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Difference between revisions of "1990 IMO Problems/Problem 6"
 
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We first commit to placing <math>1^2</math> and <math>2^2</math> on opposite sides, <math>3^2</math> and <math>4^2</math> on opposite sides, etc. Since <math>2^2-1^2=3</math>, <math>4^2-3^2=7</math>, <math>6^2-5^2=11</math>, etc., this means the desired conclusion is equivalent to<cmath> 0 = \sum_{n=0}^{994} c_n \omega^n </cmath>being true for some permutation <math>(c_0, \dots, c_{994})</math> of <math>(3, 7, 11, \dots, 3981)</math>.
 
We first commit to placing <math>1^2</math> and <math>2^2</math> on opposite sides, <math>3^2</math> and <math>4^2</math> on opposite sides, etc. Since <math>2^2-1^2=3</math>, <math>4^2-3^2=7</math>, <math>6^2-5^2=11</math>, etc., this means the desired conclusion is equivalent to<cmath> 0 = \sum_{n=0}^{994} c_n \omega^n </cmath>being true for some permutation <math>(c_0, \dots, c_{994})</math> of <math>(3, 7, 11, \dots, 3981)</math>.
  
−
Define <math>z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}</math>. Then notice thatz=3ω0+7ω199+11ω398+15ω597+19ω796ω5z=23ω5+27ω204+31ω403+35ω602+39ω801ω10z=43ω10+47ω209+51ω408+55ω607+59ω806\vdotswithin=and so summing yields the desired conclusion, as the left-hand side becomes<cmath> (1+\omega^5+\omega^{10}+\cdots+\omega^{990})z = 0 </cmath>and the right-hand side is the desired expression.
+
Define <math>z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}</math>. Then notice that <cmath>z=3ω0+7ω199+11ω398+15ω597+19ω796ω5z=23ω5+27ω204+31ω403+35ω602+39ω801ω10z=43ω10+47ω209+51ω408+55ω607+59ω806\vdotswithin=</cmath>and so summing yields the desired conclusion, as the left-hand side becomes<cmath> (1+\omega^5+\omega^{10}+\cdots+\omega^{990})z = 0 </cmath>and the right-hand side is the desired expression.
  
 
This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [https://aops.com/community/p17584021]
 
This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [https://aops.com/community/p17584021]
  
 
== See Also == {{IMO box|year=1990|num-b=5|after=Last Question}}
 
== See Also == {{IMO box|year=1990|num-b=5|after=Last Question}}

Latest revision as of 13:54, 30 January 2021

Problem

Prove that there exists a convex $1990\text{-gon}$ with the following two properties: (a) All angles are equal. (b) The lengths of the $1990$ sides are the numbers $1^2, 2^2,3^2,\dots, 1990^2$ in some order.

Solution 1

Let $\{a_1,a_2,\cdots,a_{1990}\}=\{1,2,\cdots,1990\}$ and $\theta=\frac{\pi}{995}$. Then the problem is equivalent to that there exists a way to assign $a_1,a_2,\cdots,a_{1990}$ such that $\sum_{i=1}^{1990}a_i^2e^{i\theta}=0$. Note that $e^{i\theta}+e^{i\theta+\pi}=0$, then if we can find a sequence $\{b_n\}$ such that $b_i=a_p^2-a_q^2, (p\not=q)$ and $\sum_{i=1}^{995}b_ie^{i\theta}=0$, the problem will be solved. Note that $2^2-1^2=3,4^2-3^2=7,\cdots,1990^2-1989^2=3979$, then $3,7,\cdots,3979$ can be written as a sum from two elements in sets $A=\{3,3+4\times199,3+4\times398,3+4\times597,3+4\times796\}$ and $B=\{0,4\times1,4\times2,\cdots,4\times 198\}$. If we assign the elements in $A$ in the way that $l_{a1}=3,l_{a2}=3+4\times199,l_{a3}=3+4\times398,l_{a4}=3+4\times597,l_{a5}=3+4\times796$, then clearly $\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}=0, (j\in\{1,2,3,4,5\})$ Similarly, we could assign elements in $B$ in that way ($l_{b1}=0,l_{b2}=4,\cdots$) to $e^{i\frac{2\pi}{5}}$. Then we make $b_i$ according to the previous steps. Let $i\equiv j$ (mod 5), $i\equiv k$ (mod 199), and $b_i=l_{aj}+l_{bk}$, then each $b_i$ will be some $a_ p^2-a_q^2$ and $\sum_{i=1}^{995}b_ie^{i\theta}=\sum_{j=1}^5\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}+\sum_{k=1}^{199}\sum_{i=1}^{5}l_{bk}e^{i\frac{2\pi}{5}}=0$ And we are done.

This solution was posted and copyrighted by YCHU. The original thread for this problem can be found here: [1]

Solution 2

Throughout this solution, $\omega$ denotes a primitive $995$th root of unity.

We first commit to placing $1^2$ and $2^2$ on opposite sides, $3^2$ and $4^2$ on opposite sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent to\[0 = \sum_{n=0}^{994} c_n \omega^n\]being true for some permutation $(c_0, \dots, c_{994})$ of $(3, 7, 11, \dots, 3981)$.

Define $z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$. Then notice that \begin{align*} z &= 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796} \\ \omega^5 z &= 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801} \\ \omega^{10} z &= 43 \omega^{10} + 47 \omega^{209} + 51\omega^{408} + 55 \omega^{607} + 59 \omega^{806} \\ &\vdotswithin{=} \end{align*}and so summing yields the desired conclusion, as the left-hand side becomes\[(1+\omega^5+\omega^{10}+\cdots+\omega^{990})z = 0\]and the right-hand side is the desired expression.

This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]

See Also

1990 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Question
All IMO Problems and Solutions
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