enter image description here I have tried to figure out by calculating the area but I got same results for these, so where is gone the hole?

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  • i'm sure this has been asked before, but anyway: note that the dark green and red triangles have different slopes – citedcorpse Jun 17 '13 at 17:52
  • https://en.wikipedia.org/wiki/Missing_square_puzzle – iostream007 Jun 17 '13 at 17:55
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    An oldie but a goodie -- this goes back at least to the days of Sam Loyd [ http://en.wikipedia.org/wiki/Sam_Loyd ] (who Martin Gardner did much to rekindle interest in). I have always liked this as an example of why you should not _only_ resort to a picture when working out geometry. There is a "swindle" taking place here that is difficult to spot "by eye". (If you superimposed the two triangles, you would just be able to see the cheat.) – colormegone Jun 17 '13 at 18:15
  • @RecklessReckoner Nice link! – Shuhao Cao Jun 17 '13 at 18:27

2 Answers2


When calculating the total area of those four shapes in square units, placement doesn't matter.

The perceived total area of the first triangle is $13*5/2 = 32.5$.

The perceived total area of the second triangle is one less, $31.5$.

The red triangle has area $3*8/2 = 12$. The green triangle has area $5*2/2 = 5$. The yellow polyomino has area $7$ and the green polyomino has area $8$. Thus the total area in both cases is $32$.

As was noted in the comments, the apparent discrepancy is caused by the two triangles not having the same slope, causing the diagonal to bend in or out slightly, while appearing at first glance to be relatively straight.

  • Note, incidentally, that the legs of the green and red triangles and the large "assemblage" triangle all have lengths which are Fibonacci numbers. I don't know if this is _essential_ to the construction, but other larger (and so still subtler) versions can be made in that way. – colormegone Jun 17 '13 at 18:26
  • @RecklessReckoner: using Fibonacci numbers is a nice way to get slopes that almost match because the ratio of successive ones is very close to $\phi$, alternately above and below. – Ross Millikan Jun 17 '13 at 18:29
  • The triangles appear to all use $ \ F_{n} \ $ and $ \ F_{n+2} \ $ , which still maintains the "limiting slope" property you describe (though not to $ \ \phi \ $ ) , as well as guaranteeing the fit of the pieces (the rectangle is in fact a "golden rectangle"). [This is really closer than I've looked at this problem before. It seems like Lucas numbers would also work here...] – colormegone Jun 17 '13 at 18:37

This is a tikz code snippet for the drawing in $\LaTeX$ for everyone to grab and try it yourself:

\tikzset{polygon/.style = {thick, line join=bevel, fill opacity=0.6}}
{\draw [polygon, fill=red] #1 -- ++(8,0) -- ++(0,3) -- cycle;}
{\draw [polygon, fill=teal] #1 -- ++(5,0) -- ++(0,2) -- cycle;}
{\draw [polygon, fill=yellow]#1 -- ++(2,0) -- ++(0,1) 
-- ++(3,0) -- ++(0,1) -- ++(-5,0) -- cycle;}
{\draw [polygon, fill=green]
#1 -- ++(5,0) -- ++(0,2) -- ++(-3,0)
-- ++(0,-1) -- ++(-2,0) -- cycle;}

\draw [gray](0,0) grid(13,11);





Notice in the \newcommand part for the \redshape and \tealshape, the redshape drawing sequence starts with the left bottom corner, goes right for 8 units then goes up for 3 units (-- ++(8,0) -- ++(0,3) part), while the tealshape starts from the bottom left corner, goes right for 5 units and then goes up for 2 units(- ++(5,0) -- ++(0,2) part).

Once we defined all the shapes in the \newcommand part, we don't modify them, and just rearrange them in the \tikzpicture part.

These two shapes have different slopes, hence the big combined shape is not a triangle, you can't use the triangle formula for the big shape, and say, hey, there is a square missing! Instead, you have to add the area of each single shape: red is $8\times 3/2$, teal is $5\times 2/2$, yellow occupies 7 grids, and green occupies 8 grids, and you will find that the area is conserved.

Shuhao Cao
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