Home > Uncategorized > SEEMOUS 2012 Problem 2

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.

  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: