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## SEEMOUS 2012 Problem 2

Let $a_n>0, \ n \geq 1$. Consider the right triangles $\Delta A_0A_1A_2,\Delta A_0A_2A_3,...,\Delta A_0A_{n-1}A_n,...$ as in the figure. (More precisely, for every $n \geq 2$ the hypotenuse $A_0A_n$ of $\Delta A_0A_{n-1}A_n$ is a leg of $\Delta A_0A_nA_{n+1}$ with right angle $\angle A_0A_nA_{n+1}$, and the vertices $A_{n-1}$ and $A_{n+1}$ lie on the opposite sides of the straight line $A_0A_n$; also $|A_{n-1}A_n|=a_n$ for every $n \geq 1$.)

It is possible for the set of points $\{A_n | n \geq 0\}$ to be unbounded, but the series $\sum_{n=2}^\infty m(\angle A_{n-1}A_0A_n)$ to be convergent? Here $m(\angle ABC)$ denotes the measure of $\angle ABC$.

(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 $n$ such that $\sum_{k=n} \angle A_kA_0A_{k+1}< \alpha < \pi/2$. Consider the triangle $A_nA_0C$, such that $A_{n+1} \in A_nC$ and $\angle A_nA_oC=\alpha$.  Then we can prove that all points $A_k,k \geq n+1$ are inside the triangle $A_nA_0C$. To see that, notice that all points are inside the angle $A_nA_0C$, because of the choice of $n$. First, since $\angle A_{n+1}A_0A_{n+2}=\pi/2$, it is obvious that $A_{n+2}$ lies inside the triangle $A_nA_oC$. From here we may continue inductively, choosing $C_p =A_{n+p}A_{n+p+1} \cap A_0C$, and proving that $A_{n+p+2} \in \Delta A_{n+p+1}C_pA_0 \subset \Delta A_0A_nC_0$.

There is an equivalent analysis formulation of the problem. Notice that the boundedness of $\{A_n\}$ can be reduced to the boundedness of the segments $A_0A_n=\sqrt{a_1^2+...+a_n^2}$. Also, each angle can be calculated: $\angle A_{n-1}A_0A_n=\arcsin \displaystyle \frac{a_n}{\sqrt{a_1^2+...+a_n^2}}$. The question is equivalent to this: Does the convergence of this series implies the convergence of $\sqrt{a_1^2+...+a_n^2}$?

To prove the problem via this second approach, denote $\ell_n^2=a_1^2+...+a_n^2$. The hypothesis equivalent, via a comparison criteria to

$\displaystyle \sum_{n=2}^\infty \sqrt{1-\frac{\ell_{n-1}^2}{\ell_n^2} }< \infty$

Which implies that

$\displaystyle \sum_{n=2}^\infty 1-\frac{\ell_{n-1}^2}{\ell_n^2}< \infty$

We note that since $\lim\limits_{x\to 1} \frac{\ln x}{x-1}=1$, and by the comparison criterion it follows that

$\displaystyle \sum_{n=2}^\infty - \ln\left( \frac{\ell_{n-1}^2}{\ell_n^2}\right)< \infty$

But the partial sums of this series are in fact equal to $s_n=\ln \ell_n^2-\ln \ell_1^2$. The convergence of $s_n$ implies the boundedness of $\ell_n$, and therefore the set $A_0,...,A_n,...$ must be bounded.

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