| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1968.1 | | POLAR::WALSHM | | Wed Apr 19 1995 11:02 | 8 |
| The six points are collinear. Specifically, we can define Ai as being
(i,0), and Bj as being (-j,0).
What level is the Concours? This doesn't seem a contest-level
problem... of course, I suppose maybe if I tried to *prove*
something... ;)
-- Matt
|
| 1968.2 | A proof | WIBBIN::NOYCE | The brakes still work on this bus | Wed Apr 19 1995 11:27 | 6 |
| Draw the hyperbola between A1 and A2 that passes through all the points
P where A2P = A1P+1. Also the hyperbola between A2 and A3 that passes
through the points Q where A3Q = A2Q+1. The three Bj's are on both
hyperbolas, and no two Bj's coincide. Since most hyperbolas intersect
in only 1 or 2 points, this is possible only if both hyperbolas are
the degenerate ones that arise when A1A2=1 and A2A3=1.
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| 1968.3 | re .1 & re .2 | BATVX1::ESANU | Au temps pour moi | Wed Apr 19 1995 13:44 | 8 |
| Thank you for your elegant solution. I think that there still is to be
proven that the 2 hyperbolas cannot intersect in 4 different points, which
is the general case if you consider for each hyperbola both branches.
The Concours General is at high-school senior level ("Terminale" in
French).
Mihai.
|
| 1968.4 | | WIBBIN::NOYCE | The brakes still work on this bus | Wed Apr 19 1995 16:34 | 5 |
| I think a hyperbola has two branches if it satisfies
abs( A1P - A2P ) = x
but I was asking for the single branch that drops the "abs".
Think of a LORAN map...
|
| 1968.5 | Sorry! | BATVX1::ESANU | Au temps pour moi | Thu Apr 20 1995 05:29 | 6 |
| Absolutely right, I mixed up the definition of the hyperbola (with 2 branches)
with that of one branch of the hyperbola...
Splendid solution!
Mihai.
|