Recently, two teenage girls — Calcea Johnson and Ne'Kiya Jackson — gave a presentation to the American Mathematical Society where they demonstrated a new proof of the Pythagorean Theorem. This shocks me greatly. You know, I'm the same old boy as them, yet I can only gaze up at their achievements.

Their proof uses trigonometry. This is another fact that shocks me. Because many of trigonometric identities and laws depend on the Pythagorean Theorem, and so a number of mathematicians have suggested that any proof of the theorem using trigonometry is *circular logic*.

However, this point of view has been increasingly questioned in recent decades, and a few trigonometric proofs of the Pythagorean Theorem have been developed and verified to be correct.

In this article, I'll explain how Johnson and Jackson proved the Pythagorean Theorem using simple trigonometry.

## What is the Pythagorean Theorem?

The Pythagorean Theorem is over 2500 years old, which was proposed by the ancient Greek mathematician Pythagoras. It says:

If a right triangle has base lengths $a, b$ and hypotenuse length $c$, then these values satisfy $a^2 + b^2 = c^2$.

During our middle school, we learned about the Pythagorean theorem and knew some of its integer solutions, such as $3^2 + 4^2 = 5^2$ and $5^2 + 12^2 = 13^2$ etc. These integer solutions are often known as *Pythagorean Triples*.

Trigonometry is based on right triangles. As shown in the following diagram.

we have $\sin \alpha = \frac{a}{c}$ and $\cos \alpha = \frac{b}{c}$. So $\sin^2 \alpha + \cos^2 \alpha = \frac{a^2 + b^2}{c^2} = 1$. Therefore the popular identity $\sin^2 \alpha + \cos^2 \alpha = 1$ is derived from the Pythagorean Theorem and any proof of the Pythagorean Theorem that uses this identity or any identity derived from it is indeed invalid due to circular logic.

## The traditional proof of the Pythagorean theorem

There are hundreds of methods to proof The Pythagorean theorem. According to the book "The Pythagorean Proposition" written by Elisha Scott Loomis, it mentions a total of 367 ways to prove it. Among them, the most intuitive and simplest is the graphical proof method.

**proof**: Draw a square with side length $a + b$. Inside this square, draw four copies of the right triangle, forming a square with side length c inside the larger square.

The area of the larger square can be computed in two ways: by applying the area formula to the larger square or by adding the area of the smaller square to the area of the four triangles. Thus, it must be true that

$$

(a + b)^2 = c^2 + 4 \cdot \frac{ab}{2}= c^2 + 2ab

$$

Expanding the expression on the left and canceling a common term on both sides yields

$$

a^2 +2ab+ b^2 = c^2 +2ab

$$

Therefore, $a^2 + b^2 = c^2$ is true.

$\blacksquare$

## The new proof

The core idea of this new proof can be summarized in the following diagram.

Let's start with the top left right triangle with sides $a,b$ and $c$ (hypotenuse), and follow the three geometric steps from this original right-angled triangle:

- We reflect in the side of length b, to form the top right equivalent triangle.
- We extend a line perpendicular to the side of length c in the original triangle.
- We extend a continuous line from the hypotenuse of the reflected triangle.

When our extended lines from steps 2 and 3 meet, we form a new larger right angled triangle of hypotenuse length $C$ and sides $A$ and c as pictured.

Within this larger right angled triangle, we draw a series of smaller and smaller, similar right angled triangles as pictured, forming an infinite sequence of similar triangles of decreasing size.

Now we can investigate how to derive the lengths $A$ and $C$ using this infinite sequence of similar triangles.

### Deriving lengths of the sides of the smaller triangles

Firstly, let's derive the side lengths of the light blue triangle in the following diagram.

Obviously, one of its right-angle sides is $2a$ and it's similar to the original triangle in the top left corner. Let the length of the other right-angle side be $x$ and the length of the hypotenuse be $y$, so we have

$$

\frac{x}{2a} = \frac{a}{b} \qquad \frac{y}{2a} = \frac{c}{b}

$$

Thus, we derive:

$$

x = \frac{2a^2}{b} \qquad y = \frac{2ac}{b}

$$

Next, let's investigate the pink triangle next to the light blue one as pictured.

We know that the pink one is sililar to the light blue one and one of its sides is length $\frac{2a^2}{b}$. Therefore the hypotenuse of the pink triangle (a segment of side length C) is $ \frac{2a^2c}{b^2}$ and the other side is $\frac{2a^3}{b^2}$.

This process can be continued, but it becomes apparent that each of the smaller similar triangles decrease by a factor of $\frac{a^2}{b^2}$. This means that length of $A$ is a geometric series with first term $\frac{2ac}{b}$ and the common ratio is $\frac{a^2}{b^2}$. Similarly, the length of $C$ starts with $c$ and is also a geometric series with first term $\frac{2a^2c}{b^2}$ and the common ratio is $\frac{a^2}{b^2}$ as well.

### Calculating $A$ and $C$

According to the formula for the sum of a geometric series,

$$

S_n =\begin{cases}

na_1 & \text{if } q=1\\

\frac{a_1(1-q^n)}{1-q} & \text{if } q \ne 1

\end{cases}

$$

we have to discuss two different cases respectively.

#### $q=1$, that is $a=b$

In this case, lines $A$ and $C$ are parallel and will never intersect. In fact this particular case makes proof very easy.

Consider the two congruent triangles at the top. These two isosceles right triangles form a larger isosceles right triangle, as shown in the diagram below. We have

$$

\frac{a}{c} = \frac{c}{2a}

$$

This gives $c^2 = 2a^2 = a^2+a^2$ as required.

#### $q \ne 1$, that is $a \ne b$

The case where $a \ne b $ can be uniformly considered as the case where $a < b$. If $a > b$, without loss of generality, we can swap $a$ and $b$ by rotating the triangle $90^\circ$ to transform it into $a < b$. Thus, $q = \frac{a^2}{b^2} < 1$. Substitute in the first term and the common ratio to obtain:

$$

\begin{aligned}

A &= \frac{\frac{2ac}{b}\Big(1-\big(\frac{a^2}{b^2}\big)^n\Big)}{1-\frac{a^2}{b^2}}\\

C &= c+\frac{\frac{2a^2c}{b^2}\Big(1-\big(\frac{a^2}{b^2}\big)^n\Big)}{1-\frac{a^2}{b^2}}

\end{aligned}

$$

According to the Cauchy convergence criterion, when $q < 1$, this geometric series is convergent.

$$

\begin{aligned}

\lim_{n \to \infty} A &= \frac{\frac{2ac}{b}}{1-\frac{a^2}{b^2}} = \frac{2abc}{b^2-a^2}\\

\lim_{n \to \infty} C &= c+\frac{\frac{2a^2c}{b^2}}{1-\frac{a^2}{b^2}} = \frac{c(b^2+a^2)}{b^2-a^2}

\end{aligned}

$$

### The ingenious part

Right now, the result can also be proven through algebraic operations, but it involves a significant amount of calculation. The most ingenious part of this proof is its clever use of the sine rule.

We take the ratio of A to C:

$$

\frac{A}{C} = \frac{2abc}{b^2-a^2} \cdot \frac{b^2-a^2}{c(b^2+a^2)} = \frac{2ab}{a^2+b^2}

$$

This is equal to $\sin(2\alpha)$. Apply the sine rule to the top icoceles triangle, we have

$$

\frac{\sin(2\alpha)}{2a} = \frac{\sin(\beta)}{c}

$$

Substitute in $\sin(2\alpha) = \frac{2ab}{a^2+b^2}$ and $\sin(\beta)=\frac{b}{c}$ to obtain:

$$

\frac{2ab}{2a(a^2+b^2)} = \frac{b}{c^2}

$$

which simplifies to:

$$

\frac{b}{a^2+b^2} = \frac{b}{c^2}

$$

Because $a,b,c \ne 0$, therefore $a^2+b^2 = c^2$.

$\blacksquare$