $B'$ such that the perpendicular bisectors of $\overline{AB}$ and $\overline{AB'}$ intersect on the circle. Then, if the measure of the angle between two points chosen in the problem is strictly less than $m\angle BAB'$ , the perpendicular bisectors will intersect on the circle.

Let $C$ be the intersection of the two perpendicular bisectors, let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AB'}$ . Then, $m\angle ADC=m\angle AEC=90^\circ$ Also, since $\overline{AC}$ is a radius, $AC=2AD=2AE$ so both $\triangle ADC$ and $\triangle AEC$ are 30-60-90 triangles, and $m\angle BAB'=120^\circ$ .

Now, back to the problem. WLOG, pick one point to be $A$ . Then, if $B$ is within $120^\circ$ of $A$ on the circle, the condition will be satisfied. There are $10=\left\lfloor\frac{31}{3}\right\rfloor$ satisfactory points $B$ on either side of $A$ , so the probability is $\frac{10+10}{30}=\frac{2}{3}$ , and the answer is $2+3=\boxed{5}$ .

let the points A and B be chosen so that the perpendicular bisectors of OA and OB intersect on the circumference of the circle. Let this point be P. Join OP. let PM and PN be perpendicular to OA and OB respectively. Then the triangles MOP and NOP are congruent. Extend OB , let the tangent
at P meet OB at K. ∠NPK=90-∠NPO. IN triangle PNK , ∠NKP = 90-∠NPK = ∠NPO.

therefore triangles NPK and OPN are similar. so, OP/PK = ON/NP
But ON = 1/2 *OP so PK= 2 *NP or sin∠NKP = 1/2 , since ∠NKP is acute, ∠NKP=30∘

⇒ ∠AOB=120∘ , so if the point lies on the circumference such that ∠AOB<=120∘ the the perpendicular bisectors of OA and OB intersect strictly inside the circle

therefore for a particular point A , the point B should lie in the arc which subtends an angle less than or equal to 120 ∘ on either side of A.

consider a circle of circumference 30 such that 31 points are placed on it at a distance of 1 .

For point A the length of the arc in which point B should lie is ( 240/360)*30 =20
which contains 20 points excluding A.

for A , 30 different points can be taken at random out of which 20 satisfy the condition Therefore the probability = 20/30 = 2/3

a+b = 2+3=5

consider an equilateral triangle ,it is formed by the intersection of perpendicular bisectors of 3 equidistant point in the circle.in this problem if we consider the 31 points,for a single point 10 ponts will not bisect this inside the circle probability=(31C2-10*31/2)/31C2=2/3

We construct a second circle $\Gamma'$ centered at $O$ with half the radius of the original circle $\Gamma$ . Let $S = \{A_i\},i = 1,2,\ldots,31$ be the set of the $31$ equally-spaced points. Let the midpoint of $OA_i$ be $A_i'$ and $S'=\{A_i'\},i = 1,2,\ldots,31$ . Note that the perpendicular bisector of $OA_i$ touches $\Gamma'$ at $A_i'$ and hence we can form a bijection between $S$ and $S'$ that maps $A_i$ to $A_i'$ so that this will allow us to focus on $S'$ .

Lemma: Let $L: S' \times S' \rightarrow {inside, outside}$ If $\angle A_i'OA_j' <120 ^\circ$ then the point of intersection in the the problem lies inside the circle.

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