Sources: Pierre Tougne (Pour la Science) (puzzles 1,2,3,8,13), Philippe Fondanaiche (puzzles 4,5,11,12), or derived of many collections of mathematical puzzles
Scott Purdy (spurdy@cogent.net) sends the following:
If you start with thief 1 or thief N-1 for an even N, there is a general
arrangement which always works.
1, N-2, N-4, ..., 2, N, N-1, N-3, ..., 3
and its converse
N-1, N-3, ..., 1, N, N-2, ..., 2
I don't think a generalizable pattern exists for beginning with thief 2.
[Does anyone else have a comment?]
What are the terms following the first terms of these sequences?
1. 11111001110, 2202000, 133032, 30443, 13130, .....
2. 238, 918, 1998, 3478,......
3. 24, 70, 118, 258, 494,......
Sequence is 2.07.17
2.17.27
2.27.37
2.37.47
2.47.57
2.57.67
a = 135 b = 888 c = 975 A = 47952 = 24*1998 P = 1998 But I haven't found a proof yet, why this must be the minimum...Update September 9, 2008: Bojan Basic sent this proof of minimality.
In the first place, we are
going to prove the following assertion: of all integer-sided triangles with
sides a <= b <= c, where c is fixed, the triangle
with the smallest area is the one with a = 1, b = c, whose
area is sqrt(4c^2 - 1)/4. Obviously, if we let a = x, in
order to minimize the area of the
formed triangle we must choose b as small as possible while
keeping the triangle inequality, what gives b = c + 1 - x.
If we now consider the area (given by
Heron's formula) as a function of x, we could analyze that function
by taking the first derivative—furthermore, since we are only interested in the
monotonicity, it would be convenient
to take a logarithm first (in order to simplify the calculations). The area is
sqrt((1 + 2c)*(1 + 2c - 2x)*(2x - 1))/4, its
logarithm is ln(1 + 2c)/2 + ln(1 + 2c - 2x)/2
+ ln(2x - 1)/2 +
ln(1/4), and its first derivative (with
respect to x, of course) is 2*(1 + c - 2x)/((1 + 2c
- 2x)*(2x - 1)). Now, since a is the smallest side and b
= c + 1 - x, it must hold 1 <= x <= (c + 1)/2. It is
obvious that for these values of x the first derivative is nonnegative;
therefore, the area function is increasing, and its minimum is achieved for the
smallest value of x, which is x = 1. The area is now easily
calculated, and the assertion is proven.
Suppose now that the perimeter of the triangle that we are searching for is
greater than or equal to 145*1998. Now, c >= (145*1998)/3 = 145*666, and
its area is, by the assertion above, at least sqrt(4*(145*666)^2 - 1)/4 =
sqrt(37,303,059,599)/4 > 24*1998, what is worse than the result we already have.
Therefore, in order to find the triangle satisfying the given conditions with
minimum area, it is enough to consider perimeters up to 144*1998. The
established bound allows computer search. The following pseudocode finishes
within a few hours:
for
um
= 1 to 144
for a = 1 to 666*um
for
b = 999*um - a to (1998*um - a)/2
c = 1998*um - b - a;
if (area(a,b,c) is positive integer) and
(1998 | area(a,b,c))
then print(a,b,c)
It shows that there is no solution with area
less than 24*1998, and furthermore, that the solution (135, 888, 975), with
perimeter 1998 and area 24*1998, is unique.