AoPS - Problem

Source: Iran 3rd round 2009 - final exam problem 1

Tags: geometry, exterior angle, angle bisector, geometric inequality, Iran

TheOverlord

02.01.2015 15:17

Suppose $n>2$ and let $A_1,\dots,A_n$ be points on the plane such that no three are collinear. (a) Suppose $M_1,\dots,M_n$ be points on segments $A_1A_2,A_2A_3,\dots ,A_nA_1$ respectively. Prove that if $B_1,\dots,B_n$ are points in triangles $M_2A_2M_1,M_3A_3M_2,\dots ,M_1A_1M_n$ respectively then \[|B_1B_2|+|B_2B_3|+\dots+|B_nB_1| \leq |A_1A_2|+|A_2A_3|+\dots+|A_nA_1|\] Where $|XY|$ means the length of line segment between $X$ and $Y$. (b) If $X$, $Y$ and $Z$ are three points on the plane then by $H_{XYZ}$ we mean the half-plane that it's boundary is the exterior angle bisector of angle $\hat{XYZ}$ and doesn't contain $X$ and $Z$ ,having $Y$ crossed out. Prove that if $C_1,\dots ,C_n$ are points in ${H_{A_nA_1A_2},H_{A_1A_2A_3},\dots,H_{A_{n-1}A_nA_1}}$ then \[|A_1A_2|+|A_2A_3|+\dots +|A_nA_1| \leq |C_1C_2|+|C_2C_3|+\dots+|C_nC_1|\] Time allowed for this problem was 2 hours.

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Nanas

15.01.2015 18:38

If $X_1,X_2,\cdots,X_n$ are points in the plane, we denote by $|X{}_1X{}_2 \cdots X_n|$, the length of the polygonal path $X{}_1X{}_2 \cdots X_n$. In other words, \[ |X{}_1X{}_2 \cdots X_n| = |X{}_1X{}_2| + \cdots + |X_{n-1}X_n| \] Clearly by triangle inequality,\[ |X{}_1X{}_n| \leq|X_1 \cdots X_n|\] (a) By $(*)$ we have, \[ |B_i B_{i+1}| \leq |B_i M_{i+1} B_{i+1}| \qquad (1)\] We have also the following Claim, Claim. Let $ABC$ be a triangle with $X$ be a point inside it, then $|AXC| \leq |ABC|$. Proof. Let $X'$ be the intersection of $AX$ with $\overline{BC}$, then by triangle inequality, \[ |AXC| \leq |AX| + |XX'| + |X'C| \leq |AX'| + |X'C| \leq |ABX'| + |X'C| = |ABC| \] Now applying the above claim on each of triangles $M_{j+1} A_{j+1} M_j $ we get, \[ |M_j B_j M_{j+1}| \leq |M_j A_{j+1} M_{j+1}| \qquad (2)\] Now Combining $(1)$ and $(2)$, we get \begin{align*} \sum_{i=1}^{n} |B_i B_{i+1}| \leq \sum_{i=1}^{n} |B_i M_{i+1} B_{i+1}| = \sum_{j=1}^{n} |M_j B_j M_{j+1}| \leq \sum_{j=1}^{n} |M_j A_{j+1} M_{j+1}| & = \sum_{j=1}^{n} |A_j A_{j+1}|\\ \end{align*} (b) We denote by $l{}_{A{}_i}$ the exterior angle bisector of $\angle A_{i-1} A_i A_{i+1}$. We prove the inequality in the case $C_i \in l{}_{A{}_i}$ first. Now denote by $X^{(1)}$ be the reflection of $X$ about $l{}_{A{}_2}$, and $X^{(i)}$ be the reflection of $X$ about $l{}_{A{}_{i+1}^{(i-1)}}$ (The exterior angle bisector of $\angle A_{i} A_{i+1} A_{i+2}$). Now note that as $l{}_{A{}_i^{j}}$ is exterior angle bisector of $\angle A{}_{i{}_1}^{(j)} A_{i}^{(j)} A_{i+1}^{(j)} $, we have $A_{i-1}^{(j)}A_{i}^{(j)}$ is mapped under reflection about $A_{i}^{(j)} A_{i+1}^{(j)}$, whence clearly $A_1 A_2^{(1)} A_3^{(2)} \cdots A_n^{(n-1)}A_1^{(n)} \cdots A_{n}^{(2n-1)}$ is line segment (the blue line) of length $2 |A_1 \cdots A_n| $. We have $C_1 C_2^{(1)}\cdots C_n^{(n-1)} C_1^{(n)} \cdots C_n^{2n-1)}$ is a broken line (the green one) of length $2 |C_1 \cdots C_n|$. One can use the properties of composition of rotations and reflection with easy angle chasing to conclude that $A_{1}^{(2n)} \cdots A_{n}^{(2n)}$ is a translation of $A_1 \cdots A_1$, whence $C{}_1C{}_n^{(2n-1)}=C_1 C_1^{2n} \parallel A_1 A_1^{(2n)} = A_1 A_n^{(2n-1)}$. It follows that \[2 |A_1 \cdots A_n| \leq 2 |C_1 \cdots C_n|\] as required. Now we prove that we always can reduce the problem to the above case, we smooth to the above case first we let $X_i$ and $Y_i$ be the intersections of $C{}_iC{}_{i-1}$ and $C_i C_{i+1}$ each with $l{}_A{}_i$ ,respectively. Choose $C'_i \in \overline{X{}_iY{}_i}$, It is easy to see by triangle inequality, \[|C_1\cdots C_n| \geq |X{}_1Y{}_1 X{}_2Y{}_2 \cdots X_n Y_n| \geq |C'_1 \cdots C'_n|\] So we are done.

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