A BIT OF MATHS

Anything goes, but keep it seemly...

Postby re'born » Sat Oct 06, 2007 8:02 pm

udosuk wrote:
re'born wrote:Hee Hee. Fair enough, Herr Professor.
Triple click below to see what re'born wrote:If r<>1, then r=0 and there is a 1 in the nth position, an n in the 0th position and at least a 2 in the 1st position, hence at least a 1 in 2nd position. So we have 1 + n + 2 + 1 = 4 + n (the +n*r is missing since r = 0).

Nah, still can't pay you full marks for this: "If r<>1, then r=0".:D

Remembering that I was in the case where n>=5, I noted early in my proof that
Triple click below to see what I wrote wrote:The key observation for me was to note that the sum of digits must be 10. This immediately implies that for n>=5, then the number in the nth position (I label the positions as the 0th through 9th positions) is either 0 or 1.

from which it follows that if r<>1, then r=0.
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Postby udosuk » Sun Oct 07, 2007 5:45 am

My point is, since you have proved that r>1 is impossible, you only need to write "if r<1, then r=0", or even just "if r=0, then...". The way you expressed it was too clumsy. I mean I'll just deduct 1 or 2 marks (out of 100) for the lack of elegancy, compared to the 10 marks for the logical gap you didn't fill before. You should be glad that you didn't have a picky teacher like me in your school days (or did you?).:)
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Postby re'born » Sun Oct 07, 2007 9:51 am

udosuk wrote:My point is, since you have proved that r>1 is impossible, you only need to write "if r<1, then r=0", or even just "if r=0, then...". The way you expressed it was too clumsy. I mean I'll just deduct 1 or 2 marks (out of 100) for the lack of elegancy, compared to the 10 marks for the logical gap you didn't fill before. You should be glad that you didn't have a picky teacher like me in your school days (or did you?).:)

I had the misfortune growing up of always be graded quite fairly in school. Consequently, I consistently lost the sort of marks you're taking from me now (and occasionally a few more).:)
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Postby Bigtone53 » Sun Oct 07, 2007 3:29 pm

Having mentioned the UK's Sunday Times Brainteaser before, here is today's. I do not yet know the answer (because of the Rugby), although I can see the way forward, but I share it with those who do not get the paper. This is not a deliberate attempt to get away from udosuk's strict logic:D

TEASER 2350 H Bradley C Higgins
A puzzle in a magazine was flawed: it had two different answers.
The puzzle showed two circles that did not meet. It gave the radii (prime numbers of centimetres) and the distance between the centres (a whole number of centimetres less than 50).
It then said "Imagine a straight line that is a tangent to both the circles. The distance between the two points where that line touches the circles is a whole number of centimetres. How many"

What was the radius of the smaller circle?
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Postby udosuk » Sun Oct 07, 2007 4:44 pm

I think this is more a maths problem than a brain teaser. Anyway, I worked out the answer:
Triple click below to see what I wrote:The smaller circle has radius 7, the larger one has radius 17.
The distance between the centres is 26.
The 2 different answers to the flawed puzzle are 10 & 24 respectively.

The real tricky bit (after a bit of highschool geometry work) is to find the unique combination for the following constraints:
r1, r2 are both prime +ve integers with r1<r2
c is a +ve integer less than 50
a, b are different +ve integers
c^2=(r2-r1)^2+a^2=(r1+r2)^2+b^2

=> r1=7, r2=17, a=24, b=10, c=26

I hope I at least savoured a little pride for my country after the rugby match.:?:

Added later: Here is a pic showing the actual configuration.
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Postby Bigtone53 » Sun Oct 07, 2007 10:03 pm

A sure answer, udusuk, but what was the basis for your assumption that the two tangents were at right angles to each other? The picture would look different if the two circles were further apart.
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Postby udosuk » Sun Oct 07, 2007 11:51 pm

I've overlooked another possible combination:
Triple click below to see what I wrote:r1=7, r2=23, c=34, a=30, b=14

So the smaller radius is defintely 7.

If you allow the "computer science" definition of 1 being a prime, then there is another combination:

r1=1, r2=7, c=10, a=8, b=6

But for this problem I suppose we will use the "mathematical" definition that 1 is not a prime.

Bigtone53, the tangents don't have to be at right angle to each other, you can split or narrow down the circles, but the constraints remain the same.
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Postby Bigtone53 » Mon Oct 08, 2007 9:22 am

the tangents don't have to be at right angle to each other, you can split or narrow down the circles, but the constraints remain the same.


udosuk,

I am not sure that this is the case. The original faulty problem has to have two correct answers, ie the lengths of both the tangents between the circles have to be a whole number of cm. You have picked a special case where they both are but is this constraint satisfied as the circles are rolled apart? In other words, is your special case a unique answer?

There are many problems out there which can be simply solved by assuming that there has to be a unique answer or the problem would not be set (shades of UR!). Therefore picking the ideal circumstances to find an answer therefore gives you the answer. My favourite is

A solid spherical cheese has a cylinder of cheese removed from it, leaving a hole through the centre of the cheese from one side to the other with the axis of the hole going through the centre of the former sphere. The hole is 10cm long. What volume of cheese remains?
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Postby udosuk » Mon Oct 08, 2007 12:01 pm

Bigtone53 wrote:I am not sure that this is the case. The original faulty problem has to have two correct answers, ie the lengths of both the tangents between the circles have to be a whole number of cm. You have picked a special case where they both are but is this constraint satisfied as the circles are rolled apart? In other words, is your special case a unique answer?

My special case isn't the unique answer, as I mentioned to satisfy all the constraints there are 2 sets of combination: [r1=7, r2=17, c=26, a=24, b=10] & [r1=7, r2=23, c=34, a=30, b=16] (highlight between the square brackets to view them). It just so happens for these 2 sets the radius of the smaller circle remains the same, so we have a unique answer to the whole brain teaser you post.

Just have a look at this pic for the general configuration.

If you are willing to relax one of the original constraints, e.g. r1 need not to be prime, then there are other workable combinations, where the tangents aren't perpendicular to each other. For example, in this pic, the combination I used is: [r1=4, r2=11, c=25, a=24, b=20] (check carefully to see how they work out).

Perhaps next time I will post on how I searched upon all Pythagorean triples within the range 0..49 to find the only 2 possible combinations which satisfy all the constraints. (It isn't that hard, so I don't consider it as brute force.)

Bigtone53 wrote:There are many problems out there which can be simply solved by assuming that there has to be a unique answer or the problem would not be set (shades of UR!). Therefore picking the ideal circumstances to find an answer therefore gives you the answer. My favourite is

A solid spherical cheese has a cylinder of cheese removed from it, leaving a hole through the centre of the cheese from one side to the other with the axis of the hole going through the centre of the former sphere. The hole is 10cm long. What volume of cheese remains?

When solving this problem I didn't assume any uniqueness or things like that. The 2 equations I worked out are independent of whether the 2 tangents are at right angle to each other or not.

As for your spherical cheese problem, I don't think you can work out the exact answer without knowing the diameter of the hole, even if you know the radius of the sphere is 5cm, and the original volume of the cheese is 500pi/3 cm^3?:?:
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Postby Bigtone53 » Mon Oct 08, 2007 2:07 pm

As for your spherical cheese problem, I don't think you can work out the exact answer without knowing the diameter of the hole, even if you know the radius of the sphere is 5cm, and the original volume of the cheese is 500pi/3 cm^3?


udosuk,

As it happens you can. Note that the length of the hole in the sphere after the removal of the cylinder is not the same as the diameter of the sphere.
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Postby mikejapan » Mon Oct 08, 2007 3:14 pm

How do you know the cheese is solid? It might look solid, but have holes inside that you can't see. It sounds like a Dutch cheese to me and they are famous for having holes, while at the same time looking solid.

Aren't you rather assuming the cheese is solid and therefore picking the ideal circumstance?

I think it's very important to know the name of the cheese in question before making rash assumptions about solidity.

It could be a very runny Wensleydale.
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Postby mikejapan » Mon Oct 08, 2007 3:18 pm

Or cottage cheese.
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Postby mikejapan » Mon Oct 08, 2007 3:30 pm

It wouldn't matter how runny the cheese was if you put the remaining cheese in a full bath and calculated the volume of water displaced.
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Postby Bigtone53 » Mon Oct 08, 2007 7:00 pm

1. This being a precise scientific matter, obviously the solid cheese went through an ultrascan before the start, as well as an Archimedes-type volume/density analysis.

2. I suspect that it is Swiss cheeses that have the big holes

3. If the cheese was very runny, perhaps some of it gets displaced along with the water when dunked in the bath.
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Postby wintder » Tue Oct 09, 2007 12:36 am

Oops, frivolous thread, sorry.
Last edited by wintder on Tue Oct 09, 2007 12:13 am, edited 1 time in total.
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