Can someone please solve this?

The slope of the two triangular sections is different, even though it may not show up visually in this diagram. If however you draw a straight line from the bottom left to th etop right of the composite shape then in the top diagram the intersection point of the two triangles will be below below this line (i.e. below th etrue hypotenuse). If you repeat the exercise on the lower composite then the intersection point is above the true hypotenuse. In effect the area of the empty space has been displaced into a "bulge" in the hypotenuse.

Willy.

Actually the hypotenuse angle of the green triangle is not the same as the hypotenuse of the red one, so there maybe something in that inaccuracy.

Although each overall picture "looks" like a right angled triangle, each overall picture is a quadrilateral. And they are not the same size.

The red triangle and the dark green triangle "look" as if they might be similar triangles but they ain't. The ratios of the two shorter sides are 2/5 and 3/8.

Nice optical illusion though. And nice colours

Cheers

Don

Clever.

I've just printed a copy of it so I can tease my brother with it later today.

You might have seem this already on the Forum, I can't remember.

The inner square just touches the circle.
The outer square is tangential to the circle.

What is the ratio of the area of the inner square to the outer square. With proof or reasoning.

Cheers

Don

I'm not going to reveal the answer, but you can just 'see' it if you look in the right way.

quote:

you can just 'see' it

I know what you mean

Cheers

Don

Looks like 1:2.

If you rotate the inner square its corners will lie on the halfway points of each of the larger square’s edges.
You are left with 4 corner areas of the bigger square. 2 of these add to a square of ¼ (1/2 length x 1/2 length) the area of the larger square – you have 4 triangles, so the area left not covered by the smaller square = ½ the larger.

Well – that looked quite good when I wrote it.

Adam -
Were you ever on 'Ask the Family' as a yoof? Bet you're a whizz in the pub quiz team!

James

P.S. I'm only jealous. Can't do these things for toffee. But I do know what a liripip is.

quote:

Well - that looked quite good when I wrote it.

I guess this is the sort of picture both you and Alexander had in mind ?

Nice words Adam...far better than my attempt when I saw the puzzle first time,,,and we all think we can eloquently describe in words, the sound of our various hifi systems.....

Cheers

Don

An old classic but still fiendish.

You have 12 coins (individually identifiable) and a beam balance. All the coins bar one weigh the same.

Making only 3 weighings, identify the odd coin and say whether it is heavier or lighter than the rest.

For bonus points – create a table which will, elegantly, allow you to read off the answer immediately you have been told the results of your three weighings by A.N. Other who selects a random coin to be heavy or light.

This is something I worked out with my father many years ago - can't say it was my idea to look to base 3 for the solution. We never bothered to make it really elegant but I've never seen the solution elsewhere.

He also taught me how to always win at Nim - but my first Google reveals this to be old hat.

Can anyone remember the answer to Duncan Fullerton's clasic teaser...

"how many 1 inch diameter (incompressible) spheres can be packed into a box measuring 5 inches x 5 inches x 10 inches."

All dimensions are precise (unlike my spelling).

Cheers

Don

quote:

An old classic but still fiendish.

Adam, I am ruling myself ineligable on this one, having set the same problem [in a slightly different guise] about 4 years ago on this ere forum...

...but the elegant table....Fraimed in Base 3...interesting

Cheers

Don

Did you get an answer?

It is quite difficult (apart from the necessary revelation of how to organise the weighings) to put the intuitive solution into words.

I'll donate (later) our solution for the good of mankind - and mainly because New Scientist didn't take up the offer.

quote:

Did you get an answer?

It was one of the first few puzzles in the "Brain Teaser" thread that ran to over 2,000 posts. IIRC Paul Ranson came up with the solution pretty quickly, even though the question didn't reveal that only three weighings were necessary and simply asked "what is the least number of weighings...". [several people could sort the problem in four weighings including Paul, who at first didn't believe it was possible in three!.]

I published a "chart" or a "process diagram" which showed how to do the weighings and which revealed all the possible outcomes. I still have it.

As you rightly say, putting it into words.........

Cheers

Don

I've just transferred an old sketch in a childhood notebook to excel.

Hope it makes some sense - I have to cook now.

A less common explanation for the problem with the squares would be

- cut the squares diagonally so they are just two rectangular triangles. This allows to reduce the problem to scaling triangles.

- the operation of taking two duplicates of a triangle and placing them side by side to create a larger triangle, is identical to scaling a triangle such that the length of the side becomes the original diagonal.

So the operation of scaling a triangle with a factor diagonal/side will double the area.

Here is my solution to the 'odd ball' teaser in words only (read "coin" for "ball") as abstracted from the "Brain Teaser" thread of about 5 years ago?


Number the balls 1 to 12 (simply for reference). Divide them into 3 Groups of 4

1st Weighing.
Compare Group 1,2,3,4 with Group 5,6,7,8

If Group 1,2,3,4 is heavier than Group 5,6,7,8 then IF the odd ball is in Group 1,2,3,4 it is heavy and IF the odd ball is in Group 5,6,7,8 it is light. Go to 2A
If Groups 1,2,3,4 and 5,6,7,8 do balance, the odd ball is in Group 9,10,11,12. Go to 2P

2nd Weighing
2A Compare Group 1,2,7,8 with Group 5,9,10,11
If Group 1,2,7,8 is heavy then either 1 or 2 is heavy or 5 is light. Go to 3A
If Group 1,2,7,8 is light then either 7 or 8 is light. Go to 3B
If Group 1,2,7,8 and Group 5,9,10,11 balance then 3,4 or 6 is the odd ball. Go to 3C

2P Compare Group 1,2,3 (in fact any 3 balls from 1 to 8) with Group 9,10,11.
If Group 9,10,11 is heavy then the odd ball is 9,10 or 11 and is heavy. Go to 3P
If Group 9,10,11 is light then the odd ball is 9,10 or 11 and is light. Go to 3Q
If Group 1,2,3 and Group 9,10,11 balance then the odd ball is 12. Go to 3R

3rd Weighing
3A Compare 1 and 2 and recall 5
If 1 is heavy then 1 = odd ball
If 2 is heavy then 2 = odd ball
If 1 and 2 balance then 5 is the odd ball and is light

3B Compare 7 and 8
If 7 is light then 7 = odd ball
If 8 is light then 8 = odd ball

3C Compare 2 and 4 and recall 6
If 3 is heavy then 3 = odd ball
If 4 is heavy then 4 = odd ball
If 3 and 4 balance then 6 is the odd ball and is light

3P Compare 9 and 10 and recall 11
If 9 is heavy then 9 = odd ball
If 10 is heavy then 10 = odd ball
If 9 and 10 balance then 11 is the odd ball and is heavy

3Q Compare 9 and 10 and recall 11
If 9 is light then 9 = odd ball
If 10 is light then 10 = odd ball
If 9 and 10 balance then 11 is the odd ball and is light

3R Compare 12 with any ball
If 12 is heavy it is the odd ball
If 12 is light it is the odd ball

Hope this makes sense

Meanwhile, Adam's spreadsheet above is elegant, very elegant. And the "Brain Teaser" thread never got anywhere near such an elegant solution. Where were you when you were needed Adam ?? And whatever happened to Bam and Omer?

Cheers

Don

Omer tends to confine himself to PFM these days, Don, together with his compatriot Arye. I don't know who Bam is.

Thanks John, good to know Omer and Arye are still around.

Bam was the poster of the famous "ladder" puzzle.

I arranged to meet him at one of the Bristol HiFi shows and although I think he lived near Bristol (Yate?) he got delayed and never made it. Disapperaed from the Forum shortly afterwards.

Cheers

Don

Just noticed we are on page two. So to avoid turning back, the question was " how many 1" dia spheres can be packed into a 5" x 5" x 10" box"

I supose we could start with "at least 250" couldn't we?

Cheers

Don

I think 'Bam' is still active on DIYAudio.com.

Paul

Thanks Paul

Cheers

Don