Here is the puzzle. As ever, please do **NOT** post your answers, but do say if you think you have solved the puzzle and how long it took. Solution on Monday.

You are all alone on a desert island and want to measure exactly 3 inches (I have no idea why, but that isn’t important right now). Unfortunately, you have only an 8.5 x 11 inch sheet of paper. How can you use it to measure exactly 3 inches?

If you have not tried to solve the puzzle, have a go now. For everyone else the answer is after the break.

“Dog-ear” the page by folding one corner across to the opposite side. The area below the dog-eared portion will measure 8.5 x 2.5. Reopen the paper and fold this 8.5 x 2.5 portion up into the page. The area that remains uncovered measures 8.5 x 6; it can be folded in half to produce a height of 3 inches.

Did you solve it? Any other solutions?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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I had a slightly simpler one: Fold the page in half so it is an 8.5×5.5 rectangle. Then dog-ear that rectangle, folding the corner across diagonally to leave a three-inch strip at the end.

Same here.

Yep, me too!

Me too!

Me four.

ditto

Moi aussi.

Me 27. Seems like the obvious solution.

Ditto.

Simpler and cleaner.

-1 Richard.

Ditto, but when i saw the Book “Rip it up” on the right, i thought: Rip the paper into two pieces then its easier to handle. Nowbody sayd that the paper have to stay entire.

I would be interested to read one of the Pythagorean solutions as I spent far more time thinking about that than actually getting a linear solution.

Would someone post one please?

Thanks

TMT

“Dog ear” the page as in step 1 of Wiseman’s solution. You will create a right triangle with sides 8.5″ and hypoteneuse ~12.02″. Fold the hypoteneuse in half to get ~6.01″ and in half again to get ~3.005″. There are better “ideal” solutions but this is good enough for any practical purpose, as human imprecision in making the folds will easily overwhelm the tiny difference

Simples

By Pythagoras, or simple measurement, the diagonal is 14″.

So to get 3″ simply measure using the diagonal (14″) less the long side (11″).

Thanks

So that solution is quite close but not exact

TMT

The Pythagorus solution is incorrect, as the diagonal measures 13.901439 inches not 14 inches so you could not obtain 3 inches this way.

I took a piece of A4. Folded it along it’s diagonal and realised that one of the bits sticking out was about 3 inches. That was close enough for me.

A question for anyone.

Why do Americans not use A4 as standard?

Also – is A4 readily available in the states- I know Staples stock it? Thanks

I don’t even know what size A4 is – I’ve never seen it.

Qnyway, Richard stated “standard” 8.5X11

I made a point, like when you’re making a paper plane, so the point is on the 11 inch side. That leaves 3 * 11 uncovered…

I fold the paper so I get an 5.5 * 8.5 rectangle. Then strip along one of wings so we have a measure of 8.5 long. Then with the two parts only need to have a substraction.

Exactly 3

This was my solution as well. Much simpler than Richards.

That was my solution as well.

I was mentally folding the papers, but I gave up before solving it. It’s a good one.

As an alternative, just measure out eight lengths of the long side marking the end and the start with lines in the sand (of which a desert Island is presumably full – beaches should be suitably flat) making a distance of 88 inches. Now the same from the same start point with 10 lengths of the short side from the same starting point (85 inches) and the difference between the two end marks will three inches. There is no requirement in the problem that the length has to be “portable”, but if it is just put a fold in the paper at the appropriate point.

Variations of this scheme can be used to measure any distance in multiples of half an inch subject only to a long enough beach.

The practical will point out this method accumulates measurement errors, but we have to remember all these problems inhabit a strange, abstract world of perfect folds, infinitely thin marks and paper of zero thickness (which the approved answer requires).

My solution was essentially the same as Richard’s but arrived at in a different way:

The paper is a rectangle comprised of points A, B, C, D creating four edges where edges A->B and C->D are 8.5″ and edges A->C and B->D are 11″

Fold the paper so that A meets C and B meets D creating two new points (E & F) & a new edge E->F (also 8″)

~Two letters in brackets will indicate a common point~

(A/C)->E and (B/D)->F are now 5.5″ (half of 11″)

Fold F diagonally to meet new point G which is 5.5″ from E and 3″ (A/C)

The distance from (A/C)->(F/G) is 3″ (8.5″ – 5.5″)

QED

Oops, typo. Correction here:

Fold the paper so that A meets C and B meets D creating two new points (E & F) & a new edge E->F (also 8.5″)

You can also fold 11 inches in to 11. Than you can measure any number from an inch to 11.

Or even after “dog ear ” folding, the remaining 2.5 inches can be folded again to 1 inches each. Than you can easily measure the required half inch from the rest of the paper and you’ll get 3 inches lenght.

It’s ‘then’ not ‘than’. Just letting you know. Don’t shoot the meassenger! 🙂

Measure twice the shorter side. You will get 17 inches. Substract the longer one, you will get 6. Half that. You get 3.

For fans of Mr Pythagoras who also want their answers to be exact, here’s an absurdly complicated alternative.

Dog-ear the corner as Richard says, to leave a 2.5″ strip at the bottom. Unfold.

Fold the 2.5″ strip up, then up again, and again, and again, until only a one-inch strip is left at the top of the page. Unfold everything again.

Fold the one-inch strip down twice.

You now have a piece of paper nine inches high. Near the middle of it is a horizontal crease, which is five inches from the bottom and four inches from the top.

Pivoting at the right-hand end of this crease, fold the bottom-right corner diagonally up to meet the top edge. The point where it meets is 3″ from the top-right corner, because we have made a right-angled 3-4-5 triangle!

I tore the paper in half to end up with two 5.5″ x 8.5″ pieces. Cover one piece with the other at a right angle and you end up with two 3″ pieces sticking out.

This is the quickest answer i guess. You dont have to tore the paper though.

I did something close to Richards solution:

1) “Dog ear” it into 8.5 x 8.5 triangle+ 8.5×2.5 rectangle

2) “Dog ear” the rectangle to a 2.5×2.5 triangle and a 2.5 x6 rectangle

3) fold the rectangle in half to get a rectangle of 2.5 x 3

The other sollution is better…

1) Fold it into a 5.5 x 8.5 rectangle

2) Dog ear it into a 5.5 x 5.5 triangle and leave a 5.5 x 3 rectangle

Cut the paper in half along a diagonal. You now have two triangular pieces of paper measuring 8.5 by 11 inches. Take one and fold the 11 inch edge back on itself so it is now half as long at 5.5 inches. Put it alongside the 8.5 inch edge of the other piece of paper. The difference in their lengths is 3 inches.

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