Puzzles with numbers and things

[elvaago]

April madness is sitting directly opposite in a circle of 11 people. Because of that she's sitting an odd number of places away from Oliver. So he can never have just 6 balls with the 3 people in front of him having 5 or less.

[davis_greatest]

Because he is a male?

Actually, you are closer than you think - there is a clue to solving this in thinking about the seating arrangement.

[davis_greatest]

The reason is that April Madness is sitting exactly on the other side.

Sorry?

[Mitsuko]

Because he is a male?

[elvaago]

The reason is that April Madness is sitting exactly on the other side.

[davis_greatest]

first off d_g, thank you for the point.

You're welcome, Semih!

secondly, who picked the seating arrangements (ie. i dont know the answer, so im thinking of something to say.)

Charlie worked them out. He was going for boy-girl-boy-girl...
Charlie wasn't sure whether Vidas was a boy or a girl, the two never having met before, so you will see that Vidas might be having to pretend to be a girl to make the party go with a swing. Hope that's OK...

lastly, great thread. im sure even those who dont try to answer too often enjoy the puzzles and answers. appreciated

Good, and thank you. If anyone wants to rate the thread, you just click on Rating at the top and select the 5-stars option

[April madness]

Explain why, no matter how long they play, it is impossible for there ever to be a time when Oliver has no balls and everyone else each has 11 balls.

Maybe because April madness is cheating and passes some of her snooker balls to Oliver each time when he might run into a situation he has none of them?

[snookersfun]

secondly, who picked the seating arrangements (ie. i dont know the answer, so im thinking of something to say.)

Zeroly, wb Semih
Firstly, I like the seating arrangements but wouldn't mind any different one, I guess.
Secondly, nice to see the flurry of activity of yesterday evening, especially as they dumped so much work on my head (good until the end of the year, I am sure)
Thirdly, I love all these math jokes, run into them yesterday.
Fourthly?, I shall think about the puzzle now (I am sure it is one of those, which knot your brain up)

[Semih_Sayginer]

first off d_g, thank you for the point.

On Oliver's first birthday, which is today, we have a big party, and play a great game called Pass The Balls. Charlie, Gordon and I arrange it for him, using 99 snooker balls, and we stand and watch him play with his friends.

They all sit in a circle. Going clockwise from Oliver, we have April Madness, robert602, snookersfun, Semih_Sayginer, austrian_girl, The Statman, Vidas, elvaago, and abextra (and then we arrive back at Oliver). Oliver looks very happy.

secondly, who picked the seating arrangements (ie. i dont know the answer, so im thinking of something to say.)

lastly, great thread. im sure even those who dont try to answer too often enjoy the puzzles and answers. appreciated

[The Statman]

I will award you a point for this, The Statman.

I haven't checked all the calculations, but I more or less agree the principles, subject to the following observations / exceptions:

1) Is use of the jaws allowed? Technically, I would agree that they are cushions, but I am not sure whether elvaago's question is intended to relate literally to a physical table or to an idealized mathematical rectangle without pockets?

Well, the wording of the question was: "Picture a normal snooker table" so I did.

2) If it is a physical table, then I assume we are meant to assume that the balls rebound off cushions at the same angle as they approach them (i.e. ignore the "slide" that happens on a real table).

I pictured a real table (as I was asked) and allowed the same angle of departure from the cushion as the arrival. I cannot picture 'slide'.

3) The Statman - if the jaws are allowed, as you have used them for the shortest distance, then could you not also use them for the longest distance? That is, you could then assume that the ball starts in the jaws of one corner pocket, is struck very hard into the jaw of the diagonally opposite corner pocket, such that it bounces back out to the original corner pocket jaw, then back again, and twice more such that it has travelled 5 diagonal lengths of the table! This would then be the longest possible, but the ball would have to be struck extremely hard indeed not to enter the corner pockets.

the back of the pocket is not a cushion.

4) With your answer of the shortest distance, using the jaws, you have assumed that the ball travels 5 times between the jaws. In fact, you could make it travel only 3 jaw-widths if:

a) it starts off almost touching a jaw and is played directly against the jaw that it is almost touching (so it has travelled virtually zero distance before the first contact); and

b) it is played so gently, that it comes to a halt immediately on hitting the jaw for the 4th time, instead of continuing to travel to hit the jaw for the 5th time. This then gives you /\/ (but straight up and down) instead of /\/\/

I did state 'shortest maximum' meaning shortest distance you could get where any further would involve a fifth cushion. I had read it as requiring the maximax and the minimax, rather than maximax and minimin, as it were – I see you are right!

[davis_greatest]

Round 52 - Oliver's party: Pass The Balls

On Oliver's first birthday, which is today, we have a big party, and play a great game called Pass The Balls. Charlie, Gordon and I arrange it for him, using 99 snooker balls, and we stand and watch him play with his friends.

They all sit in a circle. Going clockwise from Oliver, we have April Madness, robert602, snookersfun, Semih_Sayginer, austrian_girl, The Statman, Vidas, elvaago, and abextra (and then we arrive back at Oliver). Oliver looks very happy.

At first, everyone except austrian_girl is given 11 balls. austrian_girl, who had her birthday yesterday, had plenty of presents so she is not given any balls.

Then, the balls are passed around according to the following rule:

Every time that someone has at least 6 balls, he or she may pass 6 balls along - giving 3 to someone next to him or her, 2 to the next person along, and then 1 to the person one further along still.

For example, Oliver can pass 3 balls to April Madness, 2 to robert602 and 1 to snookersfun.

Or The Statman could pass 3 to austrian_girl, 2 to Semi_Sayginer, and 1 to snookersfun.


They all play this game for many hours.

Explain why, no matter how long they play, it is impossible for there ever to be a time when Oliver has no balls and everyone else each has 11 balls.

[davis_greatest]

Amended scoreboard positions

I have just noticed, while catching up on this thread, that robert602 was awarded a point for round 50 - a round that I had missed completely.

Since davis_greatest's most recent half-point was for elvaago's round 49, it means that davis_greatest reached 6 points before robert602 - therefore, under the rules for ordering described in message 131 (http://www.thesnookerforum.com/showt...2-page_14.html), davis_greatest is above robert602, both being on 6 points.

SO HERE IS THE AMENDED SCOREBOARD AFTER ROUND 51

snookersfun……………………….…..21
abextra……………………………..…...11
Vidas……………………………………….9½
davis_greatest…………………..……6
robert602…………………………………6
elvaago...............................4
The Statman……………………..……2
Semih_Sayginer.....................1½

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

[davis_greatest]

Points for round 49

I am going to award points galore for round 49:

1 point to The Statman, as stated above
1 point to Semih_Sayginer, for sending me a similar answer by private message
½ point to myself, for more or less answering it too and having the patience to read and even comment on the other answers


I am also awarding The Statman a retrospective point for round 4! (The very first 4 or 5 questions on this thread I had asked on the BBC, and The Statman had answered question 4 there, so he deserves the point.)


SO HERE IS THE SCOREBOARD AFTER ROUND 51

snookersfun……………………….…..21
abextra……………………………..…...11
Vidas……………………………………….9½
robert602…………………………………6
davis_greatest…………………..……6
elvaago...............................4
The Statman……………………..……2
Semih_Sayginer.....................1½

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

[davis_greatest]

I also need to read Semih's PM before updating the scoreboard... after dinner and Strictly Come Dancing ...

[davis_greatest]

Longest

I guess we must assume that the table is of the largest possible dimensions to be within the tolerances in the rules, which is 3582mm×1791mm. We shall assume that the ball is of the smallest possible to fall within tolerances, to maximise the distance: 52.45mm.

We are allowed to hit four cushions (but I think the question meant, to four times hit a cushion, rather than four different cushions). So we can have five straight lines in the ball's trajectory, like this: /\/\/ .

Thus, the longest possible distance is from one corner pocket in a zigzag path that ends in the diagonally opposite corner pocket, roughly equivalent (in basic terms) to five lengths of 12 feet.

We shall therefore put five tables together, end to end, to form a rectangle whose diagonal represents that ball’s trajectory.

The width of this rectangle will therefore be 1791mm, minus the diameter of the ball, as the base of the ball will be half a ball’s with from the cushion at start and finish.

1791-52.45= 1738.55mm.

The length of the rectangle shall be (3582-52.45)×5 which is 17647.75mm.

Then it is a simple case of Pythagoras to find that diagonal:

1738.55² = 3022556.1025 (a)
17647.75² = 311433080.0625 (b)

a+b= 314465636.165

√314465636.165 = 17733.178963880108543858758804515

Which is 17 metres 73.32 centimetres or about 58 feet 4.81566521 inches

We could remove a millimetre to ensure that the ball does not strike a fifth cushion, though this is more than compensated for by the fact that the ball can certainly sit slightly further into the pocket than it could if the cushions met to form a true corner. So in reality a further inch or maybe inch and a half could be added.

Shortest

The shortest maximum length would probably be with the ball on one jaw of the pocket, at the narrowest part of the pocket opening where the ball can be played back and forth from one jaw to the other, perpendicular to the cushion at that point.

If we assume that the corner pocket opening is 3½ inches, which is the generally accepted measurement, then that's 88.9mm

The largest a ball can be is 52.55mm.

So the shortest maximum would be 88.9-52.55= 36.35mm, which again we will multiply by 5 to give a total of 181¾mm (or about 7.1555 inches).

We will call it 7.1554 inches to ensure that the fifth cushion isn't struck.

I will award you a point for this, The Statman.

I haven't checked all the calculations, but I more or less agree the principles, subject to the following observations / exceptions:

1) Is use of the jaws allowed? Technically, I would agree that they are cushions, but I am not sure whether elvaago's question is intended to relate literally to a physical table or to an idealized mathematical rectangle without pockets?

2) If it is a physical table, then I assume we are meant to assume that the balls rebound off cushions at the same angle as they approach them (i.e. ignore the "slide" that happens on a real table).

3) The Statman - if the jaws are allowed, as you have used them for the shortest distance, then could you not also use them for the longest distance? That is, you could then assume that the ball starts in the jaws of one corner pocket, is struck very hard into the jaw of the diagonally opposite corner pocket, such that it bounces back out to the original corner pocket jaw, then back again, and twice more such that it has travelled 5 diagonal lengths of the table! This would then be the longest possible, but the ball would have to be struck extremely hard indeed not to enter the corner pockets.

4) With your answer of the shortest distance, using the jaws, you have assumed that the ball travels 5 times between the jaws. In fact, you could make it travel only 3 jaw-widths if:

a) it starts off almost touching a jaw and is played directly against the jaw that it is almost touching (so it has travelled virtually zero distance before the first contact); and

b) it is played so gently, that it comes to a halt immediately on hitting the jaw for the 4th time, instead of continuing to travel to hit the jaw for the 5th time. This then gives you /\/ (but straight up and down) instead of /\/\/

5) If jaws are not allowed, then the shortest distance would be 2 widths of the table, in the manner I described in message 607.