XV OM - I - Problem 7

Given a circle and points $ A $ and $ B $ inside it. Find a point $ P $ on this circle such that the angle $ APB $ is subtended by a chord $ MN $ equal to $ AB $. Does the problem have a solution if the given points, or only one of them, lie outside the circle?
Suppose that point $ P $ of a given circle $ C $ with radius $ r $ is a solution to the problem (Fig. 7).
Since points $ A $ and $ B $ lie inside the circle $ C $, points $ M $ and $ N $ lie on the rays $ PA $ and $ PB $ respectively, and angle $ APB $ coincides with angle $ MPN $. Triangles $ APB $ and $ MPN $ have equal bases $ AB = MN = a $ and a common angle at the vertex $ \measuredangle APB = \measuredangle MPN = \alpha $; hence, the circles circumscribed around these triangles have the same radius $ r = \frac{a}{2 \sin \alpha} $, and point $ P $ lies on the circle with radius $ r $ passing through points $ A $ and $ B $. Therefore, the construction of the desired point is as follows (Fig. 7a). We construct circles with radius $ r $ passing through points $ A $ and $ B $. Since $ AB < 2r $, there are 2 such circles: $ K_1 $ and $ K_2 $. If $ P $ is a common point of circles $ K_1 $ and $ C $, or circles $ K_2 $ and $ C $, then $ P $ is a solution to the problem. Indeed, the rays $ PA $ and $ PB $ intersect the given circle $ C $ at points $ M $ and $ N $; from triangle $ MPN $ we have $ MN = 2r \sin \measuredangle MPN = 2r \sin \measuredangle APB $; but in triangle $ APB $ we have $ \sin \measuredangle APB = AB \colon 2r $, so $ MN = AB $.
Each of the circles $ K_1 $ and $ K_2 $ intersects the circle $ C $ at two points; if $ O_1 $ and $ O_2 $ are the centers of circles $ K_1 $ and $ K_2 $, then

since $ OA < r $, $ AO_1 = r $; similarly $ 0 < OO_2 < 2r $.
The problem thus has 4 solutions.
Let us examine whether the above reasoning remains valid for any position of points $ A $ and $ B $. If point $ P $ is a solution to the problem, then points $ M $ and $ N $ lie on the lines $ PA $ and $ PB $, but not necessarily on the rays $ PA $ and $ PB $, so angle $ MPN $ coincides either with angle $ APB $ or with the adjacent angle to angle $ APB $. The conclusion that triangles $ APB $ and $ MPN $ have the same circumradius remains valid, hence point $ P $ lies, as before, on the circle with radius $ r $ passing through $ A $ and $ B $. Conversely, if point $ P $ is a common point of the given circle and a circle $ K $ with radius $ r $ passing through $ A $ and $ B $, then this point, provided it is different from $ A $ and $ B $, is a solution to the problem, as we conclude in the same way as before.
For such a point $ P $ to exist, the following conditions must be met:
1. There must exist a circle with radius $ r $ passing through points $ A $ and $ B $; this is the case when $ AB \leq 2r $. If $ AB < 2r $, there are two such circles $ K_1 $ and $ K_2 $ with centers $ O_1 $ and $ O_2 $. If $ AB = 2r $, there is one such circle: $ K_1 = K_2 $, $ O_1 = O_2 $.
2. There must exist a common point of circle $ K_1 $ or $ K_2 $ with the given circle $ C $, different from $ A $ and $ B $.
The table below gives the number of solutions for different cases of the position of points $ A $ and $ B $.

We propose to the Reader to provide a detailed justification of the above data for each of the cases 2-6 and to make the appropriate drawings.