Brain Teaser No 1

3rd time lucky?

£2.33... or 2 and 1/3 pounds.

Matthew

3rd time lucky?

Full marks for trying and, more importantly, persisting, but my arithmetic generates £4.

Now i've checked the text of the original teaser and i'm pretty sure that i haven't made any typos or errors in translation (the original was written in euros !!!!!!!!)

I'm pretty sure that Paul R has a copy of the original and would have made a comment if I'd got the text wrong. Any comment Paul ???

So, if you can see your way to four quid, fine ?

If not, i'll post my soultion and we'll see if I made a typo (I hope i didn't 'cos this would be most irritating to all who gave it some thought and most embarrasing for me)

Cheers

Don

Sorry, I've not seen this puzzle before. I haven't got past seeing similar triangles.

Paul

Don,

I disagree! £4 is too high.

OK, my proof....

Triangle of size 15,14,13
The perpendicular height from the 15cm edge to the opposite corner can be found using pythagorian principles (n=perpendicular height, m is distance from 15,14 corner)

14^2 = n^2 + m^2
13^2 = n^2 + (15-m)^2

therfore

m = (14^2 - 13^2 + 15^2)/30 = 42/5
and
n = 56/5

The centre of the large circle will sit on the point where lines from all three corners meet and from each corner the line must be at an equal angle with both sides (hope this is clear)

Therefore the gradient of lines from corners between 15 and 14 sides and 15 and 13 sides are as follows

15 and 14

Gradient = tan (arctan((56/5)/(42/5))/2) = 1/2

15 and 13

Gradient = tan (arctan((56/5)/(33/5))/2) = 4/7

Intercept can be found be solving following (X is distance along 15 line)

(1/2) X = -(4/7) * (X-15)

X = 15 / (7/8 + 1) = 8

There fore distance from 15 line and radius of large circle is (1/2)*8 = 4

To find size of smaller triangles we must subtract the circumference from the perpendicular distance found above for the 15 edge but also for the 13 and 14 edges.

For 13 and 14 edges the distance is

from 13 line 168/13

and from 14 line 12

so ratio between large coin and smaller coins are

56/5:(56/5-8)
12:(12-8)
168/13:(168/13-8)

so if the large coin equals 1 then the other coins equal 2/7, 1/2, 8/21 which totals 7/6 so the smallest coin is worth (2/7)/(13/6)*28 = 48/13 which is £3.69, another answer once again! But I don't get £4

Let me have your comments.

Matthew

Let me have your comments.

Bloody well done mate....!!!!

Hmmmmmm........you have a small error........

Now first, let me make it clear that my solution is similar in principle to yours, BUT and a very big BUT, it avoids all the akward number juggling!!!

Sooooo..... to find your small error, I had to check your solution line by line. And guess what? the small error is in the 2nd off last line....Arghhhh!

Oh! BTW when you said To find size of smaller triangles we must subtract the circumference I presume you meant to write "diameter"

So, back to that small error. When you wrote so if the large coin equals 1 then the other coins equal 2/7, 1/2, 8/21 which totals 7/6
I think you will find the correct numbers are 1/3 instead of 1/2 and the total then becomes 1.

The rest of your arithmetic will then generate £4

Now, I still think you did a bloody good job. I made numerous errors when following your method and it just shows how careful you really need to be, and just how careful you were.....just 1 little slip....

Assuming we reach agreement on the answer, you have the choice of finding the more simple solution, or letting me demonstrate it. It occupies about six lines of geometric ratios and one line of arithmetic that wouldn't make a 10 year old sweat. i'm sure that Paul R is just dying to use those similar triangles, which in principle, is what you've done.......

Let me know if you agree with the £4

Cheers

Don

The envelope switching problem........On the face of it, both part 1 and part2 are simple 50/50 choices.

OTOH, being told that the first choice holds £100, means that the other hand holds either £50 or £200, which seems to suggest that the prospects on switching improve to £125 whereas the prospects on not switching remain at £100.

Any thoughts ?


Well, i had a few thoughts whilst soaking in the bath.

It seems to me that as soon as you're told how much is in the envelope that you first choose, you obviously have more information than before. Is this extra information of any use?. Well I think it is and I oulined my thoughts before, and have repeated them above.

I've just knocked up a little spread sheet. it confirms that with a straight choice (part 1) you can stick, change or make random second choices and your chances of winning are the same....50/50. But when told there is a specified sum of money in the envelope first chosen....well, your prospects do seem to improve if you always change your mind and swap envelopes.

But then excel is excel.....any body else agree?

Cheers

Don

The envelopes aren't well defined.

Imagine if the envelopes contain 50 and 100, and you're told what the one you pick contains, which in this case happens to be 100. You're no better off for knowing unless you also know that the envelopes contain 50 and 100. Being told the envelope you hold contains 100 only appears to add more information.

For Don's analysis to work we are assuming that the contents of the second envelope randomly vary between 50 and 200, and I don't think this assumption is supported by the wording of the problem.

I think.

It's perhaps worth a trawl around the Internet for more cogent analysis...

Paul

For Don's analysis to work we are assuming that the contents of the second envelope randomly vary between 50 and 200, and I don't think this assumption is supported by the wording of the problem.


I've re-read the question (part 2) that Belsizepark posted. I have assumed we are not dealing with a riddle nor a play on words. Two facts seem clear to me. First, one envelope contains twice as much money as the other. Second, the envelope I have first chosen, contains £100. I am pretty sure this implies that the other envelope contains either £50 or £200. At present I am not certain what the probability is of the second envelope containing £50, or the probability of it being £200. I am still developing my spreadsheet to ensure the choices are realistic.

Any question about chances has to be based on the idea of "if I do this often enough, what are my prospects...". However, the question doesn't state, and IMHO doesn't imply that the envelopes always contain £50/£100/£200.... Hence, "if I do this often enough...." then the next time i could be told my first choice contains £10, or £50 or £200 or...... anything ie £x.

My gut feeling is that its best to switch.....

Cheers

Don

I have two piles of envelopes, one pile contain 50, the other 100. I'm sitting in a room with two doors. At one door is a queue of people. They enter the room one at a time, I show them two envelopes and they choose one. I tell them what that envelope contains (50 or 100) and offer them the chance to take the other.

In this case it makes no difference whether they swap or not, knowing what is in one envelope doesn't help.

In Belsizepark's puzzle can always be satisfied by envelopes containing 50 and 100 or 100 and 200, there is no (given) reason to suppose that there is a mixture of the two, so the strategy of swapping is based on an assumption.

Better the 100 in the hand etc.

Paul

so the strategy of swapping is based on an assumption

Given your senario Paul, I am inclined to agree.

However, since it makes no difference whether you switch or not, you might as well switch, just in case we haven't spotted some crucial assumption........you see, in MY room with two doors the two piles of envelopes contained £100 and £200 respectively......

Cheers

Don

I shall probe the bowels of the Internet tomorrow in search of some learned mathematical justification. Or not.

But basically, tonight, I think that the knowledge that the envelope you've been handed contains £100 doesn't add any information on which to base a swap/no swap decision.

Paul

Don,

I agree, 1 glaring error gives the wrong anser and £4 is the rather elegant value.

No time to devious a elegant solution just know so if you would like to post your solution please do so.

Matthew

Part 2 (as published by Belsizepark)

Tom holds 1 envelope in each of his hands, and tells Harry that he can have either one of them at his choice. Harry picks one and then Tom tells Harry that one envelope holds twice as much money inside it as the other. Tom tells Harry that inside the envelope Harry has picked is £100.00. Tom then offers Harry the chance to switch envelopes. The question is should Harry switch and does it matter?


Paul R said

But basically, tonight, I think that the knowledge that the envelope you've been handed contains £100 doesn't add any information on which to base a swap/no swap decision.

I agree. I have developed my little sreadsheet to simulate the second part of Belsizepark's teaser. The prospects of winning are the same as the first part. ie it makes no differene whether you stick with your first choice or change your mind.

I will check the spreadsheet first, and explain it concept in a later post

Cheers

Don

I will check the spreadsheet first, and explain its concept in a later post


Switching envelopes (Belsizepark) Part two

My spreadsheet works like this

Column A generates random numbers between 1 and 100
Column B generates random numbers 1 or 2. ie 1 or 2
Column C generates the opposite of column B ie 2 or 1
Column D generates contents of Envelope L ie Column A x Column B
Column E generates contents of Envelope R ie Column A x Column C
We now have Envelopes L and R with twice as much in one as in the other up to£100/£200 respectively and we don't know which is which.

The next three groups of columns work in similar ways to each other. They represent choosing an envelope at random, then "always sticking with that envelope"; "always swapping that envelope"; "randomly choosing whether to stick or change" In each case they work as follows :-

Column X generates a random number 1 or 2. If 1, we chose Envelope L, else we chose envelope R
Column Y decides whether we 'stick', or 'change'
Column Z works out the value of our chosen envelope

Each version of Column Z is totalled (the spreadsheet runs 1,000 rows of simulations) and this gives our prospects based on 'always stick' 'always change' or 'random change'

If it doesn't matter what you do, all prospects should be about £75,000 (ie about half the contents of all the envelopes, which in turn average about £150,000)

All runs so far (about a dozen) yield prospects between £70k and £80k or a 'win' probability of about 0.5

So, within the limits of 1,000 simulations and the above logic, it matters not whether you always stick, always change or randomly change, your prospects of winning are 50/50

Paul, I think this means I agree with you.

Copies of the spreadsheet available by e-mail. No rush....

Cheers

Don

First just a quick recap of the question

The value of each coin is proportional to its radius and the total value of the four, in pounds is 28.

The presentation pack is cleverly designed as a triangle with sides 13, 14 and 15 cm. The largest coin just fits in the box touching all three sides.

There are three pieces of very thin card, each just touches the large coin and is parallel to a side of the box. This creates three smaller triangles, one in each corner.

The three remaining coins fit into the corner triangles such that each just touches the three sides of its triangle.

I was showing some friends the coins the other night, (along with my 6 hours of holiday video and 252 slides in an interesting AV night.......) when in shear desperation it seemed, one of my friends asked how much the smallest coin was worth in ££££.

Then as Mattew said

No time to devious a elegant solution just know so if you would like to post your solution please do so.


Its too big (in KB ) to get in one attachment so there are two bits

Cheers

Don

Cheers

Don

Don,

Nice solution. There seems to be some pretty large steps in that calclation with out the proof, I'm sure somewhere along the line somebody has proved them though.

Have a great New Year, keep those brain teasers coming.

cheers

Matthew

There seems to be some pretty large steps in that calclation with out the proof,

On reflection, perhaps I see what you mean, eg

Z = 0.5(PR)

Imagine the 3 triangles formed, each with an apex at the center of the large coin. The "height" of each such triangle is R and the bases are a; b; c in turn. This makes Z = 0.5PR easier to see and the rest of the first part of the solution should be straightforward.

Tell me if there are any others that aren't too obvious. I will look at part two in a moment.

Have a great evening and a good new year.
Cheers

Don

Here's a little warm up teaser for these frosty new year days.

The picture shows a regular, uniform cross (lengths and widths of each of the four arms are identical). In how many ways can I make two straight cuts to create four identical pieces (equal area, same shape).

Cheers

Don

Lots and lots.

Matthew

Countable ways.

Paul

Paul,

Was that a question?

Matthew

No.

I think that if you put a 'cross hair' over the centre of the cross and rotate it to any angle each quadrant will contain a similar part of the cross.

I could count them, but it might take some time.

Paul

Paul,
It's been a very long time since I thought about these things, but doesnt countable mean there's a mapping to the set of integers. Now, if any angle will work, that's the set of reals, which I thought were not countable. I'm probably wrong on this though - just musing on math here.
cheers,
Dan

Dan,

I think you're right. And that's what I wanted to write in the first place. It's been one of those afternoons.

Damn.

Paul