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I came across a puzzle where we need to determine the error in the following proof.

False Theorem. $420 > 422$

Proof. We will demonstrate this fact geometrically. We begin with a $20 \times 21$ rectangle which has area $420$. Now we cut along the diagonal and slide the upper piece parallel to the cut until it has moved exactly $2$ units leftward. This leaves a couple stray corners which are $2$ units wide and just over $2$ units high. Finally, we snip off the two corners and place them together to form an additional small rectangle. Now we have two rectangles, a large one with area just over $(20 + 2) \times 19 = 418$ and a small one with area just over $2 \times 2 = 4$. Thus, the total area of the resulting figure is a bit over $418 + 4 = 422$. By conservation of area, $420$ is equal to just a little bit more than $422$.

Can someone please explain what the error is in the above argument ?

Thank You!

Harish Chandra Rajpoot
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John
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    If the numbers don't fit, the pieces probably don't fit as nicely as you think ;) – N. S. Apr 26 '12 at 19:09
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    The stray corners are under two inches, if my quick arts and crafts project is accurate... – The Chaz 2.0 Apr 26 '12 at 19:11
  • @John Have you tried cosntructing it? Which diagonal do you cut through? This is not really a proof but a vague visualization, and there is a catch in this diagonal cut. – Pedro Apr 26 '12 at 19:11
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    @John The trick is probably that the slope is $1,05$ which is very near to $1$. This probably tricks the eye when seeing the contruction. – Pedro Apr 26 '12 at 19:15
  • Even i am thinking there is an error in the statement "..which are 2 units wide and just over 2 units high". The length of the smaller rectangle would be less than 2 units is my intuition. Is that correct ? – John Apr 26 '12 at 19:17
  • See also the famous Fibonacci-based dissection paradoxes, e.g. see here. – Bill Dubuque Apr 26 '12 at 19:18

2 Answers2

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From your description, you’ve started with the rectangle oriented so that the $21$ unit side is vertical. You’ve slid the upper triangular half $2$ units to the left, so that it rises a little more than $2$ units, and you’ve snipped off the triangular projections at upper left and lower right. Your large rectangle loses two units of width, so its area is a bit more than $(21+2)\cdot 18=414$. The area of the small rectangle is a bit over $4$, so altogether you’ve a bit more than $418$, which is no contradiction.

In fact the small rectangle is $2\times 2.1$ units, so its area is $4.2$, and the large rectangle is $23.1\times 18$ units, so its area is $415.8$, and the total is still $420$, as it should be.

Brian M. Scott
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  • Thanks for the reply brian.But the 21 unit side is horizontal... – John Apr 26 '12 at 19:26
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    @John: Then the large rectangle is $19\times\left(20+\frac{40}{21}\right)$, with area $\frac{8740}{21}$, and the small rectangle is $2\times\frac{40}{21}$, with area $\frac{80}{21}$, and again all’s well. Your description is inaccurate, however, because the extra vertical bit is less than $2$, not more than $2$. I chose the orientation that I did precisely in order to match your description. – Brian M. Scott Apr 26 '12 at 19:32
  • I was wondering why my comment earlier didn't inspire the "ahah moment"... until I looked back and saw that I forgot to say which side is less than 2 inches! – The Chaz 2.0 Apr 26 '12 at 19:43
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Consider the statement

"if the numbers I give you are correct, area is not conserved."

We'll show that this statement is true via contradiction. Observe that this statement and a similar statement claiming that the area is conserved cannot both be true.

Let $L_1, W_1$ be the height and width of the original rectangle and let $L_2$ be the height of the triangles formed from sliding the upper half of the original rectangle $W_2$ units horizontally along its diagonal.

The negation of our consequent is that we "conserve area," that is

$L_1W_1 = (L_1 + L_2) * (W_1 - W_2) + L_2W_2$.

Expanding the RHS of the equation gives us

$L_1W_1 = L_1W_1 - L_1W_2 + L_2W_1$.

Then

$L_1W_2 = L_2W_1$, or $\frac{L_1}{W_1} = \frac{L_2}{W_2}$, i.e. the triangles formed by the diagonal of the original rectangle and the "stray corners" must be similar triangles.

If we plug in the numbers from our antecedent, however, we get

$\frac{20}{21} = \frac{L_2}{2}$, which is absurd since $L_2 > 2$.

hiroshin
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