Modern Pythagorean Theorem, Part 1

The Geometric Relationship

The Babylonians knew the triangle, be it acute, obtuse, or the most useful form, the right triangle (2000 BC).

The three interior angles of a triangle add to 180°, which is the result of work done in Euclid's Elements (300 BC), though much of that work could have been done by students of the Pythagoras school, of the same era. The proof lies in considering not only the interior angles, but also their supplementary angles. So, we construct the Sum of the angles of a triangle with an arrow on the line whose segment is a side, or edge, of the triangle. If a path is drawn around the triangle we know that a vector pointing in the forward direction turns one whole turn upon completing the path, which travels along the three edges, and stops at the three vertices, where it rotates to face the subsequent edge's direction.

Figure 1. When a line (dashed here) crosses another line obliquely, it makes two angles with the bisected line, one acute (θ) and one obtuse (φ) which sum to 180°—half the angle subtended by a circle (360° around).

If each change of direction at a vertex is protracted, it has the angles drawn from the intersection of the extended side with the next side following the included angle, with such supplementary angles of the intersecting sides as depicted in Figure 1 (above). So for the three turns at a vertex, we have:

$$\theta_i=180^\circ - \phi_i$$

Adding this three times gives for the exterior angles:

$$ \sum_1^3 \phi_i = 360^\circ $$

To reiterate, if the diagram were endowed with three arrows for the orientation of the walker, as travelling the perimeter, then upon returning to the original position the walker arrow will have rotated some amount three times for a sum total of one whole rotated turn. Plugging into the sum for the three supplementary angles:

$$ \sum_1^3 \theta_i = 3(180^\circ)-360^\circ $$

So we have the interior angles of a triangle sum to $180^\circ$.

Right Triangles

The right triangle has two edges meeting at a 90°-angle, or right-angled vertex. A right-angled vertex, is sometimes indicated by a small square (as demonstrated below). The relationship of the three sides is known by the following widely-known formula, the Pythagorean Theorem:

$$\begin{equation} \label{Pythagorean} a^2 + b^2 = c^2 \end{equation}$$

Where $c$ is the hypotenuse, from the Ancient Greek hypo + teino, or less than + to stretch or span, with the original meaning being that of the lesser measure of a skin, or any physical covering measure susceptible to gravity and width effects, and the distance of a perfectly straight line spanning the gap. $a$ & $b$ are the two edges of the right-angled vertex, as shown in Figure 2 (below).

The relation for edges of a right triangle, \eqref{Pythagorean}, is useful anytime there is decomposition of a position into orthogonal components, such as body centered frame, or lab frame—each of the three at right-angles to the others, as Cartesian coordinates within a , with Euler angles on a sphere, in 3D vector space.

Figure 2. A right triangle, with edges a and b, and hypotenuse c, where the right-angle is indicated by the small square at $90^\circ$ vertex.

The area of a rectangle is the width times the height. The area of a square, with four equal edges, and four right angles, and edge $x$, is $x^2$ ($x$-squared). The right-triangle theorem is geometrically a relationship of areas.

The area of the right-triangle is half the multiplicative product of the two non-hypotenuse edges. By separating a rectangle into two halves you are left with two identical right-triangles:

Figure 3. A rectangle with a diagonal drawn, separating it into two identical right triangles. The area of the rectangle is $A=ab$, and the area of each triangle is half that, $A=ab/2$. This entails the Right-Triangle Area Theorem.

Which accounts for the factor of a half in the formula for the area of a right triangle: $A=ab/2$. This method works when the shape is a parallelogram also, then the area is the length of the base times that of the height. The diagonal of a parallelogram bisects the shape, and so the area of a general triangle is $A_\triangle=\frac{1}{2}\text{base}\times\text{height}$.

Figure 4. A parallelogram with a diagonal $d$, and vertical sides of the end right triangles, labelled `h`, drawn. The diagonal bisects the shape into two mirror-image triangles. The area of each triangle is $\frac{1}{2}bh$, as the area of the parallelogram is $bh$.

Regular Polygons

A regular polygon is a shape that has a finite number of sides, each of which is the same length, $s$. An $n$-sided regular polygon, has an angle between adjacent sides as the same for all $n$ vertices. A regular polygon has area which depends on the number of sides and the length of the side.

Equilateral Triangle

The area of an (equilateral) triangle is one-half times base times height, $\frac{1}{2}bh$, all we need is the height of the equilateral triangle:

$$ b^2 = h^2 + (\frac{b}{2})^2 $$

Because we want the positive solution for $h$, and the relation of squares permits negative lengths as solutions, we indicate the positive solution with the square root symbol, also called a radical sign:

$$h=\sqrt{b^2-(b/2)^2}$$

Now the area of the equilateral is $A=\frac{\sqrt{3}}{4}b^2$. For the $n=3$ case, the area of the equilateral triangle (the smallest regular polygon) is of a quadratic form of side length—like the old square.

Really, the Pythagorean Areas are applicable for all regular polygons not just squares, where the geometric particulars are the same for each type of regular polygon to yield the side-squared terms in the recognizable relationship.

$$ \frac{1}{4}na^2\cot{(\pi/n)} + \frac{1}{4}nb^2\cot{(\pi/n)} = \frac{1}{4}nc^2\cot{(\pi/n)} $$

Area of a Regular Polygon

The area of a regular polygon is quadratic in its side-length. Each side, of length $b$, is the base for an isosceles triangle-wedge of the regular polygon. The inscribed angle subtended by the base is $\theta=2\pi/n$. The height of the triangle is determined by the angle $\theta$ combined with the side length.

Figure 5. Decagon Area Diagram: A wedge section of a decagon is depicted with subtended angle, $\theta=(2\pi)/10$, and an overlain perpendicular from the center to the base of the isosceles, bisecting it into two mirror-image right triangles.

For an $n$-sided regular polygon, the angle subtended by an isosceles wedge, with base $b$ in Figure 5 (above), is $\theta=2\pi/n$. So $a=\frac{b}{2}\cot{(\pi/n)}$, where the cotangent is simply one-over-tangent, and the area of the wedge is thus $A_w=\frac{b^2}{4}\cot{(\pi/n)}$. For the area of the $n$-gon, it is $n$ such wedges, and we have the following result:

$$ \begin{equation}\label{area-ngon} A_{ngon}=\frac{1}{4}ns^2\cot{(\pi/n)} \end{equation} $$

Figure 6. Regular triangles (equilateral) with areas $A_a$, $A_b$, and $A_c$, where the sides are $a$, $b$, and $c$ respectively. The areas satisfy the Pythagorean relation, $A_a + A_b = A_c$, because the trigonometric factors cancel, leaving the squared-side relation.

Figure 7. Regular squares with areas $A_a$, $A_b$, and $A_c$, where the sides are $a$, $b$, and $c$ respectively. The areas satisfy the Pythagorean relation, $A_a + A_b = A_c$, because the trigonometric factors cancel, leaving the squared-side relation.

Figure 8. Regular pentagons with areas $A_a$, $A_b$, and $A_c$, where the sides are $a$, $b$, and $c$ respectively. The areas satisfy the Pythagorean relation, $A_a + A_b = A_c$, because the trigonometric factors cancel, leaving the squared-side relation.

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