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#32. Duller stuff (6)

I would suggest proving that

divides

for all integers *m* and *n*. Which means showing that

*r* and *s* intersect at two points such that at each point of intersection the tangents to the circles are mutually perpendicular. Show that the centres of the circles and the points of intersection lie on a third circle, and find the radius of this circle.

**Nehushtan**- Replies: 2

I posted this in another thread but here it is for anyone who might wish to have a go.

It doesn't matter.

Let's try a little exercise:

Oh well, never mind.

**Nehushtan**- Replies: 0

So what possible transformations are isometries of the plane? You might think that there were a whole bunch of them: rotations, reflections, and translations. (A reflection followed by a translation is sometimes called a glide reflection.) In fact the picture is simpler than that: it turns out that reflections and translations can be built up from rotations alone! A translation in a certain direction is simply a reflection in two axes perpendicular that direction, while a rotation about a point O is a reflection in two axes through O.

Hence any translation is a reflection in two axes perpendicular to the direction of translation whose distance apart is half the distance to be translated.Hence any rotation is a reflection in two axes through the centre of rotation whose angular separation is half the angle to be rotated through.

The number of days he goes for *at least one* of the activities is

so the number of days he goes for none of them is 366 minus that.

**5. HAM-SANDWICH THEOREM**

It is always possible to slice a three-layered ham sandwich with a single cut of a knife in such a way that each layer of the sandwich is divided into two exactly equal halves by the cut.

The ham-sandwich theorem can be proved using the Borsuk–Ulam theorem.

**4. BORSUK–ULAM THEOREM**

Have you made a typo somewhere?

Olinguito wrote:

I learn that it is possible to make a tetradecahedron with just regular hexagons and squares.

It is also possible to make a truncated icosahedron with regular pentagons and hexagons.

Euclid

Euler

Abel

Galois

Cantor

Poincaré

mrpace wrote:

My answer is <0>, <12>, <20>, <32>

Is this correct?

Looks good to me. As a subgroup of the given group, <32> = <4> is cyclic of order 15, and a cyclic group of order 15 has precisely four subgroups.

Primenumbers wrote:

, all possible prime factors will = where n= prime and m=any multiple..........

This is just another way of stating Fermat's little theorem, nothing new.

The statement is clearly true for *n* = 2.

If *n* is an odd prime then