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$$\frac{\pi}{4} = 2 \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{7} \;.$$ Is there some geometric construction that explains this beautiful equation (known as Hutton's formula)? Perhaps a "proof without words" figure that makes it self-evident?


Here is Figure 1a from the reference Henry provided:

Nelsen, Roger B. "Proof Without Words: The Formulas of Hutton and Strassnitzky." Mathematics Magazine 86 5 (2013): 350-350.
          Fig1a

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    $\begingroup$ Roger Nelsen gave a proof without words in Math Magazine 86 (2013), 350. See www.jstor.org/stable/10.4169/math.mag.86.5.350. $\endgroup$ – Henry Cohn Apr 12 '14 at 22:12
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    $\begingroup$ What a lovely proof! $\endgroup$ – Lucia Apr 13 '14 at 3:53
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All of these Machin-like formulas have proofs-without-words. First, notice that there is a one line proof using the Gaussian integers: $$(3+i)^2 (7+i) = 50 + 50 i.$$ Taking arguments of both sides proves the result (modulo $2 \pi$).

Now, plot the products $$3 \times 3 \times 7=63,\qquad (3+i) \times 3 \times 7=63+21 i,$$ $$(3+i)^2 \times 7 = 56+42i,\qquad (3+i)^2 (7+i) = 50+50 i.$$ Here the $3$'s and $7$'s are the real parts of $3+i$ and $7+i$.

Each consecutive pair of complex numbers forms a right triangle with the third vertex at $0$: $$(0,63, 63+21i),\ (0, 63+21i, 56+42i),\ (0, 56+42i, 50+50 i).$$ Draw each of those triangles and you have a proof without words.


         
          (Image added by J.O'Rourke)

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    $\begingroup$ Thank you, David, this is a beautifully clarifying viewpoint! $\endgroup$ – Joseph O'Rourke Apr 13 '14 at 16:19
  • $\begingroup$ You're welcome! Sorry for the missing scalar factors in the first version; added now. $\endgroup$ – David E Speyer Apr 13 '14 at 17:49
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A similar argument was found by pappus on the French forum les-mathematiques.net.

enter image description here

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