Puzzles with numbers and things

[snookersfun]

thank you, very generous!
I was not really fully awake this morning (tons of spelling mistakes as well), some stupid air-raid alarm woke us at 5:30am (just false alarm, we usually never have the real ones, but wakes you up just as well)

btw. number of games had to be even also, as otherwise one would have to divide an uneven number by an even one, as numbers of matches (starting from 2 players) are u,u,e,e,u,u,... and number of possible outcomes are 2,2,3,3,4,4,...
am I right?

P.S. nice little bit of algebra

[davis_greatest]

Yes, the number of matches has to be even, because of the pattern you have spotted. In fact, there are 6 matches; and each of the 3 possible scores (4-0, 3-1 and 2-2) occurs twice.

[snookersfun]

matches

Here is something a bit different again:

Using 12 matchsticks one can outline a cross-shaped object encompassing 5 unit areas (1x1 each). Find a way, using the same 12 matchsticks, to delineate a 2-dimensional form, which encompasses just 4 unit areas. (w.o. using the extra matchsticks for inside grids or similar)

[davis_greatest]

Hmmm... not sure about this one I briefly wondered whether the perimeter of a Star of David might work, but that's too big (area 3 x square root of 3, I think).

I have A solution, but it's far from elegant:

It is a parallelogram. The long sides are 5 matchsticks, and the short sides are 1 matchstick. If you position it so that the width is 5 and the height is 0.8 (which means the acute angle between the sides will be arcsin(0.8), or about 53.13 degrees), it would give an area of 4.

But I imagine you want something more elegant than that?

[davis_greatest]

By "height", I mean of course perpendicular height.

[snookersfun]

right, not bad, but I was looking for something 'simpler', where you won't have to worry a lot about your angles as well (but some angles appart from 90 are not bad...)

[snookersfun]

Hmmm... not sure about this one I briefly wondered whether the perimeter of a Star of David might work, but that's too big (area 3 x square root of 3, I think).

looks like you got the right area for the star. all this geometry stuff does really let s.o. feel rusty, doesn't it.

Anyway you should have earned the point with your parallelogram, but I still would like to see the other solution.
So, a hint here: think triangle

[davis_greatest]

Certainly a 6-point star is possible, if you spread it out the right amount. If you compress it, so the middle all meets at a single point, it will have zero area, whereas if you open it to form a star of David-type shape (a regular hexagon with an equilateral triangle attached to each side) it will have area 3^(1.5) (= about 5.196). So, as you open it up from the zero area star, at some point it must hit area of exactly 4.

One of the first shapes I thought about was a triangle, of sides 3, 4 and 5. However, this has area 6. Any other clues?

[davis_greatest]

I suppose with the 3-4-5 right-angled triangle, you could move 3 of the matchsticks (the 2 forming the right angle and one other matchstick touching one of those two), to "cut off" the corner by a rectange of 2 units. Not that easy to describe but that would also give a shape of area 4.

I'm not sure whether that is simple enough either, though?

[snookersfun]

I suppose with the 3-4-5 right-angled triangle, you could move 3 of the matchsticks (the 2 forming the right angle and one other matchstick touching one of those two), to "cut off" the corner by a rectange of 2 units. Not that easy to describe but that would also give a shape of area 4.

I'm not sure whether that is simple enough either, though?

That's it exactly, I knew you wouldn't let me down Well, it is simple in so far, as one doesn't have to measure out distances or angles
So, you tell me, how many points do you get now?

[davis_greatest]

Excellent! I think I'll settle for just the 1 point on this occasion.

SO HERE IS THE SCOREBOARD AFTER ROUND 34

snookersfun……………………….…..16
Vidas……………………………………….8½
abextra……………………………..…...5½
robert602…………………………………5
davis_greatest…………………..……4
elvaago...............................1

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[Robert602]

Here's one lifted from last week's homework:

n snooker fans travel to the snooker nirvana that is Pontins Prestatyn. When they arrive, n matches are about to get underway. They decide to split up and each go to watch a different match, so that afterwards they may update each other on all of the match scores. Once all of the matches finish, lost as they are amongst a maze of snooker tables and unable to meet up, they go about calling each other on their mobile phones, sharing all of the scores they've seen or learnt so far. What is the fewest number of calls that can be made in such a way that, at the end, all of the fans know every score?

(I don't actually know the answer to this one yet, but I will tomorrow. I'm fairly certain it doesn't require any special knowledge)

And if anybody likes number theory/discrete maths (I dislike it with some intensity ) there's plenty more where that came from.

[Alex0paul]

Surely the least is 1 phone call because there could have been 2 people watching two games?

[Robert602]

Sorry, that was perhaps a bit unclear. I'm looking for the least number of calls for a given number (call it n) of fans/matches. Nice try though .

(Oh and I should probably also mention, no conference calls, multi texting or any of that newfangled stuff )

[davis_greatest]

Surely the least is 1 phone call because there could have been 2 people watching two games?

Alex... I haven't read the question properly yet but from a very quick skim I think it is asking for the fewest in terms of n. So the answer would be a function of n. 1 phone call works if n is 2, but you need to give an answer that would work for any n.