A New Proof of the Pythagorean Theorem Inspired by the Work of New Orleans High School Students

My favorite proof of the Pythagorean Theorem is the one I am going to present here. This proof builds upon the innovative work of a group of New Orleans high school girls, whom I will refer to as the New Orleans girls in this article. The proof is an inspired approach that avoids infinite sums and the Law of Sines, but instead relies on the properties of trigonometric functions and similar triangles. Let's dive deep into the details and solution.

Inspiration from the New Orleans Girls' Proof

The New Orleans girls' proof, which received significant attention, revolves around finding the values of sin 2α using both infinite series and the Law of Sines. To create a more straightforward and concise version, I decided to reengineer their approach by using the tangent function and focusing on the coordinates of certain points.

Reengineering the Proof

To eliminate the infinite sums and the Law of Sines, we will use the tangent function to derive the relationship between the sides of a right triangle. Let's start by looking at the key figure and rotating it appropriately. We will also introduce a new point G with coordinates different from the previous version in the literature.

Identifying Key Coordinates

First, let's look at the coordinates of point A. We know that the coordinates are A0, c. Now, we focus on the coordinates of points D and F. Since triangles ABC and BDG are similar, the ratios of their corresponding sides are equal:

The scale factor is given by h / a f / b 2a / c.

Point D has coordinates ( (2ab) / c, (2a^2) / c ). Since line AD connects points A and D, its equation can be expressed as:

((2ab) / c) y - c ((2a^2) / c - c) x

Point E is the x-intercept of this line, which we need to find:

-2abc (2a^2 - c^2) g

Solving for g, we get:

g (2abc) / (c^2 - 2a^2)

With the coordinates of E determined, we now have:

tan 2α g / c (2ab) / (c^2 - 2a^2)

Deriving the Pythagorean Theorem

The Pythagorean Theorem, a^2 b^2 c^2, can be derived from the tangent function. We will use the argument of a complex number to double its angle. For a complex number b ai, the argument is α arg(b ai). When we square this complex number, we get:

arg((b ai)^2) 2α arg(b^2 - a^2 2abi)

The tangent of this angle can be calculated as:

tan 2α (2ab) / (b^2 - a^2)

Equating the Two Expressions

By equating the two expressions for tan 2α, we get:

(2ab) / (c^2 - 2a^2) (2ab) / (b^2 - a^2)

Cancelling out the common factor, we derive:

b^2 - a^2 c^2 - 2a^2

Finally, simplifying, we arrive at the Pythagorean theorem:

a^2 b^2 c^2

This proof offers a fresh perspective on the Pythagorean Theorem and showcases the power of trigonometry and similar triangles in geometric problem-solving. It is also a testament to the creative thinking of high school students and their ability to approach a well-known mathematical concept with novel methods.