Googly Eyes

Letting $R$ be the (upper) point of intersection of the two circles as shown in the diagram, we see that, by symmetry, $\angle ABR = \angle BAR = \alpha$ , (since the two circles have the same radius). Thus $\angle ARB = \pi - 2\alpha$ , and so $\angle BRP = 2\alpha$ .

Since $\Delta BRP$ is isosceles with $BR = BP$ we have that $\angle BPR = \angle BRP = 2\alpha$ , and so $\angle PBR = \pi - 4\alpha$ .

Finally, $\beta = \pi - \angle PBR - \angle ABR = \pi - (\pi - 4\alpha) - \alpha = 3\alpha.$

Thus $\boxed{\beta = 3\alpha}$ is the correct option.

Join RB

Since AR = BR

Then angle RAB = angle RBA ..................................... (1)

angle PRB = angle RAB + angle RBA ......... (exterior angle of triangle ARB)

From (1)

angle PRB = 2 angle RAB = 2 angle (alfa) ................(2)

Since BR = BP

Then

angle PRB = angle P

Then , from (2)

angle P = 2 angle (alfa) .................................................(3)

In triangle APB

angle (beta) = angle P + angle (alfa).

Using (3), then

angle (beta) = 3 angle (alfa)

If BR is radius, we make the triangle ARB with the angles A, A, H. Then we have the triangle BRP, which have the angles T, T and Y. So: B+A+Y=180. A+A+H = 180. T+H = 180. T+T+Y= 180. Then, if 2A+H = 180 and T+H=180, 2A = T. Then 4A+Y= 180 and A +B+Y = 180. Concluing, 3A=B

This is perhaps a solution that could be used in a competition under time pressure. Under no circumstances can it be used in a proof of any sort.

In the problem, we are only given information about the circles having the same radius R, which means that the side lengths and angles may be subject to change. Therefore, point P can be moved along the circumference such that PB forms a right angle with AB without changing the problem.

We can label the second point of intersection T, so the points of intersection are R and T. The circles can be moved such that RT = R, or the radii of the circles, creating equilateral triangle ART. And voila, we have 30-60-90 triangle APB, making beta 90 degrees and alpha 30, so beta = 3 alpha.

From the figure, let angle RPB = x, We know that Beta is the exterior angle so Beta=x+Alpha. ---------(1) Draw a line from B to R, We know BR = BP, therefore angles BRP = BPR = x, angle BRP is ext angle, therefore x = alpha + alpha, x=2alpha ------(2) substitue eqn (2) in (1) Beta = three alpha

simply putting the THEOREM. we know the THEOREM that "in a circle an angle (A) is made by any arch of circle at centre then it is twice (A=2B) of an angle (A) formed on circumference of that circle." BUT Here is little modification, if two Circles of same Radii are intersect in such a way that a TRIANGLE is formed and going through their RADII and one side of triangle is passes through their INTERSECTION POINT. then the arch of one circle makes an angle(B) at CENTRE of SAME Circle. and same arch makes an angle at other's centre and is equal to 3 times of angle at first circle's centre(B=3A). this is also a THEOREM.