Brain Teaser No 1

Note, the main diffrence to solution 1 is the use of the approximation that

(A^2 + B^2)^0.5 = A + (1/2A) - (1/4A)^2 + (1/8A)^3 - (1/16A)^4

Of course it should have been
(A^2 + 1^2)^0.5 = A + (1/2A) - (1/4A)^2 + (1/8A)^3 - (1/16A)^4
ie the B^2 should have been 1^2 or 1.

Silly me !!!

BTW the series is not a true series, it was just got useing Duncan F's pattented 'death by excessive writing down' and FWIW, a couple of other versions gave similar accuracy. 'Trial & Error - with the emphasis on the error part.

Cheers

Don

I've even 'cheated' with google. No joy for the general case where the ladder length is non-specific. Following the public methods through with a generalised ladder leads to a familiar looking equation.

Split the ladder in two. The solution is symmetrical so (L/2)+x and (L/2)-x will both touch the wall and floor.

d^2 = ((L/2)-x)^2 - 1
1/d^2 = ((L/2)+x)^2 - 1

So

(((L/2)-x)^2 - 1)((L/2)+x)^2 - 1) = 1

which produces a quadratic in x^2.

If we've got x in terms of L we've got d in terms of L. I'm not seeing any elegance yet.

Just to remind you there is no insight in this post of my own invention. I've been trying trigonometry and then rotating the whole problem 45degrees, so that the floor is a line y=x and the wall y=-x, and the ladder must pass through 0, 2^0.5. And getting hugely boring huge equations in terms of d rather than L.

(Apologies for the excessive parentheses, it's 'half of L'....)

Paul

I tested my "run north 100m; run south 200m; etc theory against the other option of " walk south until you find the other 'chute, then run till you catch your mate". I used 1,600m (well I still think in miles!) as the distance appart and assumed they both set off simultaneously northwards. Of course it took them longer using the zig-zag method.

So not such a good idea! even if it could be programmed using the accepted coding. Apologies to Martin P and Omer for ignoring the finesse of the programming for the moment.

Back to the drawing board!

Cheers

Don

For light relief here's a quickie for the lateral thinkers:
What does this product equal?

(x-a)(x-b)(x-c)...(x-y)(x-z)

BAM

8/10 for the equation, Don. To win the champagne you need to use less conventional maths.

Nice one Bam, unlike the wall

Cheers

Don

Is your elegant solution exact?

Matthew

This is in connection with the ladder problem. Or rather, it is the ladder problem.

Paul

quote:

---

Originally posted by Vuk's son:
Paul

PCB = 2*PAB (IIRC)

Omer.

---

yes, this is because PAC is an isoceles triangle, and PCB = PAB + APC = 2 * PAB since PAB = APC

enjoy

ken

For any given ladder (L) there is only ONE circle that will enable you to position your point P so that it satisfies the constraints of the problem.

You have to fit a 1mx1m square between P and the diameter of the circle. Two sides of that square must be co-linear with PB and PC repectively.

This has something to do with that bit of the sine rule that goes A/SinA = 2r, where r = radius of the circumscribed circle IIRC.

Does this help or hinder??

Cheers

Don

No breakthrough though.

I don't see how one will get a 'better' equation than Don's, I can find some 'easier' ways to get to it, but if the equation is right?

Paul

I don't see how one will get a 'better' equation than..........

I think this is the whole crux of a good teaser!

Mind you, Bam had better have a good solution, or else.....

Cheers

Don

I hope you are not angling for a hint, Don. Champagne = no hints.
BAM

On reflection, any old circle will do! (Lets call the diameter L becuase the diameter represents the ladder) You can draw an infinite range of different sized squares, each touching the perimeter and the diameter, with two sides extended to the ends of the diameter. By DEFINITION each square has side = 1.

Doesn't matter whether 1 = 1"; 1mm; 1cm; 1 cubit or whaterver.

Doubt if this helps, but lets keep an open mind!!

Cheers

Don

OH! and the shorter extension is what Bam is looking for! (does this make it an extension ladder?)

Whatever gave you that impression??

Cheers

Don

quote:

---

Lets call the diameter L becuase the diameter represents the ladder

---

Not in my 'model'.

The diameter represents the bit between the foot of the ladder and the crate plus the bit between the top of the crate and the top of the ladder. The ladder is 'broken'.

By rotating the triangle formed by the top of the crate, the wall and the upper part of the ladder 90 degrees we get an interesting construction that allows us to play with areas and stuff.

It may or may not be helpful.

Paul

Imagine Stallion and Vuk throwing the old gauntlet on the ground, followed by Mick Parry stepping in to try and calm things down, but somehow getting involved! We'd have a three-way shootout on our hands. Old Martin Payne could act as referee. Mind you, Martin would have to take care where he stood. Remember also, both Stallion and Mick have gun collections, but I'm not sure about Vuk - bit of a disadvantage using a camera for this type of shoot, even if it is a Leica. Come to think of it, wasn't Mick thinking about a Leica?

Anyway, back to the truel. Rather than a free for all, and to give Martin a sporting chance as referee, the rules are that each 'player?' takes one shot at a time, with a free choice of target. Clearly survival is the key to winning and based on past performance, we know that Vuk hits his target only 1 in 3 times. Mick Parry hits his 2 times out of 3. Stallion of course always hits his target, never misses. (Hey this is only a story! It might seem like a dream to Stallion and a nightmare to everybody else, but it's still only a story!)

Now to give Vuk a chance, Martin decides to give him first shot. What is Vuk's best strategy for survival?

Cheers

Don

[This message was edited by Don Atkinson on SUNDAY 20 January 2002 at 20:00.]

I'm having difficulty deciding to which eminent gentleman you are refering with the term 'old nag'. Odds on favorite is of course Stallion, in which case there might be a better strategy.

Cheers

Don

Superficially ISTM that Vuk should have a pop at Stallion, if Stallion dies Mick might miss, if Stallion lives he should probably kill Mick, and then Vuk gets another go. And then Vuk dies...

Everytime Stallion shoots someone dies, that limits the length of the truel to two goes for Stallion. So I expect a spreadsheet showing all combinations....

Paul

Classic!

Don't we need an order of shooting? Or does it work out the same either way?

As I said, we're only interested in Vuk's strategy for best chance of surviving (winning). Being the decent man he is, Martin lets Vuk have the first go. After that, its either Mick or Stallion's turn, depending on the previous toss of a coin (you will need to consider both possibilities), or the survivor, depending on what Vuk does and its outcome. The order of shooting repeats itself as often as necessary.

Cheers

Don

Apologies

Don

Aim for Mick
1/3 chance he kills him, then Stallion kills Vuk
2/3 chance he missed, then as Mick doesn't like being shot at
Aims at Vuk, 2/3 chance he hits him, 1/3 chance he misses, Stallion then shots Mick as the greater threat.
2/9 probability that Vuk survives the first round

Aim for Himself
1/3 chance he is succesful
2/3 chance he misses, either Mick or Stallion would then aim at each other because Vuk is obviously mad!
2/3 chance of surviving first round

He should definitely aim for himself. Not much hope of surving the second round, well roughly a third.

Matthew

Sequence = V,S,M
1)V tries to shoot S
To die V misses S, S shoots M, V misses S again, S shoots V [2/3x1x2/3=4/9]. Or V shoots S, M shoots V [1/3x2/3=2/9]. Or V shoots S, M misses V, V misses M, M shoots V and so on [1/3x1/3x2/3x2/3=4/81].
In total the odds of V being shot are a little over 72%.

2) V tries to shoot M
To die V shoots M, S shoots V [1/3x1=1/3]. V misses M, S shoots M, V misses S, S shoots V [2/3x1x2/3x1=4/9].
Odds of V being shot are 78%.

3) Vuk shoots the sky
To die S shoots M, V misses S, S shoots V [1x2/3x1]=2/3.
Odds of V being shot are 67%.

Sequence = V,M,S
1) V ties to shoot S
V dies if V shoots S, M shoots V [1/3x2/3=2/9]. Or V shoots S, M misses V, V misses M, M shoots V [1/3x1/3x2/3x2/3=4/81] and so on. Or V misses S, M shoots S, V misses M, M shoots V [2/3x2/3x2/3x2/3=16/81] or V misses S, M misses S, S shoots M, V misses S, S shoots V [2/3x1/3x1x2/3x1=4/27].
Odds of V being shot are slightly more than 62%.

2) V tries to shoot M
Vdies if V shoots M, S shoots V [1/3x1=1/3]. V misses M [16/81+4/27=28/81].
Odds of V being shot are slightly more than 70%.

3) V shoots the sky
Same as V misses M, so a little over 28/81=35%.

So indeed, in either scenario it looks like Vuk should waste his first shot. If on his next round there are two left he should waste his shot again otherwise aim fair and square at the survivor. So I agree with Omer.

What a cheery game!

There must be moral in here somewhere.

I haven't checked Bam's probability calcs and I don't have a copy of my previous 'probability tree' analysis to compare. Oh for more time!

Cheers

Don