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2

1

$begingroup$

Consider a polygonal line $P_0P_1...P_n$ such that $angle P_0P_1P_2=angle P_1P_2P_3=...=angle P_{n-2}P_{n-1}P_{n}$, all measured clockwise. If $P_0P_1>P_1P_2>...>P_{n-1}P_{n}$, $P_0$ and $P_n$ cannot coincide.

While the result is remarkable in itself, the following proof that appears in Titu's 'Mathematical Olympiad Challenges' is just mind blowing :

Let us consider complex coordinates with origin at $P_0$ and let the line $P_0P_1$ be the x-axis. Let $alpha$ be the angle between any two consecutive segments and let $a_1>a_2>...>a_n$ be the lengths of the segments. If we set $z=e^{i(pi-alpha)}$, then the coordinate of $P_{n}$ is $a_1+a_2z+...+a_{n}z^{n-1}$. We must prove that this number is not equal to zero.

Using the Abel Summation lemma, we obtain,
$$a_1+a_2z+...+a_{n-1}z^n=(a_1-a_2) + (a_2-a_3)(1+z) + ... + a_{n}(1+z+...+z^{n-1})$$

If $alpha$ is zero, then this quantity is a strictly positive real number, and we are done.

If not, multiply by $(1-z)$ to get $(a_1-a_2)(1-z) + (a_2-a_3)(1-z^2) + ... + a_{n}(1-z^n)$. This expression cannot be zero. Indeed, since $mid zmid=1$, by the triangle inequality, we have,
$$mid (a_1-a_2)z + (a_2-a_3)z^2 + ... + a_{n}z^n mid < (a_1-a_2) + (a_2-a_3) + (a_3-a_4) + ... + a_{n}$$

The conclusion follows.

I was wondering if one can prove this synthetically, without using complex numbers. Any alternate proof or suggestion in the direction an alternate proof is welcome.

geometry complex-numbers contest-math euclidean-geometry summation-by-parts

share|cite|improve this question

edited Mar 20 at 19:12

Anubhab Ghosal

asked Mar 20 at 19:04

Anubhab GhosalAnubhab Ghosal

1,23819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This can be reduced to the case where $n $ is prime, though I am not sure how it helps
  $endgroup$
  – N.S.JOHN
  Apr 6 at 3:20
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Consider a polygonal line $P_0P_1...P_n$ such that $angle P_0P_1P_2=angle P_1P_2P_3=...=angle P_{n-2}P_{n-1}P_{n}$, all measured clockwise. If $P_0P_1>P_1P_2>...>P_{n-1}P_{n}$, $P_0$ and $P_n$ cannot coincide.

While the result is remarkable in itself, the following proof that appears in Titu's 'Mathematical Olympiad Challenges' is just mind blowing :

Let us consider complex coordinates with origin at $P_0$ and let the line $P_0P_1$ be the x-axis. Let $alpha$ be the angle between any two consecutive segments and let $a_1>a_2>...>a_n$ be the lengths of the segments. If we set $z=e^{i(pi-alpha)}$, then the coordinate of $P_{n}$ is $a_1+a_2z+...+a_{n}z^{n-1}$. We must prove that this number is not equal to zero.

Using the Abel Summation lemma, we obtain,
$$a_1+a_2z+...+a_{n-1}z^n=(a_1-a_2) + (a_2-a_3)(1+z) + ... + a_{n}(1+z+...+z^{n-1})$$

If $alpha$ is zero, then this quantity is a strictly positive real number, and we are done.

If not, multiply by $(1-z)$ to get $(a_1-a_2)(1-z) + (a_2-a_3)(1-z^2) + ... + a_{n}(1-z^n)$. This expression cannot be zero. Indeed, since $mid zmid=1$, by the triangle inequality, we have,
$$mid (a_1-a_2)z + (a_2-a_3)z^2 + ... + a_{n}z^n mid < (a_1-a_2) + (a_2-a_3) + (a_3-a_4) + ... + a_{n}$$

The conclusion follows.

I was wondering if one can prove this synthetically, without using complex numbers. Any alternate proof or suggestion in the direction an alternate proof is welcome.

geometry complex-numbers contest-math euclidean-geometry summation-by-parts

share|cite|improve this question

edited Mar 20 at 19:12

Anubhab Ghosal

asked Mar 20 at 19:04

Anubhab GhosalAnubhab Ghosal

1,23819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This can be reduced to the case where $n $ is prime, though I am not sure how it helps
  $endgroup$
  – N.S.JOHN
  Apr 6 at 3:20
  

  
  
  
  

add a comment | 

$begingroup$

Consider a polygonal line $P_0P_1...P_n$ such that $angle P_0P_1P_2=angle P_1P_2P_3=...=angle P_{n-2}P_{n-1}P_{n}$, all measured clockwise. If $P_0P_1>P_1P_2>...>P_{n-1}P_{n}$, $P_0$ and $P_n$ cannot coincide.

While the result is remarkable in itself, the following proof that appears in Titu's 'Mathematical Olympiad Challenges' is just mind blowing :

Let us consider complex coordinates with origin at $P_0$ and let the line $P_0P_1$ be the x-axis. Let $alpha$ be the angle between any two consecutive segments and let $a_1>a_2>...>a_n$ be the lengths of the segments. If we set $z=e^{i(pi-alpha)}$, then the coordinate of $P_{n}$ is $a_1+a_2z+...+a_{n}z^{n-1}$. We must prove that this number is not equal to zero.

Using the Abel Summation lemma, we obtain,
$$a_1+a_2z+...+a_{n-1}z^n=(a_1-a_2) + (a_2-a_3)(1+z) + ... + a_{n}(1+z+...+z^{n-1})$$

If $alpha$ is zero, then this quantity is a strictly positive real number, and we are done.

If not, multiply by $(1-z)$ to get $(a_1-a_2)(1-z) + (a_2-a_3)(1-z^2) + ... + a_{n}(1-z^n)$. This expression cannot be zero. Indeed, since $mid zmid=1$, by the triangle inequality, we have,
$$mid (a_1-a_2)z + (a_2-a_3)z^2 + ... + a_{n}z^n mid < (a_1-a_2) + (a_2-a_3) + (a_3-a_4) + ... + a_{n}$$

The conclusion follows.

I was wondering if one can prove this synthetically, without using complex numbers. Any alternate proof or suggestion in the direction an alternate proof is welcome.

geometry complex-numbers contest-math euclidean-geometry summation-by-parts

share|cite|improve this question

edited Mar 20 at 19:12

Anubhab Ghosal

asked Mar 20 at 19:04

Anubhab GhosalAnubhab Ghosal

1,23819

$endgroup$

share|cite|improve this question

edited Mar 20 at 19:12

Anubhab Ghosal

asked Mar 20 at 19:04

Anubhab GhosalAnubhab Ghosal

1,23819

share|cite|improve this question

edited Mar 20 at 19:12

Anubhab Ghosal

asked Mar 20 at 19:04

Anubhab GhosalAnubhab Ghosal

1,23819

asked Mar 20 at 19:04

Anubhab GhosalAnubhab Ghosal

1,23819

asked Mar 20 at 19:04

Anubhab GhosalAnubhab Ghosal

1,23819