Monday, 26 April 2010

I finally worked it out

This puzzle has bugged me for years. Take a right angled triangle and split it up like this, call it triangle1

Now rearrange the pieces like this to give triangle2

The overall triangle is still the same size 13 by 5, none of the pieces has changed, so where does the hole come from?

Not convinced? Here’s an animation.

Draw it out yourself and cut it up with scissors. It works. It is true, or so it seems. Of course it's a subtle trick. It all looks convincing to the eye and, given the accuracy of drawing on graph paper and cutting with scissors, it works on paper.

The solution? The easy way to see it is to superimpose triangle1 on top of triangle2.

Hey! there's a bit left stuck out along the hypotenuse - that bit is the trick. Here's a blowup:

In truth, triangle1 is a smidgeon smaller than a real 13 by 8 triangle, triangle2 a gnat's hair bigger. That slender difference, surprisingly, adds up to 1 and that's the size of the hole.

Just as a last observation, look at the whole (integer) numbers needed to make these shapes.

The main triangle is 13 by 5
red is 8 by 3
blue is 5 by 2
yellow and green have sides of 1, 2, 3 and 5
the hole is 1 by 1 but really zero


Where have I seen that before? Of course, now that's interesting!
svg code for images adapted from here. See also Missing square puzzle.


Duncanr said...

You've 'lost' me on this one, I'm ashamed to admit

Andy Holroyd said...

Teehee! I did say it had bugged me for ages. I thought drawing it might be clearer than showing the math but look at it this way:

By area

yellow + green = 15

red = 8/2*3 = 12

blue = 5/2*2 = 5

total = 32 for triangle1 or 33 for triangle2 with the hole.

the overall triangle is 13 by 5 = 13/2*5 = 32.5

so triangle1 is half a unit smaller than 13 by 5 and triangle2 half a unit bigger. The difference = 1, the size of the hole.

By angles

Look at the sharp end of the three triangles (the smallest angle). For 13 by 5 it's inv(tan(5/13)) = 21.04°

red = inv(tan(3/8)) = 20.56° - too small

blue = inv(tan(2/5)) = 21.80° - too large

so in triangle1 the hypotenuse kinks inward where red and blue meet and in triangle2 it kinks outward by the same amount. That outlines a slim parallelogram with sides equal to the lengths of the hypotenuses of red and blue. This is the overlap I showed in the post.

By Pythagoras' theorem

hypotenuse red = sqrt(8^2 + 3^2) = 8.54

hypotenuse blue = sqrt(5^2 + 2^2) = 5.39

smallest angle = 21.80° - 20.56° = 1.24°

Area of parallelogram = tan(1.24°) * 5.39 * 8.54 = 0.99 which is 1 if you work to enough decimal places.

The differences are small enough so you don't notice, but large enough to add up to 1, the hole.