Brain Teaser No 1

My next one falls into the dead easy class.

a) which is smaller 973x973 or 972x974 ?

b) elaborate !

Cheers

Don

Take a number n, in this case 973, and consider the product (n-d)x(n+d) = n^2 - d^2. And this is always < n^2. Another way to look at it is that for a given length of perimeter an exact square has the largest area.

BAM

Matthew T posted his solution to Bam's goat with a (neat?) formula and a bit of itteration in Excel. Well done Matthew!!

My attempt at this has produced the following, which I hope will eventually lead to a more elegant formula for R.

First, my value for R = 1.15872847 to 8 dp (I put the xtra decimal in, just to show it was my own work! and not just a copy - sad really!)

Second, in the following formula y=0.5R*[(R^2 - 4)^0.5]

pi/2=ArcSin(y) + R^2*ArcSin(y/R) - y

this is the formula that I used in excel to itterate for R (although it doesn't look like Matthew's formula, my apologies if it is) working is all in radians.

Third, y = half the length of the common chord. It is also the abscisa of the upper intersection of the two circles, assuming the grass field is centred at (0,0) and the rope is pinned at (1,0)

Fourth, (a fact which doesn't feature in my iteration is), x=1-[(R^2)/2] where x is the point at which the y ordinate cuts the x axis.

Fifth, (another fact that doesn't feature in my itteration), the triangular areas, within the overlapping segments, are made up from issoceles triangles. The base of each triangle is R and the equal sides are of unit length. Of course, the area of the two triangles is equal to y (which is a fact that does feature in my formula)

Now I don't pretent my formula is elegant and it still requires itteration for a solution. But perhaps it might just inspire others to have a crack at finding an elegant solution for R??

Or do we just want Bam to post the bloody solution and be done?? perhaps if my name was Stallion I would post a poll!!

Cheers

Don

[This message was edited by Don Atkinson on FRIDAY 15 March 2002 at 23:25.]

i suspect what bam is after is a CLOSED expression that doesnt require an iterative solution to get the length -- but right now i cannot see my way to this...

enjoy

ken

Neat little equation IMHO.

One important suggestion; no, one slight suggestion; no, one trivial suggestion.....the second term will be negative, for values of R near the solution. You could put a minus sign outside, then reverse the terms inside.

As I said, trivial. but then, I've never liked adding negative numbers when you could take away positive ones. Don't know why, trivial!! But it does show that we are dealing with areas that are less than the difference between two circles (I think).

Cheers

Don

suggestion makes sense don, especially as we know before hand that R should be greater than the radius of the field. in fact the expression even looks better with your suggestion. but still, we are nowhere near the elegant expression for R that Bam is after... problem being that the expression is doubly implicit in R (in the sense that alpha is a function of R and sin(2 *alpha) is also a function of R, and then there are those sq terms in R itself... phew!!!

enjoy

ken

That's probably happened before....

The other day on Radio 2 some vaguely Christian speaker mentioned in the context of the 'debate' regarding that school in the North that's teaching 'Creationism', that 'we all know what happened to Galileo'. Well, I don't. I thought it happened to Copernicus.

I worked out an equation for the rope to hang BAM by using the sum of two sectors less the overlapping pair of Isoceles triangles. It isn't particularly pretty as the posts above show. And I don't think it prettifies.

Where next? Perhaps something with an elegant solution that persists outside of Bristol would be cheering.....

Paul

according to bam, the solution definitely "prettifies" --- my guess is that we need to make use of extra information in the problem that cancels out the complicated relationship between sin of an angle and the angle itself. i have played around with the fact that (0.5pi) = arcsin (1), but this didnt get me anywhere...

enjoy

ken

BUT please not "verbify"!!!

quote:

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Originally posted by bam:
Ken, please don't encourage the Bush's of this world
explain, explicate, expound; exemplify, illustrate; characterize, delineate, depict, interpret, limn, picture, portray, render

BUT please not "verbify"!!!

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ha ha ha. sorry Bam, its paul ranson's fault!!! he got me started with "prettify"

i also hate "securitize", which i meet many times in product specifications. yuk!!!

enjoy

ken

-0.5R*sqrt(4-R^2) + (R^2)*invcos(0.5R) + invcos(1-0.5R^2) = PI/2

Which may well be the same thing as Ken's expression when you swap sines for cosines.

quote:

---

Originally posted by bam:
I'd better point out that I do not have the sort of elegant solution for the goat problem that I'm expecting of you for the ladder problem. What I have is an equation involving the length of rope, two inverse trig functions, pi, one square root and a few squares. I don't think it can be solved exactly. The main thing is that it is an expression that only has the length of rope as a variable (assuming a unit circle).

---

ah, thats good. all the expressions we have come up with are equivalent with a bit if trig. i somehow thought there was a bit of magic to remove the double implicit nature of the equation and generate a closed solution.

nice problem still...

enjoy

ken

quote:

---

ha ha ha. sorry Bam, its paul ranson's fault!!! he got me started with "prettify"

---

Sorry. I'm mortified.

Although 'prettify' is a perfectly acceptable dictionaried* word. And 'securitize' has a meaning, although probably not the one Ken C sees.

*I had to check this wasn't in the dictionary before using it....

Paul

A great king, whos hifi system you have improved by planting a solid gold earth rod in the moat of his castle, has decided to reward you by giving you a chance to marry one of his 10 daughters (personally I blame the estrogen in the rivers).

You will be presented with the daughters one at a time and, as each daughter is presented, you will be told the daughter's dowry (which is fixed in advance). Upon being presented with a daughter, you must immediately decide whether to accept or reject her (you are not allowed to return to a previously rejected daughter).

But there is a catch. You didn't think it would be this easy to marry a king's daughter did you? The king will only allow the marriage to take place if you pick the daughter with the highest dowry.

What is your best strategy and what is your chance of getting hitched?

well, live and learn...

enjoy

ken

I concede that Bam's equation is far more elegant than mine, and leave ken c and Mattew T to decide for themselves how elegant their solutions are, compared to Bam's.

I typed an error in my posted solution when describing y. The roots should have been of 4 - R^2, otherwise you are trying to find negative roots - apologies. Fortunately I got it right in excel. I feel even worse about this than normal, after suggesting slight improvements to ken c!!

FWIW, I have checked the four elements in Bam's equation against the four elements in my equation. The numerical value of each term is identical, although the terms are in a different order. If Bam's terms are A/B/C/D then mine are D/C/B/A where D = pi/2.

Nice little teaser Bam, which could have had a profound effect on world mathematics, if you had popped it before revealing the true crudidity (definitely not in the dictionary!) of your solution.

Lets hope we find that truly, stunningly, elegant formula for the ladder before you meet with any mishap! and lets hope the formula truly is stunningly elegant or the mishap might …………

Cheers

Don

A man is condemned to death and told he will be hanged (obviously with BAM's goat tether...) one day in the next week. However if he can predict which day the hangman will knock on the door he will get a pardon.

The man's lawyer gets all excited and tells the doomed chap 'You realise they can't hang you! If you're alive on Thursday then the execution must be the next day, and you'll know. So Friday's out, but if Friday's out and you're alive on Wedneday then they must have scheduled Thursday!' etc etc.

So the man goes off to his condemned cell to wait out the week and eventual freedom. And is most surprised to be hanged on Wednesday morning.

I don't see how our young hifi advisor can bend the odds to his favour, his best chance is random. 1 in 10.

Paul

Paul, a random selection strategy is not the optimum strategy.

A moment's simulation shows that discounting 3 or 4 of 10 and taking the next one that has a larger dowry than any of those 3 gives a win rate of about 0.399.

Paul

no worries, i didnt even notice the "mistake"

enjoy

ken

A bit of homework shows this to be the case. In my original geometry I opted for the Sine of angles rather than the Cosines.

The attachment explains all!!

Well, sort of!!

Cheers

Don

code:

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StrategyRandom (5) successes : 362880, rate : 0.1
StrategyBiggestAfterNth (0) successes : 362880, rate : 0.1
StrategyBiggestAfterNth (1) successes : 1026576, rate : 0.282897
StrategyBiggestAfterNth (2) successes : 1327392, rate : 0.365794
StrategyBiggestAfterNth (3) successes : 1446768, rate : 0.39869
StrategyBiggestAfterNth (4) successes : 1445184, rate : 0.398254
StrategyBiggestAfterNth (5) successes : 1352880, rate : 0.372817
StrategyBiggestAfterNth (6) successes : 1188000, rate : 0.327381
StrategyBiggestAfterNth (7) successes : 962640, rate : 0.265278
StrategyBiggestAfterNth (8) successes : 685440, rate : 0.188889
StrategyBiggestAfterNth (9) successes : 362880, rate : 0.1

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'StrategyRandom' obviously just always picks the same princess, pretty dumb and only 'random' if you consider that the order the princesses are presented in changes with each test. 0.1 success rate is as expected.

'StrategyBiggestAfterNth' lets 'N' go by, noting the largest dowry and then picks the next princess with a larger one. '0' always picks the first princess, and therefore the biggest dowry 10% of the time, 9 only picks the last princess if the last princess has the largest dowry, therefore it is always right but only gets married one time in 10.

So, where do the other numbers come from? Is the assymmetry intuitively obvious?

(And is there a better strategy? Is there a way of determining what result the optimal strategy would manage without knowing what that strategy is?)

Paul

Whatever strategy you adopt does not affect that probablity?

cheers

Nigel