| 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 |
1. What is the smallest number of squares of a chess board that can be marked
in such a manner that
(a) no two marked squares may have a common side or a common vertex, and
(b) any unmarked square has a common side or a common vertex with at least one
marked square?
Indicate a specific configuration of marked squares satisfying (a) and (b) and
show that a lesser number of marked squares will not suffice.
2. Prove that a�pq + b�qr + c�rp <= 0, whenever a, b and c are the lengths of
the sides of a triangle and p + q + r = 0.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1998.1 | solution to #1 | DECADA::YODER | MFY | Mon Sep 18 1995 09:51 | 9 |
Consider condition (b) first. A marked square can be considered to "cover"
itself and all its neighbors, that is, all squares that can be reached in a king
move. Give the board a coordinate system with (1,1) and (8,8) as opposite
corners; consider those squares whose x- and y-coordinates are both == 1 (mod
3). A marked square can cover at most 1 of those, and there are 9 in all, so we
need at least 9 marked squares. (The squares are {1,4,7} x {1,4,7}.)
If we mark exactly those 9 squares, we cover the entire board, and no 2 of the 9
are neighbors. So this solves part (a).
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| 1998.2 | solution for #2 | DECADA::YODER | MFY | Mon Sep 18 1995 11:38 | 22 |
We can find two of p,q,r whose product is >=0 (if one of them is 0 this is
immediate, otherwise there must be two with the same sign). WLOG let these two
be p and q. By the triangle inequality
a <= b+c
2 2 2
a <= b + 2bc + c
2 2 2
a pq <= pq(b + 2bc + c )
2 2 2 2 2
From 0 <= (bq-cp) we get 2bcpq <= b q + c p ; using this,
2 2 2 2 2
a pq <= b (pq+q ) + c (pq+p )
2 2 2
a pq <= b (-qr) + c (-pr)
and the desired inequality follows immediately.
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