Puzzles with numbers and things

[snookersfun]

Robert602 is back and The Statman enters the race

HERE IS THE SCOREBOARD AFTER ROUND 50, well done Robert602 with round 49 to be decided by d_g or Elvaago. (Thanks for joining us T S, looking good there)


snookersfun……………………….…..21
abextra……………………………..…...11
Vidas……………………………………….8½
robert602…………………………………6
davis_greatest…………………..……5½
elvaago...............................4
Semih_Sayginer.....................½

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

[snookersfun]

well done! A bit shorter would have been from here:
Multiplying both through by 504 (7x8x9 of course)
7x+8y+9z=2352 (small typo)
9x+8y+7z=2016

addition of these two equations to yield:
16(x+y+z) = 4368
therefore x+y+z=273

[Robert602]

I just did a few clumsy simultaneous equations, there's probably a more elegant way.

From the point of view of the journey there
Let z = the distance downhill
y = the distance on the flat
x = the distance uphill

Then we're looking for x+y+z, and we have:

x/56 + y/63 + z/72 = 4
x/72 + y/63 + z/56 = 4 2/3 (since uphill has now become downhill, and vice versa)

Using time taken = distance travelled / speed, and adding them all up.

Multiplying both through by 504
7x+8y+9z=2353
9x+8y+7z=2016

Subtracting one from the other
2z-2x=336

Rearranging:
z=168+x

Plugging back into 7x+8y+9z=2352

7x+8y+(9x+9*168)=2352
16x+8y=840
y=105-2x

Therefore x+y+z = x+(105-2x)+(168+x) =273

[The Statman]

Can I post a question even if I have no clue about the exact answer? :-)

If so, then here goes.

Picture a normal snooker table.

Without using any sort of spin, what is the largest distance you can cover between hitting the white ball, the white ball hitting 4 cushions and then coming to a halt? Tell me where to put the white ball and describe the trajectory.

Similarly, what is the smallest distance you can cover?

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.

[snookersfun]

yes, indeed. Are you putting up a short solution?

[Robert602]

Is it 273cm?

[snookersfun]

Here is another one, although Elvaago's is still open:

Statman's woodlouse (loved it)

The woodlouse left the snookertable and proceedes on a 'hilly' surface. It moves downhill at 72 cm.p.h. (cm per hour), on the level at 63 cm.p.h., and uphill at only 56 cm.p.h. The woodlouse takes 4 hours to travel from place A to B. The return trip (along the same way) takes 4 hours and 40 minutes. What is the distance between point A and B.

[davis_greatest]

I'd have thought the shortest trajectory must be having the cue ball starting touching a side cushion in the corner (do we ignore corner pockets?), playing the ball directly into the side cushion so it hits it after travelling distance zero, then hitting the end cushion at the same end of the table, but travelling as near as is possible to parallel with the ends of the table. Then the total distance travelled will just be twice the width of the table.

[davis_greatest]

I wouldn't be surprised if the largest distance would be hitting the ball straight up and down the length of the table several times.
This is not supposed to be an attempt at an answer...

Obviously up and down but not parallel to the side cushions (and not hitting them) would give a longer trajectory than parallel to the side cushions.

PS I've been thinking recently about asking a question similar to this, but haven't thought what exactly yet!

[snookersfun]

I wouldn't be surprised if the largest distance would be hitting the ball straight up and down the length of the table several times.
This is not supposed to be an attempt at an answer...

[elvaago]

Can I post a question even if I have no clue about the exact answer? :-)

If so, then here goes.

Picture a normal snooker table.

Without using any sort of spin, what is the largest distance you can cover between hitting the white ball, the white ball hitting 4 cushions and then coming to a halt? Tell me where to put the white ball and describe the trajectory.

Similarly, what is the smallest distance you can cover?

[davis_greatest]

snookersfun extending her lead

HERE IS THE SCOREBOARD AFTER ROUND 48

snookersfun……………………….…..21
abextra……………………………..…...11
Vidas……………………………………….8½
davis_greatest…………………..……5½
robert602…………………………………5
elvaago...............................4
Semih_Sayginer.....................½

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

[davis_greatest]

Congratulations, snookersfun

25 rows is the correct answer!

Your proof by PM also looked correct. Here, though, is a quick way to do it in your head without all that algebra:

Let n = the number of rows in the triangle of reds in Big Maxisnooker = the number of reds in the last row of Big Maxisnooker

Each ape won 9 frames, including 9 clearances of the colours at the end - these will cancel out so let's ignore them and just consider how many reds each ape potted.

In the first frame of Monstersnooker, Charlie made a total clearance, and the number of reds he potted was 9 times the number of reds in Little Maxisnooker, plus one. In the 8 frames of Little Maxisnooker, Gordon potted all the reds. Therefore, after 9 frames, the number of reds that Charlie had potted more than Gordon was equal to the number of reds in a frame of Little Maxisnooker, plus one.

In the 2nd half of the match, the reverse happened. First, Gordon made a total clearance in Monstersnooker, and the number of reds he potted was 9 times the number of reds in Big Maxisnooker, plus one. In the 8 frames of Big Maxisnooker, Charlie potted all the reds. Therefore, in the second set of 9 frames, the number of reds that Gordon had potted more than Charlie was equal to the number of reds in a frame of Big Maxisnooker, plus one.

Combining the two halves of the match, we see that the total number of reds that Gordon had potted more than Charlie was equal to the number of reds in a frame of Big Maxisnooker minus the number of reds in a frame of Little Maxisnooker, which is just the number of reds in the last row of Big Maxisnooker, or n.

If Charlie and Gordon had potted all blacks (no pinks), then if Gordon had potted n more reds than Charlie, Gordon would have scored 8n more points (each red + black is worth 8 points). However, we know that they scored the same number of points. Every pink potted (instead of a black) reduces the points scored by one. Gordon potted 200 pinks more than Charlie.

Therefore 8n = 200.

So n = 200 / 8 = 25.


(As you say, the lowest break is 24x25/2 x 8 + 27 = 2427.)

Scoreboard to follow...

[snookersfun]

it doesn't end...
2427 for the lowest breaks and 25 for the number of rows in BigMax

I'll PM the corrected proof as well

[snookersfun]

next inspiration... on way to work... I still calculate like crap....
2627
and now, I really check up all my stuff