[bonanova]

I just bought a circular glass top for my coffee table and I wanted to measure its precise size.

I grabbed my measuring tape; but alas, with use all its numbers had rubbed off, and it could not be read.

I also have a calibrated square whose side is finely adjustable from arbitrarily small up to 1 foot.

With the square, I determined the diameter was greater - by several multiples - than 2^1/2 feet: the square's largest diagonal length.

I thought for a moment then, on a hunch, I speed dialed Morty's and described my problem to Alex.

Well, he said, do ya still have the box the thing came in?

Yes, I replied, and that would give me its diameter; but if I could measure the box, I could certainly ...

Ahem! I heard over the phone, as if I was not supposed to speak while Alex was thinking.

What I was going to say was ...

Following Alex's advice, I made a single measurement with my square ruler and found the radius was exactly 3 feet!

What was the reading on my [square] ruler?

Edited December 3, 2008 by bonanova
clarifying that I used my square ruler

[Guest]

Therefore, AB = 1.2426

That is the hypotenuse of a triangle formed with 2 sides of length BZ, which we'll call L

L**2 + L**2 = 1.2426

L**2 = 0.6213

L = 0.78822

If hypotenuse is 1.242 and one side is L then

L²+L²=1.242² not 1.242

so the result should be 0.878, not 0.788

but still I'm not sure which is read on the scale of a square ruler.

If it is diagonal, the result is 3(2^1/2-1)

If it is side length, the result is 3(1-2^-1/2)

[Guest]

You make an interesting case, but (I think) are lucky in your analysis that your mistake at the start (in red) is then fixed in your next step!

Let's call the original Box A, your "1/2" box B, your "1/4" box C and your "1/8" box in the corner D

A has sides of length x

B therefore has sides = sqrt( 2 . (x/2)² ) = x/√2. It has an area which is half of the original, but not half the sides

C has sides = sqrt( 2. (x/2√2)² ) = x/2. It has an area which is 1/4 of A and half the sides of the original A.

D takes up the space in the corners of C in A (with C exactly in the centre) and has sides = x/4 and an area which is 1/16 of A.

Actually, you were correct up to this point, but the diagonal of box D will be x/2√2 (or 3/√2 feet, given x = 6ft).

Either way, your biggest problem would be that the calibrated ruler/square doesn't reach across the box, so there's no way to draw your lines accurately in the box, I'm afraid.

The diagonal of the calibrated square is (diagonal of box A - radius of table) = 3√2 - 3 = 1.2426 feet and the sides are 3.(1 - 1/√2) = 0.8787 feet.

(edit: re-worded categorical denial) First, it wasn't luck exactly that kept me in the ballpark, that's just the way my brain is wired. It seems to always compensate for my ADD issues

Secondly, the OP said there was a tape measure with no numbers. That would be accurate enough for the purpose of drawing lines if kept tight, as I mentioned.

Lastly, with a bit of adjustment, my solution works well.

To pick up where you left off, at this point you would again connect the corners to define a point in the center of square "d", then use your square to measure perpendicularly from the side of the square to that point. THAT measurement should be exactly 3/4 of one foot. And THAT would be 1/8 of your total diameter. 3/4ft x 8 = 6ft diameter. Divide by 2 and there ya go.

That was the measurement I was trying (but failed miserably, I've given myself a good talking to about the difference between comparing area and dimensions! ie square B vs. square C ) to describe. It was easy to see my mistake after coming back and reading it again.

Thanks for straightening me out.

Edited December 4, 2008 by Grayven

[Guest]

Assume that you don't have any square scale. But your neighbour is a carpenter, and he owns one. How can you solve the OP's problem by using it, but not carrying the box and tables top to his office?

[Guest]

If hypotenuse is 1.242 and one side is L then

L²+L²=1.242² not 1.242

so the result should be 0.878, not 0.788

but still I'm not sure which is read on the scale of a square ruler.

If it is diagonal, the result is 3(2^1/2-1)

If it is side length, the result is 3(1-2^-1/2)

1.242 is 2 * hypotenuse squared

0.6213 is the hyp squared

0.788 is the hypotenuse

[Guest]

You can't really see the true dimension from the diagram. It is just a sketch I draw in a hurry, not a fully scaled drawing.

If you really figure out that the diagonal of the square ruler is co-linear with the radius, then you will sure agree CD = BC.

No, you are wrong, CD does not equal BC. I wasn't saying that the diagram proved it, rather that is was so obvious that you

could tell from the diagram. I think it is easier to see visually from the updated version, where BM is shorter than MN.

Think about a circle, centered on C with radius BC. If you trace the circle by rotating B around the center, C, you will get a

much smaller circle. For CD to be on that circle, the path that B takes around C would have follow the circumference of

the larger circle. This is just not the case!

Edited December 4, 2008 by xucam

[Guest]

1.242 is 2 * hypotenuse squared

0.6213 is the hyp squared

0.788 is the hypotenuse

nobody was pointing out that 1.242 squared is 2 * hypotenuse squared. And that is where your error is.

And I'm intrigued by the neighbouring carpenter. nobody: Do you happen to have a bag full of those oh-so-useful polystyrene packing balls to measure an area with?

[Guest]

If hypotenuse is 1.242 and one side is L then

L²+L²=1.242² not 1.242

so the result should be 0.878, not 0.788

but still I'm not sure which is read on the scale of a square ruler.

If it is diagonal, the result is 3(2^1/2-1)

If it is side length, the result is 3(1-2^-1/2)

OK, I [belatedly] see your point. I made a calculation error, and I believe that the answer is 0.878. This would be read on the square ruler

If it is side length, the result is 3(1-2^-1/2)

This will yield a negative number.

Edited December 4, 2008 by xucam

[Guest]

This will yield a negative number.

3.(1 - 2^-1/2) is shorthand for 3.(1- ¹/_√2 ) and, as √2 is larger than 1, ¹/_√2 is less than 1 and the value given is positive (and equals your 0.8787 )

to get there, we take the hypontenuse with the surds in and rearrange to find the side length, x:

2 . x² = [ 3.(√2 - 1) ]² = 1.2426²

x² = 3²/2.(√2 - 1)²

x = 3/√2.(√2-1)

x = 3(1 - ¹/_√2)

QED. But that doesn't mean you were wrong - your initial calculations to get to 1.2426 in your first post were always correct: it's just your calulation to get the side of the rule that was wrong.

Edited December 4, 2008 by foolonthehill

[Guest]

3.(1 - 2^-1/2) is shorthand for 3.(1- ¹/_√2 )

Nevermind.

[bonanova]

I had fun creating my square ruler and you guys had fun figuring out how the heck it worked.

Credit to woon to point out that it was the side, not the diagonal, of the ruler that was asked for.

nobody, FOTH and PT in that order, eventually got it.

Nice work.

woon and xucam got close, by methods I may have misunderstood ... apologies if they got it right.

But grayven's post gets honorable mention for most complicated approach and the bonanova Humor award.

[Guest]

But grayven's post gets honorable mention for most complicated approach and the bonanova Humor award.

:bow:

Thanks

[Guest]

okay, ive got it

you read the outside of the box

it probably says on it

When at first you don't succeed, look in the trash for the instructions.

Nice puzzle, Bonanova. I was following as people were trying to solve. Curse my meager math skills.

[Guest]

No, you are wrong, CD does not equal BC. I wasn't saying that the diagram proved it, rather that is was so obvious that you

could tell from the diagram. I think it is easier to see visually from the updated version, where BM is shorter than MN.

Think about a circle, centered on C with radius BC. If you trace the circle by rotating B around the center, C, you will get a

much smaller circle. For CD to be on that circle, the path that B takes around C would have follow the circumference of

the larger circle. This is just not the case!

xucam, I now sincerely appologize to you. I know where am I going wrong already.

Since the curve does not meet the square ruler edge perpendicularly, it cannot be the radius, so they are not equal.

Once again, thanks you for bring up this to make me awake.

[Guest]

I made a single measuremnet

1

[Guest]

And I'm intrigued by the neighbouring carpenter. nobody: Do you happen to have a bag full of those oh-so-useful polystyrene packing balls to measure an area with?

Really, Grayven's solution is imaginative. But it is in fact the solution of a problem as how do answer to bonanova, when you have a small straight ruler, not a square ruler.? If a square ruler exists, does it make sense to make a cumbersome measurement?

I had make a modification as neighbouring carpenter , but with Grayven's method, we don't need a square ruler anymore, and my version becomes obsolete.

After putting glass table into the box, stretch the numberless tape diagonally on the box, mark the point on the tape where it intersects the circle. Then go to your neighbour and measure it by the square ruler.

[bonanova]

Grayven can use the square ruler as a straight ruler,

since It's calibrated according to the length of its side.

But its utility as a square comes into play by setting it in the box

corner, and expanding it until it touches the edge of the circle.

That ensures you are measuring along the box diagonal.

[Guest]

My solution did actually use the square as well. It wasn't clearly stated, I think, but you would need it to find the point on the box that would create a perpendicular line with the point that was measured. In fact, I was careful to include the use of the square, since Bononova went to the trouble to put it in there.

For that matter, I think my solution is what Bononova really was after. At least on a subconscious level, or why would he have mentioned the tape in the first place?? Yeah, if it wasn't supposed to be used, you just wouldn't have had one at all. So clever, you outwitted yourself, even! BRILLIANT!

Seriously though, I HAD visualized the solution bononova had in mind, but ruled it out because I wasn't confident I could show the math correctly. Alternately, I complicated the method to reduce the math to simple multiplication, essentially using the box folding method without any actual "folding, spindling, or mutilating" Hmmm, this all gives me an idea...