Here is a interesting picture with two arrangements of four shapes.


How can they make a different area with the same shapes?

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    $\begingroup$ i.imgur.com/nA53dlx.gif $\endgroup$
    – BCLC
    Commented Dec 27, 2015 at 14:33
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    $\begingroup$ Similar: ocf.berkeley.edu/~wwu/cgi-bin/yabb/… $\endgroup$
    – jamesdlin
    Commented Dec 28, 2015 at 22:38
  • $\begingroup$ Anybody have a link to that "infinite chocolate video"? $\endgroup$ Commented Dec 28, 2015 at 22:45
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    $\begingroup$ @HagenvonEitzen: Here it is: youtube.com/watch?v=dmBsPgPu0Wc $\endgroup$
    – Atul
    Commented Dec 30, 2015 at 6:32
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    $\begingroup$ Straight away, I can see the triangles in the second image aren't really triangles, as the hypotenuse isn't perfectly straight. $\endgroup$
    – Bradman175
    Commented Sep 8, 2016 at 2:26

8 Answers 8


This is a famous physical puzzle that can be tied to the fibonacci series.

To answer the question as posed, the issue is that the two slopes are different ($\frac25$ vs $\frac38$). Note that all those numbers are in the fibonacci series ($1,1,2,3,5,8,13,21,\ldots$).

Successive fractions are closer approximations to $\varphi$, alternating between above and below. Diagrams like this can be generated by making a square with sides equal to a number in the fibonacci series (in this question 8), then dividing it into two rectangles with widths of the two fibonacci numbers that make up the first one chosen (3 and 5).

Cut the smaller one down the diagonal, and cut the bigger one down the middle at a diagonal, such that the width of the diagonal cut is the next smallest number (2 in this case). Note that this will leave a trapezoid, whose small parallel size matches the original small rectangle's smaller side (3 in this case), and whose larger parallel size matches the original larger rectangle's smaller side (5 in this case).

Since $\frac25\approx\frac38$, and from the above constructions, the pieces can be rearranged into a rectangle (as shown), the area of which will always be one away from the original square, but will look approximately correct, since the slopes almost match.

Edit: Since this answer received so many up-votes (thank you!), I suppose people are very interested in it, so I thought I'd draw up a few images!

1,1,2,3: $3\times3 = 9 = 10 = 2\times5$


1,2,3,5: $5\times5 = 25 = 24 = 3\times8$


2,3,5,8: $8\times8 = 64 = 65 = 5\times13$ (The OP's example)


3,5,8,13: $13\times13 = 169 = 168 = 8\times21$


5,8,13,21: $21\times21 = 441 = 442 = 13\times34$


A comment from @EricJ. prompted a discussion that may be worth bringing up here:

I'm not claiming that all such puzzles are based on the Fibonacci series. Just that all Fibonacci numbers can generate these diagrams. There are several characteristics of Fibonacci numbers that make this work.

  1. One is that the square of a Fibonacci number alternates between being one more and one less than the product of the numbers on either side.
  2. There's the slope thing I mentioned already, meaning that our construction will give us two slopes that are approximately equal. And
  3. There's an argument that the overall construction can be done based on each number being the sum of the previous two.

The latter two points can best be understood by studying the construction in detail. The first point can be proved by induction:

RTP: $f_k^2=f_{k-1}f_{k+1}\pm1$ for $k\geq1$

We're indexing so that $f_0=0$ and $f_1=1$.

Step 1: $k=1$: $f_1^2=f_0f_2+1$ can be verified by substitution.

Step 2: Assume it's true for $k$. So $f_k^2 = f_{k-1}f_{k+1}\mp1$. (I'm using $\mp1$ because I'm expecting it to alternate, and so in step 3 I'll prove it for $\pm1$)

Step 3: We need to show that $f_{k+1}^2 = f_{k}f_{k+2}\pm1$. So here goes:

\begin{eqnarray} f_{k}f_{k+2}\pm1 & = & f_{k}(f_{k}+f_{k+1})\pm1\\ & = & f_{k}^2+f_kf_{k+1}\pm1\\ & = & (f_{k-1}f_{k+1}\mp1)+f_kf_{k+1}\pm1\\ & = & f_{k+1}(f_{k-1}+f_{k})\\ & = & f_{k+1}(f_{k+1})\\ & = & f_{k+1}^2 \end{eqnarray}

I used the definition of the Fibonacci series twice ($f_{k+2} = f_k+f_{k+1}$ and $f_{k-1}+f_k=f_{k+1}$) and the assumption of step 2.

This means that when you do the above construction the areas will always differ by just 1 (alternating above and below each time).

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    $\begingroup$ Excellent answer! I'm familiar with these puzzles but never heard of the Fibonacci connection. In fact, I didn't even realize there was an algorithm for generating such shapes. $\endgroup$ Commented Dec 25, 2015 at 16:16
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    $\begingroup$ This is why I never trust "proof by demonstration" pictures on math.stackexchange.com. $\endgroup$
    – jpmc26
    Commented Dec 27, 2015 at 7:46
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    $\begingroup$ After seeing this question I realized why my Faster-Than-Light Time-Travelling Ferrari 488 wasn't quite working (I mean, it'd go backwards in time faster-than-light Just Fine - but it kept coming back as a Ford Fiesta!) and I'd JUST gotten done fixing it up and launched it on a test run when I read this stupid answer! And THEN the Ferrari came back - but this time it came back as a bicycle WITH A DINOSAUR RIDING IT! So, OK, I whacked the dino (his name's Fred, BTW - nice guy. Funny ol' thing, life...) with a frying pan, and now I'm re-fixing the FTL engine. So t'anks fer nuttin'!! :-) $\endgroup$ Commented Dec 29, 2015 at 12:22
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    $\begingroup$ I think he's just saying that he was doing something impossible and now can't anymore because it's been proved to be impossible. What I don't get is why so many up votes? I mean I thought I was being clever, but not that clever! $\endgroup$
    – Dr Xorile
    Commented Jan 4, 2016 at 7:43
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    $\begingroup$ Oh, yeah? WELL! I just upvoted your answer, just to make it an even 100! SO THERE!!!! :-) (And as an aside to @ghosts_in_the_code - I wasn't really saying that this answer is stupid - I referred to it as "stupid" in an ironic sense that means "The answer has proved me wrong! What a stupid answer!" - i.e. I'm the real dummy here. Hopefully we're all good now...). My comment is actually an homage to three movies: Back To The Future, ET, and Caddyshack. And the Hitchiker's Guide To The Galaxy trilogy - all five books. :-) $\endgroup$ Commented Jan 4, 2016 at 16:25

The diagram is misleading, as it hides a gap in the middle of the second configuration.

This is what we actually get if we rearrange the shapes in question. Notice that the diagonal “bows” slightly, leaving some extra space between the shapes – this is where the extra unit of area creeps in.

enter image description here enter image description here

But you shouldn’t trust me any more than the person who drew the original picture!

As we see here, pictures can be misleading – so my diagram isn’t proof that the original diagram was wrong. This just gives an intuitive sense of where the extra space has come from.

For a proper proof, consider the gradients:

  • The gradient of the blue trapezium is $5/2 = 2.5$
  • The gradient of the red triangle is $8/3 = 2.666...$

Since the gradients don’t match, we can’t arrange them side-by-side like this without some blank space between them. But because they’re close, the eye can be tricked into thinking they form a single continuous line, and doesn’t notice the slope on the triangle changing midway down.

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    $\begingroup$ I love these puzzles - there's a good philosophical moral of not accepting things as they're presented to you. $\endgroup$
    – dwjohnston
    Commented Jan 6, 2016 at 4:51

The image on the right cheats: the pieces don't actually fit together perfectly, there's a gap in between. To prove it, we can calculate the size of the gap, by calculating the size of a triangle, formed by:

  • the longest side of the yellow triangle: $a = \sqrt{3^2+8^2}$
  • the inclined side of the trapezoid: $b = \sqrt{2^2+5^2}$
  • the diagonal of the rectangle on the right: $c = \sqrt{5^2+13^2}$

The area of this triangle can be calculated using Heron's formula:

$$ A = \sqrt{s(s-a)(s-b)(s-c)} $$


$$ s = \frac{1}{2}(a+b+c) $$

Substituting the values into the formula gives exactly 0.5 for $A$. There are two such triangles, so that's a total 1 = the expected discrepancy.


It's a misleading diagram. In reality, the angles do not match up- the larger interior angle of the orange triangle is about 69.5 degrees, whereas it's 68.2 for the grey quadrilateral. (Correct me if I'm wrong- dusting off my trig here.) In the diagram with area 65, the orange areas are actually quadrilaterals. If you look closely, you can see that they have a slight inflection where they meet the other orange section. So that extra area comes from expanding them just a bit.


The triangles don't have the same slope; you can see that the large diagonal line through the "larger" rectangle bends. It's covered up by the thick lines around the triangles, but there is a very thin hole that has a total area of one square - the same square that supposedly "appeared out of nowhere".


Just enlarge the image and you'll see the answer.

enter image description here


Simple answer:

Those shapes (in orange) at the right side of picture, are not triangles at all! they are two quadrilaterals. an thus they have area greater than visually expected. so there is no equity here. They are different and thus have different total area.


The picture of the bottom rectangle is misleading, because it fools people into incorrectly assuming the width of the triangles to be exactly 3 units.

The real width can be easily calculated - it's a fraction of the total width, defined by the height of the point on the diagonal, or at exactly 8/13th of 5, ie. 3.076923077 (and not 3), q.e.d.

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    $\begingroup$ According to the puzzle statement, the triangles in both diagrams are identical, and the shape of the triangles is defined in the 8x8 configuration to be exactly three units by eight units. The error is well illustrated in alexwlchan's answer, and mathematically explained in several others: The pieces don't actually fit together in the 5x13 configuration. There's a thin trapezoidal gap between them, which is hidden by the thick, black, and not-quite-straight diagonal line drawn in the original 5x13 illustration. $\endgroup$ Commented Dec 30, 2015 at 16:40

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