# Area of a Triangle on a Sphere

My good friend who runs the wonderful blog In The Margins gave me this problem quite some time ago, and I honestly enjoyed completing it quite a bit. Thought I’d go ahead and share it with the world!

Problem:
“Find the area of a triangle on a sphere without using Calculus, given just its angles $\mathbf{\alpha, \ \beta, \ \text{and } \gamma}$.”

I recommend pausing here and trying this yourself first, it’s quite fun!

Solution:
First, make sure you know what a great circle is. It’s the intersection of a plane (“paper”) and a sphere (“orange?”) where the plane goes through the center of the sphere. Look at the image to the right for an example (for instance, the Earth’s equator is a great circle in a sense!)

Begin by first drawing a triangle on a sphere with arbitrary angles $\alpha, \ \beta, \ \text{and } \gamma$. Afterwards, extend the sides of the triangle to form great circles. In doing this, you should notice that the same triangle appears on the back of the sphere.

Now denote the side between $\gamma$ and $\alpha$ as 1, $\alpha$ and $\beta$ as 2, and $\beta$ and $\gamma$ as 3. Now only looking at side 1, shade in the hemisphere not containing the triangle, and the repeat the process with the other two sides.

The numbers in circles represent how many times that particular partition was shaded over.

Let’s denote the following:
$A_1 :=$ Set of Shaded Points from The Hemisphere of Side $1$
$A_2 :=$ Set of Shaded Points from The Hemisphere of Side $2$
$A_3 :=$ Set of Shaded Points from The Hemisphere of Side $3$

Note then, that this follows:
$|A_1| = |A_2| = |A_3| = 2 \pi r^2$
$|A_1 \cap A_2 \cap A_3|=A_T$ (where $A_T$ is the area of the triangle)
$|A_1 \cup A_2 \cup A_3| = 4 \pi r^2 - A_T$
$|A_1 \cap A_2| = 2 \alpha r^2$ (this, along with the following, are called antipodal digons)
$|A_2 \cap A_3| = 2 \beta r^2$
$|A_3 \cap A_1| = 2 \gamma r^2$

Looks like we’re in a good situation to employ the Inclusion-Exclusion Principle!

Inclusion-Exclusion Principle for 3 Finite Sets:
Let $A$, $B$, and $C$ be three finite sets. Then
$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| .$

Which means we must have
$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$

$\implies 4 \pi r^2 - A_T= 6 \pi r^2 - 2r^2(\alpha + \beta + \gamma) + A_T$
$\implies A_T = r^2(\alpha + \beta + \gamma - \pi)$

Thus we have shown the following:

Proposition:
The area of a triangle $A_T$ on a sphere of radius $r$ with angles $\alpha, \ \beta, \ \gamma$ is $A_T = r^2(\alpha + \beta + \gamma - \pi) .$

And there we have it! There’s our area. Oh! And also!

Corollary:
The sum of the interior angles of a triangle on a sphere must exceed $\pi$.

Isn’t that so cool? Being used all your life to the sum of the interior angles of a triangle being precisely 180 degrees — how do you feel seeing that now it’s mandatory it exceed 180 degrees?