Solutions by H Reuvers

Problem 2000-4/A (Open problem, by Frits Beukers)

Let a, b, c be integers such that the symmetric matrix
0 a b
a 0 c
b c 0
has three integer eigenvalues. Prove, or give a counter example to, the following statement: either abc = 0 or (a2 - b2)(b2 - c2)(c2 - a2) = 0.

Solution:

The proposed statement is false. Here follows how I found a counter example.
The characteristic polynomial is -x3 + x (a2 + b2 + c2) + 2abc = -(x - x1)(x - x2)(x - x3), where x1, x2, x3 are its zeroes.
So x1 + x2 + x3 = 0, x1x2 + x1x3 + x2x3 = -(a2 + b2 + c2), x1x2x3 = 2abc.
I wrote and ran a Pascal program that seeks integers x1, x2, x3, a, b, c (in the range from -200 to 200) that satisfy these three equations.
I found x1 = -190, x2 = 55, x3 = 135, a = -125, b = -99, c = -57.

Problem 23, september 2001 (Open problem)

Does there exist, for any positive integer N, a positive integer k whose square x is the sum of m ≥ N distinct powers of 3?

Partial solution:

To explore the case, I wrote a Pascal program to find such k for m up to 20. I found for m = 20:

18761382 = 1 + 3 + 32 + 34 + 36 + 37 + 38 + 39 + 310 + 313 + 314 + 316 + 317 + 318 + 319 + 320 + 322 + 323 + 325 + 326.

Furthermore I found: 3 solutions for m = 19, 7 solutions for m = 18, 17 solutions for m = 17, 23 solutions for m = 16, 18 solutions for m = 15, 24 solutions for m = 14, 90 solutions for m = 13, 47 solutions for m = 12, 27 solutions for m = 11, 89 solutions for m = 10, 63 solutions for m = 9, 48 solutions for m = 8, 18 solutions for m = 7, 0 solutions for m = 6, 187 solutions for m = 5, 19 solutions for m = 4, 0 solutions for m = 3, 13 solutions for m = 2, 14 solutions for m = 1.
These are all 708 solutions that can be found with a home computer.

If we ignore the solutions x that can be divided by 9 and thus reduced to x/9, we have: 1 solution for m = 20, 2 solutions for m = 19, 5 solutions for m = 18, 11 solutions for m = 17, 15 solutions for m = 16, 9 solutions for m = 15, 9 solutions for m = 14, 31 solutions for m = 13, 17 solutions for m = 12, 9 solutions for m = 11, 22 solutions for m = 10, 14 solutions for m = 9, 10 solutions for m = 8, 4 solutions for m = 7, 0 solutions for m = 6, 28 solutions for m = 5, 2 solutions for m = 4, 0 solutions for m = 3, 1 solutions for m = 2, 1 solutions for m = 1.
So these are the 191 primitive solutions within the capacity of the computer.
For instance, for m = 4 we have 162 = 1 + 3 + 32 + 35 and 37882 = 1 + 32 + 33 + 315.

In general, there are 3n distinct squares that are at most 32n, and 22n sums of distinct powers of 3 that are at most 32n.
So I thought the number of solutions that are at most 32n must be (4/3)n. Then for n = 14 we get 56, so 708 and 191 are quite a bit more than expected.
Moreover, I found a solution with m = 20.
Therefore I think the answer to the open problem is 'yes'.

Star problem september 2003

Determine the maximal area of a rectangle that can be covered by six circular disks of unit diameter.

Numerical solution:

The rectangle with side lengths 3√2/2 and 2√2/2 can be partitioned into six squares with side length √2/2 whose circumscribing circular disks have unit diameter and cover this rectangle with area 3.

To find a rectangle with an even greater area, I wrote and ran a Pascal program. The listing is as follows:

program cover;
var m,n:integer;x,y,x1,x2,x3,x4,x5,x6,y1,y2,y3,y4,y5,y6:real; gevonden:boolean;
begin
for m:=0 to 300 do
repeat
x1:=1*sqrt(2)/4+(3*m/10000)*(-1+2*random); y1:=1*sqrt(2)/4+(2*m/10000)*(-1+2*random);
x2:=3*sqrt(2)/4+(3*m/10000)*(-1+2*random); y2:=1*sqrt(2)/4+(2*m/10000)*(-1+2*random);
x3:=5*sqrt(2)/4+(3*m/10000)*(-1+2*random); y3:=1*sqrt(2)/4+(2*m/10000)*(-1+2*random);
x4:=1*sqrt(2)/4+(3*m/10000)*(-1+2*random); y4:=3*sqrt(2)/4+(2*m/10000)*(-1+2*random);
x5:=3*sqrt(2)/4+(3*m/10000)*(-1+2*random); y5:=3*sqrt(2)/4+(2*m/10000)*(-1+2*random);
x6:=5*sqrt(2)/4+(3*m/10000)*(-1+2*random); y6:=3*sqrt(2)/4+(2*m/10000)*(-1+2*random);
n:=0;
repeat
n:=n+1; gevonden:=false;
x:=(3*sqrt(2)/2+2*m/10000)*random; y:=(2*sqrt(2)/2+3*m/10000)*random;
if ((x-x1)*(x-x1) + (y-y1)*(y-y1)) > 1/4 then
if ((x-x2)*(x-x2) + (y-y2)*(y-y2)) > 1/4 then
if ((x-x3)*(x-x3) + (y-y3)*(y-y3)) > 1/4 then
if ((x-x4)*(x-x4) + (y-y4)*(y-y4)) > 1/4 then
if ((x-x5)*(x-x5) + (y-y5)*(y-y5)) > 1/4 then
if ((x-x6)*(x-x6) + (y-y6)*(y-y6)) > 1/4 then gevonden:=true;
until gevonden or (n=1000000);
if not gevonden then
begin
writeln(m:5);
writeln(x1:13:10,y1:13:10,' ',x2:13:10,y2:13:10,' ',x3:13:10,y3:13:10);
writeln(x4:13:10,y4:13:10,' ',x5:13:10,y5:13:10,' ',x6:13:10,y6:13:10);
writeln; readln
end
until not gevonden
end.

This yields a rectangle with (rounded) side lengths 2.1233 and 1.4172 and (rounded) area 3.0092.

Then I tried to extend the rectangle further in the direction (1,0).
This yields a rectangle with (rounded) side lengths 2.1235 and 1.4172 and (rounded) area 3.0095.
The (rounded) centers of six covering circles are:
(0.3560,1.0614), (1.0638,1.0623), (1.7703,1.0596),
(0.3508,0.3525), (1.0589,0.3523), (1.7683,0.3518).

Star problem september 2002

Let x=(x1,x2,...,xn) be a random variable such that for each i holds P(xi=1) = P(xi=-1) = 1/2.
Let v be a vector of unity in Rn.
Calculate the probability that the inner product x.v is smaller than 1.

Numerical solution:

I wrote and ran a Pascal program for n = 1, 2, ... , 6 wherein a random vector of unity and a random vector x as above are generated, and the inner product x.v calculated, one million times.
I found the probabilities are:
For n = 1 prob = 0.5, for n = 2 prob = 0.75, for n = 3 prob = 0.7916, for n = 4 prob = 0.8028, for n = 5 prob = 0.8073, for n = 6 prob = 0.8095, for n = 7 prob = 0.8099, for n = 8,9,10 prob = 0.8100.
I think for larger n we always have prob = 0.8100 (rounded in 4 decimals).