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Topic: Comments on some IMO problems
Replies: 6   Last Post: Jul 24, 1996 3:14 PM

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Waldemar Pompe

Posts: 9
Registered: 12/12/04
Comments on some IMO problems
Posted: Jul 21, 1996 5:04 PM
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Thank you for posting here the IMO problems! I solved only the first
day and doubt if I try the second, coz I'm not very impressed.
I refer to Wonder Worm's comments and solutions.

1. Let ABCD be a rectangular board, with AB = 20 and BC = 12. The board is
divided into 20x12 squares. Let r be a given positive integer. A coin can
be moved from one square to another iff the distance between the centres of
the two squares is sqrt(r). The task is to find a sequence of moves taking
the coin from the square with A as a vertex to the square with B as a vertex.

a) Show that the task cannot be done if r is divisible by 2 or 3.
b) Prove that the task can be done if r = 73.
c) Can the task be done when r = 97?

>c) The answer is no. 97 = 9^2 + 4^2 is the only way of expressing 97 as
> the sum of two (non-negative) squares, since it is prime. So we can
> only change coordinates by (+-4, +-9) or (+-9, +-4). I constructed a
> table for each point showing the points reachable from it that were
> legal (i.e. inside the given rectangle). Actually, not for each point,
> but for each reachable point. There are only 60 such points, and (20,1)
> is not one of them.


I also made a diagram for this part, but mine contains only
58 possible reachable fields! Here it is:


D -------------------- C
| X X X X |
|X X X X X X |
| X X X X |
|X X X X X |
| X X X X |
| X X X X X X| 12
| X X X X |
| X X X X X|
| X X X X |
|X X X X X X |
| X X X X |
|X X X X X X |
A -------------------- B
20

where X denotes the reachable field for the coin in part (c).
I don't claim it's correct. Has anyone made a similar diagram
to compare?

> I was quite disappointed with this as a question; part a yields to standard
> technique, while parts b and c can be done fairly mindlessly and reasonably
> quickly by the aforementioned table idea (although it is easier to construct
> the solution in case b). Questions which come in parts are always (IMHO)
> a bad idea for olympiads, and 4 cases is too many.


I entirely agree. I learned nothing from this problem and
had no pleasure "working" on it... I wonder how many IMO
contestants made the above diagram during the competition
and how many coordinates enjoyed correcting such papers!

> 3. Let S = {0, 1, 2, ...}. Find all functions f defined on S taking their
> values in S such that f(m + f(n)) = f(f(m)) + f(n) for all m and n in S.


> Let f(1) = a. Then f(a) = f(f(1)) = f(1) = a.

[...]

> In general, m is not equal to n, so we must have a = 0 or 1. These
correspond
> to the functions f(x) = 0 for all x, and f(x) = x for all x, both
of which are
> easily seen to be solutions.

I'm afraid that's not all... How about f(0)=0, f(2k-1)=f(2k)=2k
for k=1,2,3,... ? Generally, fix a positive integer a and let
r denote the remainder of division n by a. Then
f(n) = n + ag(r) - r,
where g: {0,1,...,a-1} --> S is any function with g(0)=0,
is also a solution!

Contrary to all geometry ignorants, I'll say that I really
enjoyed solving problem 2, which I find beautiful!
I found a nice solution, and my friend found another,
also very nice. Since I'm kinda tired of sitting at the
computer today, I'm not posting these solutions now. If there is
a need of seeing them, I'll do it in a couple of days -- just
let me know.

Waldemar Pompe








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