The Higher-Dimensional Pythagorean Theorem

December 12, 2021 Shreyas Comments 2 comments

1. Introduction

$a^2 + b^2 = c^2…$ We have all heard of the familiar Pythagorean theorem, relating the side lengths of a right triangle. But, have you heard of De Gua’s Theorem? An interesting generalization of the Pythagorean theorem into 3 dimensions, this theorem is very similar result for “right tetrahedrons” where three of the edges are mutually perpendicular. Suppose, in such a tetrahedron $ABCD$ we have

$\begin{equation*} AD \perp BD,\ BD \perp CD,\ CD \perp AD \end{equation*}$

Denote the area of $X$ as $[X]$ . Then,

$\begin{equation*} [ABD]^2 + [ACD]^2 + [BCD]^2 = [ABC]^2 \end{equation*}$

In today’s post, we will investigate whether a similar result holds for an arbitrary number of dimensions. But before doing that, we must start by proving the case for 3 dimensions…

2. The 3-D Case

We start by placing the points of the tetrahedron on a coordinate system. Suppose

$\begin{equation*} D = (0,\ 0,\ 0),\ A = (a,\ 0,\ 0)\ B = (0,\ b,\ 0),\ C = (0,\ 0,\ c) \end{equation*}$

At first, I thought of using Heron’s formula to solve for $[ABC]$ given its side lengths with the Pythagorean theorem, but quickly realized it would be extremely tedious and very difficult to generalize to higher dimensions. Thinking in a more geometrical sense, I thought of projecting the triangle $ABC$ down onto one of the planes of the coordinate system, say the x-y plane. The resultant triangle is identical to $ABD$ and thus its area should be easy to calculate. But how can we use this to find $[ABC]$ ?

The key part is realizing that projecting any planar figure onto another plane multiplies its area by a particular constant: $\cos(\theta)$ , where $\theta$ is the dihedral angle between the two planes.

Example of projection of an ellipse onto a circle. The area of the ellipse is multiplied by the cosine of the angle between the two planes containing the ellipse and the circle

If you don’t believe this, prove it for simple rectangles first, and then think of breaking up general shapes into many thin rectangular slices.

How should we go about finding this dihedral angle? Let us drop a perpendicular from $D$ onto $AB$ , and call the foot of this perpendicular $E$ :

Now, observe that the dihedral angle between the plane of $ABC$ and the x-y plane is just $\angle DEC$ . So the ratio we seek is $\cos(\angle DEC) = \frac{DE}{EC}$ . Stepping into the two-dimensional world of the x-y plane for a moment, we see that the equation of line $AB$ is given by $bx+ay=ab$ . Thus, we can calculate its distance to the $D = (0, 0)$ as

$\begin{equation*} EC = \frac{|a*0 + b*0 - ab|}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}} \end{equation*}$

Now, because $DC = c$ and $\triangle EDC$ is right, we have

$\begin{equation*} EC^2 = ED^2 + DC^2 = \frac{a^2b^2}{a^2 + b^2} + c^2 = \frac{a^2b^2 + a^2c^2 + b^2c^2}{a^2 + b^2} \end{equation*}$

Thus,

$\begin{equation*} \frac{[ABD]}{[ABC]} = \cos(\theta) = \frac{DE}{EC} = \frac{\frac{ab}{a^2 + b^2}}{\frac{\sqrt{a^2b^2 + a^2c^2 + b^2c^2}}{\sqrt{a^2 + b^2}}} = \frac{ab}{\sqrt{a^2b^2 + a^2c^2 + b^2c^2}} \end{equation*}$

But, $[ABD] = \frac{1}{2}ab$ , so solving for $[abc]$ yields

$\begin{equation*} [ABC] = \frac{1}{2}\sqrt{a^2b^2 + b^2c^2 + a^2c^2} = \sqrt{\left(\frac{1}{2}ab\right)^2 + \left(\frac{1}{2}ac\right)^2 + \left(\frac{1}{2}bc\right)^2}\rightarrow \end{equation*}$

$\begin{equation*} [ABC]^2 = [ABD]^2 + [ACD]^2 + [BCD]^2 \end{equation*}$

as claimed.

3. The General Case

We now have a clear idea of what to try in higher dimensional cases. Given an n-dimensional Cartesian coordinate system, we set points

$\begin{equation*} D = (0,\ 0 \dots,\ 0),\ A_1 = (a_1,\ 0 \dots,\ 0)\ A_2 = (0,\ a_2 \dots,\ 0) \dots,\ A_n = (0,\ 0 \dots,\ a_n) \end{equation*}$

where $a_i > 0,\ 1\leq i \leq n$ for convenience. For simplicity, we will refer to this figure as a right n-D triangle (this is almost certainly not the actual name). We would like to prove that

$\begin{equation*} [A_1A_2\dots A_n]^2 = [DA_2A_3\dots A_n]^2 + [DA_1A_3\dots A_n]^2\dots + [DA_1A_2 \dots A_{n - 1}]^2, \end{equation*}$

where the brackets now denote the $n - 1$ dimensional equivalent of volume. In the general case, I will use the assumption that projections from one hyperplane to another in $n$ -dimensional space transform $(n - 1)$ -dimensional volumes by scaling them with a constant factor: the cosine of the “angle” between the two hyperplanes (this can probably be proved in a similar way as 3 dimensions). I will also assume that this dihedral angle can be calculated in a similar way: by drawing a suitable right triangle in a plane perpendicular to the two hyperplanes. Observe that the projection of $A_1A_2\dots A_n$ onto the first $n - 1$ axes is $A_1A_2\dots A_{n - 1}D$ , because $A_n$ is the only point that is not already in the lower-dimensional space, and it transforms to $D$ . Furthermore, note that $A_1A_2\dots A_n$ is contained within an $(n - 1)$ -dimensional hyperplane with equation

$\begin{equation*} a_2a_3\dots a_nx_1 + a_1a_3\dots a_nx_2 \dots + a_1a_2\dots a_{n - 1}x_n \end{equation*}$

This can be verified by simply plugging in the coordinates of each $A_i$ into the equation.

Now, for a moment we restrict ourselves to the $(n - 1)$ -D space formed by the $x_1,\ x_2,\dots x_{n - 1}$ axes (so we will temporarily delete the last coordinate of every point). Observe that the points $A_1,\ A_2,\dots A_{n - 1}$ pass through the $(n - 2)$ -D hyperplane with equation

$\begin{equation*} a_2a_3\dots a_{n - 1}x_1 + a_1a_3\dots a_{n - 1}x_2\dots + a_1a_2\dots a_{n - 2}x_{n - 1} = a_1a_2\dots a_{n - 1} \end{equation*}$

Verify this by plugging in the first $n - 1$ coordinates of each $A_1\dots A_{n - 1}$ and seeing that the equation is satisfied. Call this $(n - 2)$ -D hyperplane $Y$ .

Now, we can drop an altitude from $D$ onto $Y$ , call the foot of this altitude $E$ . We will now establish the length $DE$ , which is equivalent to the distance from $D = \ (0, 0\dots, 0)$ to $Y$ . We can use the general formula for the distance between a point and a hyperplane:

$\begin{equation*} DE = \frac{|a_2a_3\dots a_{n - 1} * (0) + a_1a_3\dots a_(n - 1) * (0)\dots + a_1a_2\dots a_{n - 2} * (0) - a_1a_2\dots a_{n - 1}|}{\sqrt{(a_2a_3\dots a_{n - 1})^2 + \dots (a_1\dots a_{n - 2})^2}} \end{equation*}$

$\begin{equation*} = \frac{a_1\dots a_{n - 1}}{\sqrt{(a_2a_3\dots a_{n - 1})^2 + \dots (a_1\dots a_{n - 2})^2}} \end{equation*}$

Observe that $Y$ is can also be expressed as the intersection of the $(n - 1)$ -D hyperplane $X$ containing $A_1,\ A_2\dots A_n$ and the $(n - 1)$ -D hyperplane $X_0$ containing $D,\ A_1,\ A_2,\dots A_{n - 1}$ . Furthermore, $X_0$ is also formed by the coordinate axes $x_1 \dots x_{n - 1}$ . Then, $\triangle A_nDE$ has a right angle at $D$ , because $DE\subseteq X_0$ and $A_nD \perp X_0$ as $A_nD$ is contained in the $x_n$ coordinate axis. From here, it is easy to see that the projection of $A_n$ onto $Y$ is also $E$ ; if $Y \subseteq X_0$ , then projecting onto $Y$ is the same as projecting onto $X_0$ and then onto $Y$ , and projecting $A_n$ onto $X_0$ yields $D$ .

From the above arguments we can see that the angle $\angle A_nED$ in triangle $A_nDE$ is the dihedral angle $\theta$ between hyperplanes $X$ and $X_0$ . Then,

$\begin{equation*} [DA_1A_2\dots A_{n - 1}] = [A_1\dots A_n]\cos(\theta) = [A_1\dots A_n]\frac{DE}{A_nE}\rightarrow \end{equation*}$

$\begin{equation*} [A_1\dots A_n] = [DA_1\dots A_{n - 1}]\frac{A_nE}{DE}\rightarrow [A_1\dots A_n]^2 = [DA_1\dots A_{n - 1}]^2\frac{A_nE^2}{DE^2} \end{equation*}$

We will make one final assumption: the volume of a right k-D triangle with edge lengths $a_1,\dots a_k$ is equal to $c_ka_1a_2\dots a_k$ , where $c_k$ is a constant dependent on the dimension $k$ . For example, the volume of a right tetrahedron with edges $a_1,\ a_2,\ a_3$ is $\frac{1}{6}a_1a_2a_3$ , so $c_3 = \frac{1}{6}$ . This can be justified by setting $c_k$ to be the volume of a right k-D triangle with edge lengths 1, then scaling the figure by a factor of $a_i$ along each axis $x_i$ , thus multiplying the initial volume of $c_k$ by $a_i$ at each step. Now, we use the Pythagorean Theorem of $\triangle A_nDE$ to find $A_nE$ :

$\begin{equation*} A_nE^2 = DE^2 + A_nD^2 = \frac{(a_1a_2\dots a_{n - 1})^2}{(a_2\dots a_{n - 1})^2 + \dots (a_1\dots a_{n - 2})^2} + a_n^2 \end{equation*}$

$\begin{equation*} = \frac{(a_2a_3\dots a_n)^2 + (a_1a_3\dots a_n)^2\dots + (a_1a_2\dots a_{n - 1})^2}{(a_2\dots a_{n - 1})^2 \dots + (a_1\dots a_{n - 2})^2}, \end{equation*}$

where the $i$ th squared term in the numerator consists of the product of the numbers $a_1\dots a_n$ except for $a_i$ . Now, we compute $\frac{A_nE^2}{DE^2}$ as

$\begin{equation*} \frac{A_nE^2}{DE^2} = A_nE^2*\frac{1}{DE^2} \end{equation*}$

$\begin{equation*} = \end{equation*}$

$\begin{equation*} \left(\frac{(a_2a_3\dots a_n)^2 + (a_1a_3\dots a_n)^2\dots + (a_1a_2\dots a_{n - 1})^2}{(a_2\dots a_{n - 1})^2 \dots + (a_1\dots a_{n - 2})^2}\right)\left( \frac{(a_2\dots a_{n - 1})^2 \dots + (a_1\dots a_{n - 2})^2}{(a_1\dots a_{n - 1})^2}\right) \end{equation*}$

$\begin{equation*} = \frac{(a_2a_3\dots a_n)^2 + (a_1a_3\dots a_n)^2\dots + (a_1a_2\dots a_{n - 1})^2}{(a_1\dots a_{n - 1})^2} \end{equation*}$

Multiplying this by $[DA_1\dots A_{n - 1}]^2 = c_{n - 1}^2(a_1\dots a_{n - 1})^2$ yields

$\begin{equation*} [A_1\dots A_n]^2 = [DA_1\dots A_{n - 1}]^2 * \frac{A_nE^2}{DE^2} \end{equation*}$

$\begin{equation*} = \left(c_{n - 1}^2(a_1\dots a_{n - 1})^2\right)\left(\frac{(a_2a_3\dots a_n)^2 + (a_1a_3\dots a_n)^2\dots + (a_1a_2\dots a_{n - 1})^2}{(a_1\dots a_{n - 1})^2}\right) \end{equation*}$

$\begin{equation*} = (c_{n - 1}a_2a_3\dots a_n)^2 + (c_{n - 1}a_1a_3\dots a_n)^2\dots + (c_{n - 1}a_1a_2\dots a_{n - 1})^2 \end{equation*}$

$\begin{equation*} = [A_2A_3\dots A_n]^2 + [A_1A_3\dots A_n]^2\dots + [A_1A_2\dots A_{n - 1}]^2 \end{equation*}$

as claimed, and we have finally proven the general result! So, the surprising result of De Gua’s theorem is not simply a coincidence of three dimensions, but can be extended to any number of dimensions.

2 thoughts on “The Higher-Dimensional Pythagorean Theorem”

jojo says:

September 12, 2022 at 5:21 pm

i don’t understand anything but this is a super cool idea and an absolute slay

Jack Bruns says:

November 8, 2022 at 10:34 pm

An interesting read, explained in easy to understand language.

Shreyas' blog – Ignited Minds

I love math and science. This is a blog about things I think of, or work on, in my spare time.