Puzzles with numbers and things

[snookersfun]

the answer looks quite good , do you mind explaining

[davis_greatest]

I'm terrible at these but I'll PM you the answer now!

[snookersfun]

to prevent this thread from slipping from the home page
and give late entrants a chance to gather points :

How about geometry???

I attached the picture, here is the question:
In this diagram, AB is the diameter of the circle. If AB is 10 cm and the area of the triangle is 11 cm2, find the perimeter of triangle ABC.

[davis_greatest]

Congratulations, Vidas! .....

... volleyed home a nice cross into the box from snookersfun

HERE IS THE SCOREBOARD AFTER ROUND 29

snookersfun……………………….…..14
Vidas……………………………………….8½
abextra……………………………..…...4½
robert602…………………………………4
davis_greatest…………………..……2

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

[snookersfun]

That is fine!
I thought about one player playing his 31 games and winning 16 of them, which leaves a group of 16, which will be further devided by the same criteria. Looking back my explanation wasn't quiet there (especially as I had the win or loose situation to confuse everything, see I missed the point that one can find a player which must have won the one game more) and Vidas' explanation is much clearer. So, all well

[davis_greatest]

I would, but it looks like you found some proof, at least I can't think of any other different way how to solve it.

May be it needs only some further explanation...
in other words,
among 32 players, we always can find player A who won half of the matches(16) or more.

group1: 16 players who lost to A.
consider all matches within this group only.
we can find player B who won at least half of the matches there.

group2: 8 players who lost to B.
similarly find player C with 4 wins.

group3: 4 players who lost to C, etc.
so the algorithm is clear..

Well, yes! (Of course, with 32 players, each player plays 31, so the average number of wins is 15½ (not 16) and so one player must win at least 16.)

snookersfun, was this what you meant? You seemed to be talking about finding 8 players etc who had won/lost, rather than ONE person who had beaten 8 players etc. That is a different thing entirely.

Let me know what you meant. You were certainly along the right lines but I think I am going to give the point to Vidas for the proper explanation. (Sorry.)

[Vidas]

...well it is min. of 8 wins or losses again... and it is for any player out of this group of 16 playing his 15 games against all the rest in that group.

As I said, Vidas is so much better at this... take over, will you?

I would, but it looks like you found some proof, at least I can't think of any other different way how to solve it.

May be it needs only some further explanation...
in other words,
among 32 players, we always can find player A who won half of the matches(16) or more.

group1: 16 players who lost to A.
consider all matches within this group only.
we can find player B who won at least half of the matches there.

group2: 8 players who lost to B.
similarly find player C with 4 wins.

group3: 4 players who lost to C, etc.
so the algorithm is clear..

[davis_greatest]

Haven't read the opriginal question so apologies if I've missed some critical point here....

PS The question is only 3 sentences - and at the top of this page.

[davis_greatest]

Let's assume that only the World's Top 32 qualify to make it easy.

If the World No 1 beats every body ranked worse than him, and the same with the world number 2, all the way down to the world ranked 31, who beats the world number 32 and loses to everyone else, while the world number 32 loses to everybody.

I believe in this case that I can stand all 32 players one behind the other (in world ranking order as it happens) and each man would have beaten every man behind him in the line.

Haven't read the opriginal question so apologies if I've missed some critical point here....

rambon, you are only showing that IF there is a certain outcome of matches, then you can stand at least 6 (in fact, all 32) players in line.

But I am asking you to show that you will ALWAYS be able to stand 6 players in line, no matter what the outcome of the matches.

snookersfun is very much along the right lines but either isn't explaining it properly or I am misunderstanding what she means.

[rambon]

Let's assume that only the World's Top 32 qualify to make it easy.

If the World No 1 beats every body ranked worse than him, and the same with the world number 2, all the way down to the world ranked 31, who beats the world number 32 and loses to everyone else, while the world number 32 loses to everybody.

I believe in this case that I can stand all 32 players one behind the other (in world ranking order as it happens) and each man would have beaten every man behind him in the line.

Haven't read the opriginal question so apologies if I've missed some critical point here....

[snookersfun]

...well it is min. of 8 wins or losses again... and it is for any player out of this group of 16 playing his 15 games against all the rest in that group.

As I said, Vidas is so much better at this... take over, will you?

[davis_greatest]

lets see, if I learned something here:
each player plays 31 matches-has to win or loose a minimum of 16 of those. Of the players won (or lost, one can go in both directions here) 8 must win again within this group, after that 4, 2 and 1, so that one will reach a sequence of 6 players.

What do you mean "8 must win again within this group..."?

[snookersfun]

you don't like this one, do you?
I see Vidas is now on, it is his speciality....
I go to sleep now. Happy puzzling

[snookersfun]

The questions are coming thick and fast now!

Next year, after the success of the round robin format at the Grand Prix, they decide to have a round robin format at the Crucible.

It starts with all 32 players playing each other once (round robin style).

Show that, after the round robin matches have finished, you will be able to find 6 players and stand them in a line (one behind the other) in such a way that each player in the line has beaten all the players standing behind him.

lets see, if I learned something here:
each player plays 31 matches-has to win or loose a minimum of 16 of those. Of the players won (or lost, one can go in both directions here) 8 must win again within this group, after that 4, 2 and 1, so that one will reach a sequence of 6 players.

[davis_greatest]

no excel, no Lagrange... this one was easy

The age of the younger person had to end in 9 otherwise the sum of the numbers would change by one only.
Then I took into account that each 9 turning into a 0 reduces the sum of the older person by 9, thus 2 nines by 18, which can be offset by the gain of one in the hundreds to loose a 17. Two nines at the end need another 16 to reach a sum of 34 (2x17), therfore only 88 or 97 possible, 88 being the lower possibility.

Good - almost perfect. You should also mention though that 7999 is not possible (despite 7 and 9 also adding up to 16, and 79 being lower than 88) because if the number ends in 999 (but not 9999) then adding one decreases the sum by 26.