Problem 1225

We will prove the statement by induction on the total number of lines and circles. For one line or circle, the statement is obvious. Now suppose that any map defined by $n$ lines and circles can be colored in the required way, and we will show how to then color a map defined by $n+1$ lines and circles. Remove one of these lines (or circles) and color the map defined by the remaining $n$ lines and circles. Then keep the colors of all regions lying on one side of the removed line (or circle), and replace the colors of all regions lying on the other side with their opposites.

The Sine Theorem and the First Cosine Theorem for a Trihedral Angle. Let there be a trihedral angle with plane angles $\alpha, \beta, \gamma$ and dihedral angles $A, B, C$ opposite to them. For it, the Sine Theorem (8.7) and two Cosine Theorems (8.6), (8.8) (see below) hold. After one of these theorems is proved, the others can be obtained through algebraic transformations. Let us abstract from the geometric nature of the problem and assume that we are simply given the equalities

$$\begin{aligned} & \cos \alpha=\cos \beta \cos \gamma+\sin \beta \sin \gamma \cos A \\ & \cos \beta=\cos \alpha \cos \gamma+\sin \alpha \sin \gamma \cos B \\ & \cos \gamma=\cos \alpha \cos \beta+\sin \alpha \sin \beta \cos C \end{aligned}$$

and, moreover, the quantities $\alpha, \beta, \gamma$ and $A, B, C$ are between 0 and $\pi$. Prove that

$$\frac{\sin A}{\sin \alpha}=\frac{\sin B}{\sin \beta}=\frac{\sin C}{\sin \gamma}$$

## Solution

From the first equality

$$\cos A=\frac{\cos \alpha-\cos \beta \cos \gamma}{\sin \beta \sin \gamma}$$

Hence,

$$\begin{aligned} & \sin ^{2} A=\frac{1-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma+2 \cos \alpha \cos \beta \cos \gamma}{\sin ^{2} \beta \sin ^{2} \gamma} \\ & \frac{\sin ^{2} A}{\sin ^{2} \alpha}=\frac{1-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma+2 \cos \alpha \cos \beta \cos \gamma}{\sin ^{2} \alpha \sin ^{2} \beta \sin ^{2} \gamma} \end{aligned}$$

Since these formulas transform into each other under a cyclic permutation of the variables $\boldsymbol{\alpha}, \boldsymbol{\beta}, \boldsymbol{\gamma}, A, B, C$ and this transformation does not change the right-hand side of the last equality, then

$$\frac{\sin ^{2} A}{\sin ^{2} \alpha}=\frac{\sin ^{2} B}{\sin ^{2} \beta}=\frac{\sin ^{2} C}{\sin ^{2} \gamma}$$

Since all quantities $\alpha, \beta, \gamma, A, B, C$ are between 0 and $\pi$, then

$$\frac{\sin A}{\sin \alpha}=\frac{\sin B}{\sin \beta}=\frac{\sin C}{\sin \gamma}$$