Puzzles with numbers and things

[snookersfun]

don't despair The Statman

read my last post TS, it's all there. I am sure you will leapfrog Semih for the effort, but will leave it for d_g, he is the math freak.

[davis_greatest]

read my last post TS, it's all there. I am sure you will leapfrog Semih for the effort, but will leave it for d_g, he is the math freak.

I will look, I promise - but don't think I will get a chance before Saturday...

[elvaago]

I'm fairly sure The Statman got it right there, with the most thorough explanation possible. But today has been so busy at work that I haven't had much time to do anything but, well, work!

[snookersfun]

triangles

I found another nice one, for anybody with time on his hands

What is the length of AD in the following triangle? (some right angles here and EB=BD=CD=1)

[Vidas]

I found another nice one, for anybody with time on his hands

What is the length of AD in the following triangle? (some right angles here and EB=BD=CD=1)

3
√2 - 1 ?

[snookersfun]

Well done Vidas! Can you put up the solution now and I'll give you the point

[Vidas]

added one more line(see picture).
ABC, EBA, FBD all are similar triangles,
AB=AD+1, BC=2FB.
(AD+DB)/BC = EB/(AD+DB) = FB/DB, or (AD+1)/2FB=1/(AD+1) = FB,
(AD+1)^3=2.

[snookersfun]

Vidas, very nice, here is your point then

HERE IS THE SCOREBOARD AFTER ROUND 51, with round 49 still waiting for d_g to resurface

snookersfun……………………….…..21
abextra……………………………..…...11
Vidas……………………………………….9½
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)

[abextra]

Puikus, Vidas, nice to see you back in action!

[snookersfun]

some funny stuff:
Three statisticians went rabbit hunting one day. While walking along, they scared a rabbit out of the brush and send him running. The first statistician shot and there was a puff of dust 1 yard behind the rabbit. The second shot next and there was a puff of dust one yard in front of the rabbit. The third one yelled, "We got it!!"

The Dictionary: what mathematics professors say and what they mean by it:
Clearly: I don't want to write down all the "in-between" steps.

Trivial: If I have to show you how to do this, you're in the wrong class.

It can easily be shown: No more than four hours are needed to prove it.

Check for yourself: This is the boring part of the proof, so you can do it on your own time.

Hint: The hardest of several possible ways to do a proof.

Brute force: Four special cases, three counting arguments and two long inductions.

Elegant proof: Requires no previous knowledge of the subject matter and is less than ten lines long.

Similarly: At least one line of the proof of this case is the same as before.

Two line proof: I'll leave out everything but the conclusion, you can't question 'em if you can't see 'em.

Briefly: I'm running out of time, so I'll just write and talk faster.

Proceed formally: Manipulate symbols by the rules without any hint of their true meaning.

Proof omitted: Trust me, It's true.

Biologists think they're biochemists.
Biochemists think they're chemists.
Chemists think the're physical chemists.
Physical Chemists think they're physicists.
Physicists think they're God.
God thinks he is a mathematician.

There are three kinds of mathematicians -- those who can count and those who can't.

[davis_greatest]

More answers to maths questions

Number 1....

(I've never attached a picture before - hopefully it will appear below.)

[davis_greatest]

Good - it worked!

Number 2...


[Edit - I've found out why this image didn't appear - it's got too many pixels.]

[davis_greatest]

Oh dear, number 2 didn't work...

Never mind - I won't delete that message. It obviously didn't like .gif files.

Here is number 3... Should work - another .jpg ...

[davis_greatest]

Last one

Not sure why number 2 didn't appear - it says .gif files are allowed. Here in the meantime is the last one, number 4...

[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.