What is the essence of the new proof of Pythagoras' theorem by the highschoolers Johnson and Jackson?

Question

In an article by The Guardian a new proof of Pythagoras's theorem by New Orleans students Calcea Johnson and Ne’Kiya Jackson is announced. Currently there is no complete documentation, but in this video a possible line of arguments is reconstructed from the available drawings. Central to the proof is a right-angled triangle with angle $2\alpha$ constructed from the basic right-angled triangle with angle $\alpha$. Then similarity, infinite sums and the trigonometric identity for the sine of the double angle are used. But what is the essence of this new proof? And are the infinite sums really needed?

Answer 1

Instead of building the main triangle with angle $2\alpha$ from inside we build it from outside here.

The main ideas are:

• scale the mirrored triangle to make it fit into the original triangle (aligned at the right angles), which brings $a^2$ and $b^2$ into play and
• find a division with $2 \alpha$, whose sine brings $c^2$ into play

An infinite sum is not needed. Note that the proof only works for $a \ne b$; w.l.o.g. $a < b$.

In the scaled mirrored triangle $\triangle(BCD) \sim \triangle(ABC)$ (because $\angle(DCB) = \angle(ACB) = 90°$, $\angle(CBD) = \angle(BAC) = \alpha$) we have $$ \frac{CD}{BC} = \frac{BC}{AC},$$ so $$ CD = \frac{BC^2}{AC} = \frac{a^2}{b}.$$ Analogously $$ BD = c\frac{a}{b}.$$ From $DE$ being perpendicular on $BD$ we get $\angle(ADE)=\alpha$, giving us $$AF = \frac{AD}{2} = \frac{AC-CD}{2} = \frac{1}{2}\left(b - \frac{a^2}{b}\right) = \frac{b^2-a^2}{2b}.$$ From the similarity $\triangle(ABC) \sim \triangle(AEF)$ we see $$ \frac{AE}{AF} = \frac{AB}{AC},$$ leading to $$ AE = \frac{AE \cdot AB}{AC} = \frac{c(b^2-a^2)}{2b^2}.$$

The final equation is built from two ways of computing $\sin(2\alpha)$. For the first side we use the double angle identity (for its proof only similarity is needed, see e.g. here) and take the proportions from $\triangle(ABC)$, i.e. $$\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) = 2\frac{BC}{AB}\frac{AC}{AB} = 2\frac{ab}{c^2}.$$ For the second side we consider the proportions in $\triangle(BDE)$: $$\sin(2\alpha) = \frac{BD}{BE} = \frac{BD}{AB-AE} = \frac{c\frac{a}{b}}{c-\frac{c(b^2-a^2)}{2b^2}} = \frac{\frac{a}{b}}{\frac{2b^2-(b^2-a^2)}{2b^2}} = \frac{2ab}{a^2+b^2}. $$ Combining these two results leads to the expected relation.

Answer 2

There are other aspects of Ne’Kiya Jackson and Calcea Johnson's proof of Pythagoras' Theorem, assuming it's as shown in the video above, which I believe are more clever and interesting than its non-circular use of trigonometry. The trigonometry is a curiosity, for sure, but even more novel and intriguing connections appear through geometric and algebraic variations of their theorem, which I will present in a future answer. I still need to research how closely these variations approach any of proofs in The Pythagorean Proposition by Elisha Loomis.

I am starting here with a more straightforward and direct trigonometric proof of Pythagoras' theorem, which came to me and was inspired by the work of Jackson and Johnson. It does not require the law of sines, just the basic definitions of the sine and cosine ratios appearing in a right triangle.

Simplified Proof

Given a right triangle, $\triangle{ABC}$ with legs $\overline{AC}$, $\overline{BC}$ and hypotenuse $\overline{AB}$. Prove that $\sin^2(A) +\cos^2(A) = 1$, where $\sin(A) = \frac{BC}{AB}$ and $\cos(A) = \frac{AC}{AB}$.

Drop an altitude $\overline{CD}$ from point $C$ to point $D$ on segment $\overline{AB}$, and then another altitude $\overline{DE}$ (for $\triangle{BCD}$) from point $D$ to point $E$ on segment $\overline{BC}$. Continue constructing altitudes back and forth between the two lines in this manner, constructing similar triangles, ad infinitum.

Note the triangle similarities: $$\triangle{ABC} \sim \triangle{ACD} \sim \triangle{CDE} \sim \triangle{DEF} \sim \ldots $$

Now, using our trigonometry definitions, with these similar triangles, we have: $$\sin(A) = \frac{BC}{AB} = \frac{CD}{AC} = \frac{DE}{CD} = \frac{EF}{DE} = \ldots$$ and $$\cos(A) = \frac{AC}{AB} = \frac{AD}{AC} = \frac{DF}{DE} = \ldots$$

Reconstructing our segment $\overline{AB}$ as an infinite sum of lengths $AD, DF, FH,$ etc., we have, after rearranging the $\cos(A)$ expression from above, that $AD = (AC)\cos(A)$ and $DF = (DE)\cos(A)$, and so on for each additional segment along $\overline{AB}$.

We also have that $DE = (CD)\sin(A)$ and $CD = (AC)\sin(A)$. Putting this all together reveals that $AB$ is the sum of a geometric sequence, with a ratio of $\sin^2(A)$ between successive terms.

$$DF = (DE)\cos(A) = (CD)\sin(A)\cos(A) = (AC)\sin^2(A)\cos(A)$$

So with $AD = (AC)\cos(A)$, and $DF = (AC)\cos(A)\sin^2(A)$, etc., we have an infinite geometric series for $AB$: $$AB = (AC)\cos(A)(1 + \sin^2(A) + \sin^4(A) + \sin^6(A) + \ldots)$$

Since $|\sin^2(A)|<1$ for all right triangles, the series converges to $\frac{1}{1-r}$ where $r=\sin^2(A)$, and the above equation becomes:

$$AB = \frac{(AC)\cos(A)}{1-\sin^2(A)}$$ Then $$1 = \frac{\frac{AC}{AB}\cos(A)}{1-\sin^2(A)} = \frac{\cos^2(A)}{1-\sin^2(A)}$$

And finally: $$\cos^2(A) = 1-\sin^2(A)$$

$$\sin^2(A)+ \cos^2(A)= 1$$

Answer 3

Variations on the Jackson-Johnson Proof

These variations both explicitly rely on the construction of Jackson and Johnson in their trigonometric proof. The first demonstrates that the use of trigonometry is superfluous to the proof, and the very same construction they develop can be applied in a strictly geometric sense.

The second proof, at least at first glance, appears to be truly unique to their method of construction. Attempting this approach with the Simplified Proof that I presented in my previous answer will lead to a circular result, I believe. The construction of Jackson and Johnson splendidly removes this issue of circularity, and by running the construction in reverse, the isosceles right triangle case, for which the original proof will not work, has been resolved.

A Geometric Proof

The naming of ratios using trigonometry and the use of the law of sines are not absolutely necessary for the proof as presented. The novel tiling of the triangle with all similar triangles can be applied in a strictly geometric manner, as follows.

Following the Jackson and Johnson proof, here's the diagram:

And it has been shown previously that: $$X = \frac{2ac}{b} (1+\frac{a^2}{b^2}+\frac{a^4}{b^4}+\ldots) = \frac{2ac}{b(1-\frac{a^2}{b^2})}=2ab\frac{c}{b^2-a^2}$$ and $$Z = c+\frac{2ca^2}{b^2}(1+\frac{a^2}{b^2}+\frac{a^4}{b^4}+\ldots) = c+\frac{2ca^2}{b^2(1-\frac{a^2}{b^2})}=\frac{c(b^2-a^2)}{b^2-a^2}+\frac{2ca^2}{b^2-a^2}=(b^2+a^2)\frac{c}{b^2-a^2}$$

The area of the large triangle my be found in two different ways: $$\text{Area} = \frac{1}{2}cX = c^2\frac{ab}{b^2-a^2}$$ or (summing the areas of all the similar right triangles) $$\begin{equation}\begin{aligned} \text{Area} &= ab+\frac{2a^3}{b}(1+\frac{a^2}{b^2}+\frac{a^4}{b^4}+\ldots)=ab+\frac{2a^3}{b(1-\frac{a^2}{b^2})} \\ &=\frac{ab^3-a^3b}{b^2-a^2}+\frac{2a^3b}{b^2-a^2}=\frac{ab^3+a^3b}{b^2-a^2} \\ &=(a^2+b^2)\frac{ab}{b^2-a^2} \end{aligned}\end{equation}$$

Equating the two expressions for the triangle's area gives us the equation: $$c^2\frac{ab}{b^2-a^2}=(a^2+b^2)\frac{ab}{b^2-a^2}$$ which is equivalent to: $$a^2+b^2=c^2$$

This proof does not use trigonometry, and the equating of areas serves the same purpose that the law of sines did in the original.

However, this proof still suffers from a limitation of the original construction, that it will not produce a converging geometric series when the original right triangle is isosceles.

To get around that problem, the next proof is an algebraic reformulation.

An Almost-Circular Algebriac Proof

For this variation on the proof, we perform the construction in reverse. That is we will take as given a right triangle with side lengths $c$, $X$, and hypotenuse $Z$. Construct an isosceles triangle using a segment of length $c$ along the hypotenuse, which is then split into two congruent right triangles having legs of length $a$ and $b$.

Notice that $a$ and $b$ depend only on the length of $c$ and the interior angle at that top vertex. We will demonstrate algebraically that for any such $a$ and $b$, we have $c^2+X^2=Z^2$. This is where the Jackson and Johnson construction really shines, because it relates $c$ to $Z$ in a way that enables such an approach.

For convenience, here is the drawing again: :

And the expressions for $X$ and $Z$ in terms of $a$, $b$, and $c$: $$X = 2ab\frac{c}{b^2-a^2}$$ $$Z = (b^2+a^2)\frac{c}{b^2-a^2}$$

Using these, we have: $$\begin{equation}\begin{aligned} c^2+X^2 &= c^2 + (\frac{c}{b^2-a^2})^2(2ab)^2 \\ &= c^2 + 4a^2b^2(\frac{c}{b^2-a^2})^2 \\ &= (b^2-a^2)^2(\frac{c}{b^2-a^2})^2 + 4a^2b^2(\frac{c}{b^2-a^2})^2 \\ &= (b^4-2a^2b^2+a^4 + 4a^2b^2)(\frac{c}{b^2-a^2})^2 \\ &= (a^2+b^2)^2(\frac{c}{b^2-a^2})^2 \\ &= Z^2 \end{aligned}\end{equation}$$

What makes this proof most novel and unique how it uses geometric properties of right triangles and triangle similarity in a new way, with or without the trig.