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** mnemonic **

For π = 3.14159265358979323846..., I wrote the following mnemonic:

Mum, I fear I might nevermore

be clever, sharp and smart,

although searching further knowledge for

to get decimals will reward.

** wallpaper with flowers **

CLICK HERE FOR THE PATTERN OF FLOWERS

The group of symmetry is generated by:

rotations over 60 degrees about any angle of any triangle

and over 120
degrees about the intersection point

of the perpendiculars of any triangle;

reflections in any side of any triangle

and in any perpendicular of any triangle;

translations along any side of any triangle.

The fraction green is: 2*pi/sqrt(3) - 3.

**trajectory of a billiard-ball** on a triangular table (30-60-90 graden) with mirrors

(7 July 1999:)

On my birthday I stand you some little roots: a recipe to have the calculator quickly calculate
square roots of natural numbers (which should be not too large) in about twenty signifiacant
figures.

The problem is that the calculator gives only ten figures. The solution is: you do one step
Newton-Raphson to approach the positive zero of f(x) = x^{2} - n , with as a starting point
the outcome x0 for sqrt(n) which the calculator gives.

As you know, the formula of Newton-Raphson is

x1 = x0 - f(x0)/f '(x0) = x0/2 + n/(2x0) .

So the recipe becomes:

1) Calculate sqrt(n) with the calculator, this is x0

2)
Calculate on a piece of paper x0/2 and 2x0

3) Perform the division algorithm you learned
on primary school to calculate the first twenty figures of

(n times a billion)/
(2x0 times a billion);

for this, use the following little BASIC program
(or a PASCAL variant or ... )

4) Now calculate on a sheet of paper x0/2 + n/(2x0) in twenty figures. This turns out to be a
matter of copying and a little attention about the tenth figure.

Enjoy your meal. For ruminants: why is it impossible to approach in this way cubic roots in twenty figures?

10 INPUT A

20 INPUT B

30 M = INT(B/A)

40 PRINT M

50 R = B - M*A

60 B = 10*R

70 GOTO 30

** **

In the magazine Nieuw Archief voor Wiskunde (New Maths Archive) of june 2001 you can
find the following open problem:

A billiard table has the form of a regular pentagon with side 1. Lay a billiard-ball
anywhere on the table and cue in any direction. Suppose it rolls to a distance 10.
What is the maximum number of times it hits a side of the table?

__ Answer: __ I did several tries with Cabri, and FOURTEEN turns out to be optimal.
Here is an example of such a try with fourteen hits:

On a large planar grass-field two pins are in the ground a distance d apart.
To each of both pins a goat is bound with a cord of length r_{g} and one of length
r_{k} respectivily (r_{g} > r_{k}), and
r_{g} < d < r_{g}+r_{k} (so the goats can touch each other
but not the other's pin).

You can see in the drawing to the left that the goats can meet each other in a 'domain of
conflict' which consists of two circular sectors.

In the drawing to the right you can see how you can calculate the area of such a circular
sector:

the area of the 'piece of cake' is
a_{1}*r_{k}^{2}/2,

the area of the triangle is h_{1}*b/2 =
r_{k}*cos(a_{1}/2)*r_{k}*sin(a_{1}/2) =
r_{k}^{2}*sin(a_{1})/2,

so the area of the circular sector is r_{k}^{2}*(a_{1} -
sin(a_{1}))/2.

So the total area of the domain of conflict is

opptot = r_{k}^{2}*(a_{1} - sin(a_{1}))/2 +
r_{g}^{2}*(a_{2} - sin(a_{2}))/2.

Given d, r_{k} and r_{g}, you can calculate a_{1} and a_{2}
with the equalities

(1) (b/2 = ) r_{k}*sin(a_{1}/2) = rg*sin(a_{2}/2), and

(2) d = (h_{1} + h_{2} = ) r_{k}*cos(a_{1}/2) +
r_{g}*cos(a_{2}/2).

The result is :

sin(a_{2}) = (2*(d^{2} + r_{g}^{2} -
r_{k}^{2})/(2*d*r_{g}))*sqrt(1 -
((d^{2} + r_{g}^{2} -
r_{k}^{2})/(2*d*r_{g}))^{2}),

and an analogous expression for sin(a_{1}).

The fraction of time that both goats are grazing in the domain of conflict is:

the product of opptot/(pi*r_{k}^{2}) and
opptot/(pi*r_{g}^{2}).

The probability of confrontation is directly proportional to this product divided by opptot,
so to

(a_{1} - sin(a_{1}))/(r_{g}^{2}) +
(a_{2} - sin(a_{2}))/(r_{k}^{2}) as well.

In the following little Pascalprogram, we calculate this last expression with

r_{k}=10 and a series of values of r_{g} and d:

You can see the results by clicking on results.

It turns out that the probability of confrontation is maximal when both cords have approximately
equal length and the distance between the pins is sligthly less than the mean of r_{g} and
r_{k}+r_{g}.

** A generalisation of the proposition of Morley **

Given an arbitrary triangle ABC. When you draw in each angle the three trisectrices,
and you take near each side the intersection point of the two trisectrices that are closest to that side,
then the three intersection points are the angular points of a regular triangle (see NIEUWE WISKRANT, sept 2001).

I wondered whether this is true as well if I divide the angle in n equal parts with n>3
instead of three equal parts. If I take near each side the two n-sectrices
that are closest to that side, is it true that the three intersection points are the angular points of
a regular triangle?

Of course this is true if the original triangle is regular. But in other cases?

I calculated the outcome with vector geometry. Take A(0,0) and B(1,0). Suppose that the
angle at A measures a degrees, and at B b degrees (a and b integral and ranging from 1 to 89,
so 7921 pairs).
For each n from 3 to 10 I calculated with the computer for how many pairs (a,b) the triangle
STU is (almost) regular, taking into account rounding errors.

The norm is that each pair out of the three lengths of line segments
ST, SU and TU differ less than epsilon. If eps=0.0001 I get for n=3:
7921 pairs, and for n=4 t/m 100: 1. This seems to indicate that it is only true for n>3
if both a and b are 60 degrees.

See the pascal program .

With an adapted version of this program, I derive for n=3: the ratio of areas of the STU
and ABC is at most 0.03414828, this maximum occurs when the triangle ABC is regular,
and this ratio approaches 0 if the area of ABC approaches 0.
Of course, this can be proved in an analytical way. The center of the inscribed circles of ABC
and STU do only coincide if ABC is regular.

**A nice discontinuous function**

Consider the function f from [0,1] to [0,1] that assigns (for example) to 0.325013...
0.11101101111100101110... (substitute for each figure in the original the
accessory number of ones and one 0). The function is discontinuous in each point of (0,1)
that has only a finite number of decimals unequal to 0, for instance in 0.23
(for this kind of points we choose the representation with the nines to calculate f, so
0.22999999... instead of 0.23).

The function is injective and increasing and the maximum value is

f(1) = f(0.99999...) = 0.111111111011111111101111111110... = 1111111110/9999999999.

notice that x and f(x) are either both rational or both irrational (because the decimal
representations are either both repeating or both non-repeating).

**A nice new transcendental number**

The set of algebraic numbers is enumerable and therefore has measure 0.

So almost all real numbers are transcendental. Nevertheless, it is rather difficult to give
other examples of transcendental numbers except algebraic expressions in pi or e.

To begin with, we have the proposition of Liouville,
and accessory examples like SUM(k=1,2,3,...: 10^(-k!)) = 0.11000100000000000000000100..

We get another nice example by constructing an enumeration of the algebraic numbers in (0,1),
and applying the Cantor diagonal procedure on this enumeration.

I did this with the help of an enumeration of the rational numbers that begins with
q_{1}=0, q_{2}=1, q_{3}=-1, q_{4}=2, q_{5}=-2,
q_{6}=1/2, ... , and an enumeration of the polynomials with rational coefficients
which enumerates (in lexicographical order) first all polynomials with degree 0 (or less) and
coefficient q_{0}, then all polynomials with degree 1 (or less) and coefficients
q_{0} and q_{1}, then all polynomials with degree 2 (or less) and
coefficients q_{0}, q_{1} and q_{2}, etc.

If I enumerate the algebraic numbers of (0,1) in order of their occurring as a zero of a
polynomial in the list of polynomials mentioned above (zeroes of the same polynomial in
order from small to large), and apply to the so-constructed list a_{1},
a_{2}, a_{3} ... of decimal representations of algebraic numbers the
Cantor diagonal procedure, then I find the *reuvers transcendental number*
rtn with n-th decimal equal to a_{n}(n)+1(mod 10). This number begins with
rtn = 0.71.. (I have not yet calculated more decimals).

Notice that this number is most probably not an algebraic expression in pi or e
(because these expressions form a set that is enumerable, so with measure 0).

When somebody draws a triangle, what is the probability that the triangle is acute-angled?

First solution: choose three times an arbitrary point in the plane. The probability that the third point is thus situated with respect to the other two points, that the triangle formed by these three points is acute-angled, is zero.

Second solution: choose an arbitrary angle alpha between 0 and 180 degrees, and then an arbitrary angle beta between 0 and 180 degrees such that alpha + beta < 180. The probability that the triangle with angles alpha and beta is acute-angled, is 1/2 * 2/3 * 1/2 = 1/6.

What is your opinion?

The irreducibility of the polynomial x^{3} + 2 over the rationals provides for the
unsolvability of the
equation x^{3} + 4y^{3} -2z^{3} + 6xyz = 0 for rationals x,y,z (not all
zero). This result can be extended to similar results for any irreducible polynomial with
rational coefficients.

PROOF: Let I be the ideal (x^{3}+2)Q[x]; the residue class ring Q[X]/I is a field.

For any three rationals a,b,c (not all 0), the residue class a + bx + cx^{2} + I
has a unique inverse A + Bx + Cx^{2} + I, for which we have:

(a + bx + cx^{2} + I)(A + Bx + Cx^{2} + I) = (aA-2bC-2cB) + (aB+bA-2cC)x +
(aC+bB+cA)x^{2} + I = 1 + I.

So for any triple (a,b,c) of rationals (not all 0), the system of linear equations

aA - 2 cB - 2bC = 1

bA + aB - 2cC = 0

cA + bB + aC = 0

has a unique solution (A,B,C).

So the determinant of the coefficient matrix, a^{3} + 4c^{3} -2b^{3} + 6abc,
is not 0. This proves our assertion.

When you have three integers a,b,c that are relatively prime and with a+b=c, then it doesn't
often occur that each of a,b and c has a large exponent in the prime factor decomposition.

So here is a nice challenge: find a,b,c such that the minimum of the greatest exponents in the
prime factor decompositions of a,b,c is as large as possible.

The champion in this field is, as far as I know, a person named Nitaj. He found a=3^{18}*23*2269,
b=17^{3}*29*31^{8}, c=2^{10}*5^{2}*7^{15}. So the
minimum of the maximal exponents is 8 in his triple. But I guess that he
didn't find this result with a simple little pc.

I found a relatively nice result with a simple Pascal program:
a=2^{8}*3=768,
b=7*11^{5}=1127357, c=5^{5}*19^{2}=1128125.

So in my own triple the minimum is 5. Who can find a better result than mine?

Consider a triangular lattice of points (i,j) with i and j integer, i at most 29,
j at least 0 and at most i.

A particle starts at the top (0,0) and takes 60 random steps. What is the probability that it
ends up in the bottom row (that is, in a point with i=29)?

I wrote a Pascal program to make a computer simulation. With 32767(=maxint) runs, the frequency
of the particle ending up in the bottom row was 0.00003051851.

So I daresay that the probability is about 0.00003.

Suppose someone shoots a small bullet through a face of a large cube and the bullet leaves the cube through another face. What is the probability that these two faces are opposite faces of the cube?

For this problem I wrote and ran the following Pascal program (using coordinates such that the opposite faces are y=0 and y=1):

program seitz;

const pi=3.1415926536;

var r1,r2,r3,r4,s1,s2,m,n:real;

begin

m:=0; n:=0;

while true do

begin

n:=n+1;

r1:=random; r2:=random; r3:=random; r4:=random;

s1:=r1+cos(r4*pi)/sin(r4*pi); s2:=r2+cos(r3*pi)/(sin(r3*pi)*sin(r4*pi));

if ((0 < s1) and (s1 < 1) and (0 < s2) and (s2 < 1)) then m:=m+1;

writeln(m/n:13:10)

end

end.

The outcome was 0.0745 (rounded).

Looking for new triangles with integral side lengths a,b,c and vertices with integral coordinates, I took A(0,0), B(c,0), C(p,q), and demanded (p,q,b) and (p-c,q,a) be pythagorean triples.

I wrote and ran variants of a Pascal program with the following pith:

for c:=1 to 20 do for d:=1 to 10 do for u:=1 to 10 do for v:=1 to u do

begin p:=d*(u*u-v*v); q:=2*d*u*v; b:=d*(u*u+v*v); a:=sqrt((p-c)*(p-c)+q*q);

if a=round(a) then writeln(p:10,q:10,a:13:0,b:13:0,c:10) end

I found the expected rectangular triangles, but also unexpected ones like

A(0,0), B(4,0), C(9,12), (a,b,c)=(13,15,4)

A(0,0), B(6,0), C(3,4), (a,b,c)=(5,5,6)

A(0,0), B(14,0), C(5,12), (a,b,c)=(15,13,14)

A(0,0), B(17,0), C(7,24), (a,b,c)=(26,25,17)

When solving a maths problem, it often
occurs that we "smell" the solution, like a hound that smells a hare; but we can't
catch it at the same time.

And yet, the answer seems so simple afterwards.

That makes me think of the words of Saint Paul: " .. that they might seek God, if
perhaps they might grope for Him and find Him, even though He is not far from
each one of us."