Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Olinguito wrote:

JaneFairfax wrote:

UAIOE

glutathione

quadripole

ultravioletunfashioned

mustachioed

JaneFairfax wrote:

UEAIO

cupellation

numeration

putrefaction

rubefaction

ulceration

nucleation

**Nehushtan**- Replies: 3

Physics, Chemistry, Biology, Mathematics – which one is best?

May the mass times acceleration be with you

wintersolstice wrote:

That was my solution as well.

I saw a rainbow yesterday.

#32. Duller stuff (6)

#33. Someone for every male child (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**