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.

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  1. December 9, 2012 at 8:17 am | #1

    do you have any link to IMAR mathematical competiton?

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