| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
The following delightful remark is rather ancient, it seems to me that it was made a long time ago by the Romanian mathematician Alexandru Froda, whose centennial is commemorated this year: Let f: R^2 -> R be continuous; then, on every circle in R^2 there are two diametrically opposed points A and B, such that f(A) = f(B). (Here R^2 is the cartesian product R x R).
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1887.1 | AUSSIE::GARSON | achtentachtig kacheltjes | Thu Aug 18 1994 20:14 | 9 | |
re .0
These kind of theorems are often surprising.
I think that to prove it you define a new function g to be f(A)-f(B).
f(A) = f(B) is then equivalent to g = 0. If g = 0 at some arbitrary
starting point then the theorem is proved. Otherwise g varies
continuously from some non-zero value, t, to -t as the point moves
180� around the circle and thus is 0 somewhere on the circle.
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| 1887.2 | You've got it ! (1887.1) | EVTSG8::ESANU | Fri Aug 19 1994 12:23 | 5 | |
That's it, of course. For completeness's sake, to be mentioned also that mapping a circle's point to its diametrically opposite is a continuous function. Isn't it nice in its simplicity and surprising at a first glance? | |||||