Brain Teaser No 1

Gentlemen

Oh dear! I have goofed big-time on this one....

I said the diameter of the tube was 4 inches and I intended to say that the circumference of the tube was 4 inches.

Given what I actually wrote, you are of course all quite correct, and please accept my apologies.

It was only when I saw Matthew's solution, that I noticed his inclusion of (pi) in what otherwise looked like my formula, and thus realised my mistake.

Of course, the question that I intended to set has a much neater answer (41 inches), and I am impressed that you all developed correct answers, and maintained the courage of your (correct) convictions.

Well done, and again, my apologies.

Cheers

Don

Don,

Well, I recall thinking your figure was a little out of scale. No worries - please keep them coming.

cheers,

Dan

The Cutting Edge

What are the parameters (eg shape and principal dimensions) of the shortest smooth curve that can cut the area of an equilateral triangle in half.?

(I hope this one is clear and unambiguous!)

Cheers

Don

It is a part of a circle and has length.
(pi/(4*3^0.5))^0.5 times an edge length of the triangle.

Matthew

You all knew it was bound to happen one day ?

Not sure this is equivalent to Matthew's answer, but...

I thought it should be a circular arc with radius of curvature of
sqrt(3^1.5/(2*pi)) * edge length, centered at a vertex.

Dan

Don,

In other words, imagine putting a circle in the middle of your picture of a hexagon at the top of the previous page, that has area half that of that hexagon.

cheers,

Dan

Dan,

You mean like this.....

Cheers

Don

Fritz,

Neat.

I bet Top Cat enjoyed it as well

Cheers

Don

Don,

Yes, exactly. Now is that the solution you had in mind?

Dan

Yes, exactly. Now is that the solution you had in mind?

Yes, exactly.

Cheers

Don

However

Matthew, Dan

Your answers are different to each other (by my reconning!!!!)

And mine is different to both yours.....

However, I had difficulty "reading" both your equations, ie trying to decide what terms were inside the brackets and what terms were subjected to square roots.

It's therefor possible that Dan's answer is the same as mine, because we differ by the square root of 2....

If the triangle has a unit side, my radius would be 0.643037

I calculate Dan's to be 0.909393 and Matthew's to be 0.673387 but remember, I might be "reading" your formulae wrongly. And I have no way of knowing whether my answer is right!!!!

Cheers

Don

And

I will write my calcs out neatly, scan them, and post them soon........if that helps

Cheers

Don

Don,

You're absolutely right - I missed a factor of 2 in the denominator:

r = sqrt(3^1.5/(4*pi)) = 0.643037

If triangles have sides of length b then,

Area of hexagon = 6 x sqrt(3) x b^2 / 4

which should equal 2 x pi x r^2

cheers,

Dan

How I visualised it

Cheers

Don

[This message was edited by Don Atkinson on SATURDAY 31 January 2004 at 20:40.]

quote:

---

Originally posted by Dan M:
Don,

You're absolutely right - I missed a factor of 2 in the denominator:

r = sqrt(3^1.5/(4*pi)) = 0.643037

If triangles have sides of length b then,

Area of hexagon = 6 x sqrt(3) x b^2 / 4

which should equal 2 x pi x r^2

cheers,

Dan

---

Stone Me ! That's exactly what I thought too.

Fritz Von Coldfusioninnit²

What Makes 100%???

What is 100%?

What does it mean to give MORE than 100%?

Ever wonder about those people who say they are giving more than 100%?

We have all been to those meetings where someone wants you to give over
100%.

How about achieving 103%?

Here's a little mathematical formula that might help you answer these
questions:

If:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

are represented as:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.

Then:

H-A-R-D-W-O-R-K

8+1+18+4+23+15+18+11 =3D 98%

And

K-N-O-W-L-E-D-G-E

11+14+15+23+12+5+4+7+5 =3D 96%

But,

A-T-T-I-T-U-D-E

1+20+20+9+20+21+4+5 =3D 100%

and,

B-U-L-L-S-H-I-T

2+21+12+12+19+8+9+20 =3D 103%

And look how far ass-kissing will take you.:

A-S-S-K-I-S-S-I-N-G

1+19+19+11+9+19+19+9+14+7 =3D 118%

So, one can then conclude with mathematical certainty that:

While Hard work and Knowledge will get you close, and Attitude will get you there, Bullshit and Arse-kissing will put you over the top!!!

Fritz Von Coldconfusioninnit

100% Fritz.

Cheers

Don

The length was the length of the curve not the radius of the circle.

Matthew

The length was the length of the curve not the radius of the circle.

Aaaahhhh......

Now if only we had actaully READ your answer, which, on re-reading it, is perfectly clear......

Cheers

Don

Prove that the remaining squares can be exactly covered by L-shaped objects of 3 squares each,

2^(2n) is always even (and is always an even power of 2....) which in turn might dictate that there must always be an even number of rows and columns in the grid, or some multiple of 2 rows and/or 3 columns etc.....
[2^(2n)] - 1 is always odd
the'odd' 'L' must always be adjacent to the 'missing' square
pairs of 'Ls' can make up 3x2 rectangles , or 'Ts', or 'Us'
the total number of 'L' shapes must always be odd

If the conjecture is true, then [2^(2n)] - 4 must be a multiple of 3 and 2 (ie 6)
I can visualise 'fitting' 'Ls' around a 'missing' square in a corner, or near a corner, or near an edge.

does any of this look useful???

Cheers

Don

If it were true then there would always have to be an 'x' such that,

3x = 2^2n - 1 = 4^n - 1

Or is that trivial?

The proposition is clearly true for n=1,

 * *  ** ** 
** ** *   *

And given the fourway rotational symmetry of the square would be true for n=2 if we could tile this,

  **
  **
****
****

in our little 'L's. Which we can,

  aa
  ba
dbbc
ddcc

We note that this is twice as large as our basic 'L'.
By the same symmetry argument it would be true for n=3 if we could tile,

    ****
    ****
    ****
    ****
********
********
********
********

Which again is twice as large and can clearly be tiled by these,

  **
  **
****
****

Which we demonstrated could be tiled by the basic tile.

So because we can always tile an 'L' with 'L's half its size we can always tile a square 2^n on a side except for an arbitrary spot with 'little' 'L's.

Well I'm convinced. And I think that also proves the assertion in the first paragraph...

Paul

quote:

---

Formally you still have to show that you can do this scaling regardless of the initial position of the missing square, but it sounds quite reasonable.

---

I think that's implied by the rotational symmetry. For n=1 a single tile in one of four orientations. For n=2 that tile sits in a quarter of the whole, and the whole can be rotated into the four positions, etc etc. As long as we can cover a quarter of the square with a single point missing we can cover the whole square. That whole square becomes a quarter of the next size up.

Your proof is rather better though...

Paul

I thought that if I have a square that I can completely cover except for any one spot then I can also cover a square of twice the size except for any one spot if I can completely cover the other three quarters. The other three quarters always look like the original three squares except for their scale. So if I can prove it for n and 2n then it doesn't matter what n is.

There must be an elegant way to present the argument, I guess it needs a pen and paper...

Paul

A clear solution?

A square of 2x2
The L clearer fits around the missing square

A square of 4x4
Section into 4 2x2 squares. The section with the missing square is complete with an L. The otehr three are completed by a L which bridges all of them. Highlighted below.

A square 2^n x 2^n
A square of any size can be quartered into a square of side 2^(n-1) and ultimately to 2x2. Using the same method as in a 4x4 square the quarters which do not contian the missing square are completed by a bridging L.

Matthew