Home > Olympiad > Sum of some angles in a polygon

## Sum of some angles in a polygon

Let ${A_1..A_n}$ be a convex polygon which has no two sides which are parallel. For each side ${A_iA_{i+1}}$ consider the vertex ${C_i}$ of the polygon which is the furthest away from this side. Prove that

$\displaystyle \sum_{i=1}^n \angle A_iC_iA_{i+1}=\pi.$

IMAR Contest, 2005, Juniors

Proof: Let’s call ${[A_iC_i],[A_{i+1}C_i]}$ special segments and let’s find some properties they have.

Claim 1: Each two special segments meet.

Proof: If the two segments have a vertex in common, then they surely meet. Denote ${C_{i+1}}$ the furthest away vertex from the side ${A_{i+1}A_{i+2}}$. If ${C_{i+1}=C_i}$, then the segments ${A_iC_i}$ and ${A_{i+1}C_{i+1}}$ meet.

Suppose that ${C_{i+1}\neq C_i}$. If ${C_{i+1}}$ is on the same side of ${A_iC_i}$ as ${A_{i+1}}$ then ${C_i}$ is further away from ${A_{i+1}A_{i+2}}$ than ${C_{i+1}}$, which is a contradiction.

Therefore ${[A_iC_i]}$ and ${A_{i+1}C_{i+1}}$ meet, i.e. ${A_iC_i}$ separates ${A_{i+1}}$ and ${C_{i+1}}$. The general case is treated the same way.

Claim 2: Suppose ${A_i}$ is furthest away from ${A_{m-1}A_m}$ and ${A_j}$ is furthest away from ${A_mA_{m+1}}$. Then each side on the polygonal path from ${A_i}$ to ${A_j}$ has ${A_m}$ as furthest away point.

Proof: Pick a side ${A_kA_{k+1},\ i\leq k\leq j}$ which has as further away point ${B}$. Each special segment ${BA_k,BA_{k+1}}$ must cross the segments ${A_iA_{m+1},A_iA_m}$ and ${A_jA_{m-1},A_jA_m}$. Because the polygon is convex, this cannot happen unless ${B=A_m}$.

Claim 3: The set of furthest away vertices ${B_1,..,B_k}$ is made of an odd number of points, and we have

$\displaystyle \sum_{i=1}^n \angle A_iC_iA_{i+1}=\sum_{j=1}^k \angle B_{j-\frac{k-1}{2}}B_jB_{j+\frac{k-1}{2}}.$

Proof: Denote ${B_1,..,B_k}$ the set of furthest away points numbered in counterclockwise direction.

Consider the set ${\mathcal{S}_1}$ of sides ${A_iA_{i+1}, m_1\leq i \leq p_1-1}$ such that ${B_1}$ is the furthest away vertex from all these sides. Consider the next furthest away vertex ${B_2}$ in the counterclockwise direction and ${\mathcal{S}_2}$ its set of furthest away sides, and so on for all ${k}$ furthest away vertices.

It is obvious that ${\mathcal{S}_i \cap \mathcal{S}_j = \emptyset}$ whenever ${i \neq j}$ and Claim 2 implies that the sides in ${\mathcal{S}_{i+1}}$ come after the sides from ${\mathcal{S}_i}$. Moreover, the endpoints of sets ${\mathcal{S}_i}$ are among ${B_j,\ i=1..n}$.

Suppose the endpoints of ${\mathcal{S}_1}$ are ${B_jB_{j+1}}$. Then to each of ${B_2,..,B_j}$ are associated the sets ${\mathcal{S}_2,...\mathcal{S}_j}$. And because of the second claim the endpoint of ${\mathcal{S}_j}$ is ${B_1}$. Therefore ${B_{j+1+j-1}=B_1}$. This means that ${k}$ is odd, and because of the construction of ${\mathcal{S}_i,\ i=1,k}$ the above formula holds.

Why do we have

$\displaystyle \sum_{j=1}^k \angle B_{j-\frac{k-1}{2}}B_jB_{j+\frac{k-1}{2}}=\pi ?$

Consider the polygon ${B_1..B_k}$. Denote

$\displaystyle \begin{cases} \gamma_i= \angle B_{j-\frac{k-1}{2}}B_jB_{j+\frac{k-1}{2}} \\ \alpha_i = \angle B_jB_{j-\frac{k-1}{2}}B_{j+\frac{k-1}{2}} \\ \beta_i = \angle B_{j-\frac{k-1}{2}}B_{j+\frac{k-1}{2}}B_j \end{cases}.$

Obviously we have ${\alpha_i+\beta_i+\gamma_i=\pi}$ as the sum of the angles of a triangle. Moreover, we have ${\angle B_i=\alpha_{j+\frac{k-1}{2}}+\beta_{j-\frac{k-1}{2}}-\gamma_i, }$ so summing after ${i}$ we get

$\displaystyle (k-2)\pi = k\pi -2\sum_{i=1}^k \gamma_i.$

In conclusion

$\displaystyle \sum_{i=1}^k \gamma_i=\pi$

and the proof is over.