# Euler's Polyhedron Formula

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This page uses continuous scrollytelling to present a variation of Legendre's proof for the following formula.

V - E + F = 2

For any convex polyhedron (or planar graph), the number of vertices minus the number of edges plus the number of faces is 2. Leonhard Euler discovered this in 1752 although Descartes discovered a variation over 100 years earlier.

This equation makes it easy to prove that there only 5 Platonic solids, but perhaps its real beauty lies in how it connects disparate fields of mathematics. These connections are suggested by Legendre's proof, which is purely geometrical and leverages some interesting properties of geodesic triangles.

Scroll down at your own pace, but try not to go too fast, otherwise you'll skip over the animation. The proof is divided into 5 steps:

## Part 1: Surface Area of Sphere

The surface area of a sphere can be derived with freshman calculus, but it was discovered by Archimedes long before the invention of calculus.

Archimedes realized that the surface area of a sphere is equal to the area of its smallest enclosing cylinder, which is 4πr2. This is somewhat intuitive if you think about lat-long rectangles.

At the equator, lat-long rectangles are fat and short. Closer to the poles they are tall and thin. As they move closer to the poles, they become tall at the same rate at which they become thin, so their area remains constant.

Going forward, we'll keep things simple by focusing on a sphere whose radius is 1. The surface area of this sphere is .

## Part 2: Area of a Double Lune

Great circles are lines on the surface of a sphere that divide the sphere in half. If you place your car anywhere on a sphere's surface, then continuously drive without turning the steering wheel, you'll always inscribe a great circle. Any portion of a great circle is called a geodesic path.

The region on a sphere between two great circles is called a "lune". Let's figure out the surface area of a lune bounded by θ radians.

If θ is π radians, the lune encompasses one hemisphere. Since the area of a hemisphere is , the area of the lune is .

This is one reason why radians are more elegant than degrees!

The surface area of a lune plus its antipode is . Let's call this a double lune.

## Part 3: Girard's Theorem

Spherical triangles are inscribed by geodesic lines (i.e. portions of great circles). Unlike a planar triangle, the area of a spherical triangle can be determined solely from its angles. With planar triangles, the sum of the three angles is always π radians. Not so with spherical triangles!

For example, consider the triangle that encompasses one-eighth of the sphere surface, which has three 90° angles, or π/2 radians each. Clearly these do not add up to π.

Let's figure out the area of any geodesic triangle with angles a, b, and c.

Each corner in the triangle corresponds to a double lune on the surface. We can visualize each double lune with one of the additive primary colors.

The sum of the lune areas can be visualized by adding up their respective colors.

Notice that the total area of the lunes is equivalent to the surface of the entire sphere, except that the triangle and its antipode are each counted an additional 2x times.

Recall that:

• The surface area of the unit sphere is .
• The surface area of each double lune is .

Moreover, we now know that:

• The surface area of the sphere is equal to the area of all the triangle's double lunes, minus 4x the area of the triangle.

Therefore, if the area of geodesic triangle abc is A, then:

• 4π = 4a + 4b + 4c - 4A

Or, more simply stated:

• A = a + b + c - π

This formula was independently discovered by Albert Girard (1595-1632) and Thomas Harriot (1560-1621).

## Part 4: Spherical Polygons

Now that we know how to compute the area of a geodesic triangle, can we figure out how to compute the area of a geodesic polygon?

Yes, we can! Every n-gon can be decomposed into (n-2) triangles.

The sum of all the angles in a polygon is equal to the sum of all the angles in its constituent triangles. And, we now know that each of those (n-2) constituent triangles has an area of:

<angle sum> - π

Therefore, the area of the polygon must be:

<angle sum> - (n-2) π

Stated another way:

A = (a + b + c + d + ...) - nπ + 2π.

To portray this formula visually, we've inscribed the components of the sum onto the sphere.

Note that each vertex and edge correspond to a component of the sum, as well as the face itself. This will be useful later in the proof.

## Part 5: The Conclusion

Now that we have a few tools under our belt, let's consider a convex polyhedron.

What happens when we inflate the polyhedron to meet its enclosing sphere? The sum of the areas of all the resulting geodesic polygons should be equivalent to the surface area of the sphere!

Next, apply the visual method for computing the area sum across all polygons. Each vertex contributes a total of radians, each edge contributes -2π (one for each side), and each face contributes .

Putting it all together:

• Surface area of unit sphere =
2πV - 2πE + 2πF

Or, simply stated:

• 4π = 2πV - 2πE + 2πF

Therefore:

• V - E + F = 2

Et Voilà!

The polyhedron formula is also known as Euler's Characteristic Formula because the right-hand side of the equation is actually a "characteristic" of the sphere's topology. If we were to inscribe the graph on a torus instead of a sphere, the Euler characteristic would be 0 rather than 2. 