SIMO Updates

I haven't got the time to render any major revamp or update of this homepage, hence a conglomeration of updates on one single page :)

1997 National Team Selection Test
The two tests were held on 28 Dec 96 and 18 Jan 97. Click here for the problems.

Comments

1. A deceptively difficult problem that requires no more than Menelaus's Theorem, angle-bisector theorem and similar triangles.
2. Another simple yet sophisticated problem in that it requires the use of recurrence relations. The relevant relation is rather obviously a_n = a_n-1 + a_n. A textbook problem.
3. This problem appeared previously in Mathematics Magazine and is probably inspired by problem 5 of the 1994 IMO.
4. Looks Russian! Note that a+b+c+d=0 and hence |bc-ad| = |(a+b)(a+c)|.
5. A summation of binomial coefficients: another textbook problem. Can be solved by induction.
6. A disappointing and uninteresting problem 6. Can be solved by induction.

Your contribution (to the selection test) will be greatly appreciated! :)

Reviews
Have you read A Short Proof of the Erdos-Mordell Theorem published in the Jan 1997 issue of American Mathematical Monthly? For a statement of this theorem, you could refer to the 1995 SIMO problems. An ingenious and elegant proof based on simple geometric considerations, the author began by demonstrating that

Forum
I plan to construct a forum comprising of extracts from my correspondences with various members of the SIMO community on problem-solving. Such discussions should neither be restricted to those training for the IMO nor be restrained by time and space! :)

1996 Putnam Examination
\item[B-1] Define a \textbf{selfish} set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of $\{1, 2, \ldots, n\}$ which are \textit{minimal} selfish sets, that is, selfish sets none of whose proper subsets is selfish.

On 28 Jan, Weifa wrote from UK:
Answer to B1 is the Fibonacci sequence right?

On 31 Jan, I wrote:
   The above is problem B-1 from the 1996 Putnam Examination. Let a_n denote the number of such subsets. First, note that a set S is selfish iff |S| = min S.
   Hence, a_n = sum_{n=1}^n C^{n-r}_{r-1}. (pardon me, I'm not familiar with TeX, though I do know enough to understand most of the putnam problems :)
   On the other hand, let A_n be the collection of minimal selfish subsets of {1, 2, ..., n}, so |A_n| = a_n.
   Number of subsets in A_n not containing n = a_{n-1}
   Number of subsets in A_n containing n = a_{n-2}
To see this, take any S \in A_{n-2}. (S+1) U {n} is minimal selfish. This also establishes a bijection.
   Hence, a_n = a_{n-1} + a_{n-2}.
   Upon considering the bounday conditions, we arrive at the conclusion that a_n is the fibonacci sequence.
Note a spin-off from solving this problem: we have derived an expression of the fibonacci numbers of a summation of binomial coefficients!

On 1 Feb, Dr Tay T.S. wrote:
Interesting combinatorial proof of the recurrence relation. It can also be proved easily by looking at the terms corresponding to a_{n-1} and a_{n-2} in the Pascal triangle. This method though not as beautiful as yours, usually works for this type of summations.

Afternote: The afore-mentioned summation is actually equivalent to that in problem 5 of the selection test!