Answer to the Friday Puzzle!

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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.

34 comments on “Answer to the Friday Puzzle!

  1. Ed says:

    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.

  2. The Masked Twit says:

    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

    • Pete says:

      “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

  3. Martin says:

    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″).

    • The Masked Twit says:

      Thanks
      So that solution is quite close but not exact
      TMT

    • Toby says:

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

  4. Julie says:

    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

    • Bill36 says:

      I don’t even know what size A4 is – I’ve never seen it.
      Qnyway, Richard stated “standard” 8.5X11

  5. Gib says:

    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…

  6. 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

  7. Indy says:

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

  8. Steve Jones says:

    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).

  9. Todio says:

    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

  10. Todio says:

    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″)

  11. Anonymous says:

    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.

  12. Anonymous says:

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

  13. Ed says:

    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!

  14. Anonymous says:

    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.

  15. -M- says:

    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

  16. Anonymous says:

    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.

  17. Jackie says:

    Awfully enlightening many many thanks, It appears to be like like your current readers may well possibly want even more blog posts similar to this continue the fantastic effort.

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