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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 × 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) × 19 = 418 and a small one with area just over 2 × 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 in the above argument is ?

Thank You

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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
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. – Peter Tamaroff Apr 26 '12 at 19:11
1  
@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. – Peter Tamaroff 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
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1 Answer

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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.

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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
Thanks Brian :) – John Apr 26 '12 at 19:33
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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