Brain Teaser No 1

18 layers and 16 balls in each layer.

Each layer comprises 16 balls in a regular 4x4 grid.
First layer (red) is overlaid by second layer (blue) which is offset diagonally
each PAIR of layers fully occupies the 5x5 plan area
Groups of balls form square based pyramids, but of fairly shallow height
Overall height of stack is about 11.0
The solution doesn't work but the idea is there.

You can all have a go at working out the principal dimensions !!

IF (famous words) If, we could squeeze two more balls into a couple of corners we could get 18 balls per layer and hence only need 16 layers each with a thickness of 0.60 making a stack10.0 high. See next post tomorrow, or give it a go yourself.

What do you think ??

Cheers

Don

Try a picture

Matthew,

Suspect the solution will be a confusing as the question.

Spot on mate, spot on.

Cheers

Don

A Simple one to give us all Hope

I have a sphere, radius R. A cone, base radius R and height P. And a cylinder radius R and height P.

P = 2R

What's the simplest/elegant/interesting formula you can come up with to link all three together ?

Cheers

Don

Don

Vcylinder = Vcone + Vsphere?

cheers

Dan

Another simple one

How do you arrange 6 match sticks to give you 4 triangles of equal size.

cheers

Dan

oops

Should have specified that the matches are to be arranged end to end. Hint, the was a hint!

cheers

Dan

Lynn,

Bingo! Not that I want to go down that road again...

-Dan

Dan,

Vcylinder = Vcone + Vsphere?

You don't need the ? perfect

Don

hlh

This "explorer" was a head louse perhaps?

cheers

Nigel

Lynn,

can anyone corroborate this analysis?

Your idea certainly has some merit. However, The two layers of balls as you describe, ie diaganols sqrt3, need a plan area of just under 5.3x5.3 and we only have 5x5.

I think we need to give this one a bit more thought.

Cheers

Don

The balls,

My idea is based on 16 layers each with 18 balls. However, we start with a square grid of 4x4 balls in each layer, and later on, we have to "jiggle" in an extra 2 balls per layer.

The 4x4 grid is based on a centre of ball to centre of ball distance of exactly 4(sqrt 2)/5 (four fifths of the square root of two = 1.1317)

The ball to ball diagonal is then 1.6 and the layer thickness is 0.6...precisley.

With pairs of layers off-set diagonally, the plan area needed for two layers is 4.959798

The spare room around two sides is 0.040202.

By "jiggling" the 16 balls it might be possible to squeeze two more balls into each layer....

Anybody care to try......
Cheers

Don

PS on re-reading this post I realise the terminology has rather got out of hand.......

Lynn,

Anway ... moving this week and getting married in 3;

Given that was posted on Sat 15th, how did the move go ?

Did Matthew's 288 ball problem help when it came to squeezing all your belongings into the removals van ?

And best wishes for next Saturday too. Hope all goes well.

Cheers

Don

quote:

---

And best wishes for next Saturday too. Hope all goes well.

---

I'd like to second that.

-Dan

quote:

---

Originally posted by Don Atkinson:
The balls,

My idea is based on 16 layers each with 18 balls. However, we start with a square grid of 4x4 balls in each layer, and later on, we have to "jiggle" in an extra 2 balls per layer.

---

Layers of 18, your working on the right lines, but think 3x3 rather then 4x4.

cheers

Matthew

The 288 balls.

Hope we have arrested matthew's challenge before then

I have a REAL solution........at last

16 layers at 0.6 intervals with 18 balls in each layer
Each layer comprises two 3x3 grids, interlaced diagonally
Successive layers alternate by 'handing' (perhaps its called reflection)

Each 3x3 grid is based on ball c/c of 1.6 and diagonals of 2.2427.....
The actual dimensions can all be written as proper fractions and surds

The layer interval involves a 3,4,5 triangle (more precisely a 0.6; 0.8; 1.0 triangle)
The 0.8 is half the 1.6 grid spacing and the 1.0 is the c/c distance of two touching balls

I will publish my solution in parts so that others can join in and demonstrate they also have recognised this particular solution.. Perhaps there are better solutions???

Meanwhile, congratulations to Matthew for what I think was a particularly neat teaser!!!

Cheers

Don

Don,

It sounds like you are there. It is a nice solution, I was somewhat surprised when I came upon it.

Lynn,

All the best for the 14th, hope you get better weather.

Time for some more teasers?

Matthew

288 balls

The diagram below shows the concept

Each grid (red and blue) is 3x3
The principal grid dimensions are
1.6 c/c (twice 4/5)
0.8 interlaced (4/5)
2.2427 diagonal (sqrt of twice*(sqr of twice 4/5))

Cheers

Don

Don,

That's the one.

Glad we eventually got this teaser solved.

cheers

Matthew

That's the one.

You don't get off that lightly.....you still have to watch my explanation unfold.....

288 balls

The diagram below shows

The size of the space encircled by 4 red balls is 1.2427
A blue ball fits into this space easily with a minimum clearance 0.1213
Hence the red grid and the blue grid can interlace in a common flat plane
ie a single layer

Cheers

Don

288 balls (the last bit)

The diagram below shows

Each set of four yellow balls forms a square of side 1.1213 (half the sqrt of twice*(sqr of twice 4/5)
The diagonal of each yellow square is 1.6 (primary dimension used to set up the grids)

A fifth ball can be set on top of 4 yellow balls to form a pyramid (centre of yellow ball base to centre of fifth ball) is 0.6
Note :
base diagonal is 1.6 hence half base diagonal is 0.8
slope height of pyramid is 1.0 (centre ball to centre ball)
hence by Pythagoras height is 0.6 (or observe 3/4/5 triangle)

Cheers

Don

Lynn,

Now, have you given any thought to that quadrilateral of real estate?

Not yet. But I have been busy both last week and over the weekend. Next week also looks busy until about Thursday.

This means plenty of time for others to get there before me.........

Cheers

Don

Now, have you given any thought to that quadrilateral of real estate?

OK, a few moments to myself today.

I will post details later but the general concept goes like this :-

Any iregular quadrilateral. Draw the diagonals.

Construct a parallelogram parallel to the diagonals, just touching the corners of the quad.
This //gram encloses the quad and has twice the area.

Now construct a second //gram, "concentric" with the first, passing through the mid-points of the quad's diagonals. The small //gram just touches the insides of the quad and, being one quarter the area of the large //gram, is also one half the area of the quadrilateral.

As I say, details to follow but comments welcome in the meantime.

Cheers

Don

"....quadrilateral of real estate...."

A REAL brain teaser....

Why, just WHY would a real estate agent ever want to know something like this.........

Cheers

Don

A TRIVIAL brain teaser, just to get the weekend off to a good start....

Two and half gold panners, pan two and half kilos of gold in two and a half hours. How much gold can we expect ten panners to pan in ten hours ?

Don't, please DON'T ask me where we find half a gold panner......

Cheers

Don