## SEEMOUS 2012 Problem 2

Let . Consider the right triangles as in the figure. (More precisely, for every the hypotenuse of is a leg of with right angle , and the vertices and lie on the opposite sides of the straight line ; also for every .)

It is possible for the set of points to be unbounded, but the series to be convergent? Here denotes the measure of .

(A subset of the plane is bounded if, for example is contained in a disk.)

Suppose that the series of angles converges. Then there is a positive integer such that . Consider the triangle , such that and . Then we can prove that all points are inside the triangle . To see that, notice that all points are inside the angle , because of the choice of . First, since , it is obvious that lies inside the triangle . From here we may continue inductively, choosing , and proving that .

There is an equivalent analysis formulation of the problem. Notice that the boundedness of can be reduced to the boundedness of the segments . Also, each angle can be calculated: . The question is equivalent to this: Does the convergence of this series implies the convergence of ?

To prove the problem via this second approach, denote . The hypothesis equivalent, via a comparison criteria to

Which implies that

We note that since , and by the comparison criterion it follows that

But the partial sums of this series are in fact equal to . The convergence of implies the boundedness of , and therefore the set must be bounded.

A similar approach can be formulated using a series of to characterize the sum of measures of the angles.