Prove that there exists a convex 1990-gon

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Prove that there exists a convex 1990-gon

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Source: IMO 1990, Day 2, Problem 6, IMO ShortList 1990, Problem 16 (NET 1)

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orl

#1 h Nov 11, 2005, 6:55 PM • 3 Y

Y by manuel153, Adventure10, Mango247

Prove that there exists a convex 1990-gon with the following two properties :

a.) All angles are equal.
b.) The lengths of the 1990 sides are the numbers $1^2$ , $2^2$ , $3^2$ , $\cdots$ , $1990^2$ in some order.

This post has been edited 1 time. Last edited by orl, Aug 15, 2008, 4:18 PM

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#2 h Nov 11, 2005, 7:47 PM • 2 Y

Y by Adventure10, Mango247

I think a nice idea is to consider the 1990-th roots of unity...and after that consider the vectors with sizes $1^2,2^2,\ldots,1990^2$ and multiples of the roots of unity. Drawing it would be easier to show the idea...=]

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Erken

#3 h Jun 4, 2008, 9:43 AM • 1 Y

Y by Adventure10

Ilthigore wrote:

Is there an official solution that doesn't use complex numbers? Were they 'IMO syllabus' back in 1990? :S

Actually,there is an unofficial solution that doesn't use complex numbers,but it is much longer and a bit ugly.

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orl

#4 h Aug 15, 2008, 3:48 PM • 2 Y

Y by Adventure10, Mango247

Erken wrote:

Actually,there is an unofficial solution that doesn't use complex numbers,but it is much longer and a bit ugly.

How about writing it down here please?

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YCHU

#5 h Dec 11, 2010, 12:35 AM • 2 Y

Y by Adventure10, Mango247

Use Complex numbers.
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.

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Nguyen

#6 h Mar 25, 2020, 9:23 AM

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Set f(x)=x*(x-1)*(x^10-1)/(x^2-ε(5))*(x^1990-1)/(x^10-ε(199)).
Here, ε(n) denotes e^(2πi/n).

We can claim that:
f(x) has 1990 terms with degree 1 through 1990.
The coefficients are a permutation of all 1990-th roots of unity.
f(x) is divisible by (x-1)^3.

Therefore, f(1)=f''(1)=0, summing them yields the answer.

Generalization: if n has at least s+1 distinct prime factors, then such a polygon with sides 1^s, 2^s, ..., n^s exists.

This post has been edited 1 time. Last edited by Nguyen, Mar 25, 2020, 9:25 AM

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#7 h Sep 10, 2020, 7:37 PM • 5 Y

Y by Limerent, Mathematicsislovely, v4913, chystudent1-_-, HamstPan38825

Solution from Twitch Solves ISL:

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 post has been edited 1 time. Last edited by v_Enhance, Sep 10, 2020, 7:37 PM

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#8 h Jan 11, 2022, 2:18 PM • 1 Y

Y by David-Vieta

I like the first part of this answer:

v_Enhance wrote:

Solution from Twitch Solves ISL:
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 $(3, 7, 11, \dots, 3981)$ .

But I can't follow the 2nd part, especially how we can get
$\omega^5 z = 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801}$
from
$z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$

So this is my version for 2nd part:

We have to prove that $\sum_{k=0}^{994} (4d_k-1) \omega^k = 0$ where $d_k \to \{1,2,3,4...,995\}$ and $\omega=\cos \frac{2\pi}{995} + i \sin \frac{2\pi}{995}$

(Define $\omega$ as a primitive 995th root of unity is also good, but I avoid it because there are many primitive 995th roots of unity)

Because $\omega^k=(\cos \frac{2\pi}{995} + i \sin \frac{2\pi}{995})^k=\cos \frac{2k\pi}{995} + i \sin \frac{2k\pi}{995}$
We can show that $\omega^{995}=1$ and $\sum_{k=0}^{994}\omega^k = 0$
Because $995=5 \cdot 199$ we can also show that $\sum_{k=0}^{4}\omega^{199k} = 0$ and $\sum_{k=0}^{198}\omega^{5k} = 0$

Let's make our equation simpler:
$\sum_{k=0}^{994} (4d_k-1) \omega^k = 4 \sum_{k=0}^{994}d_k \omega^k - \sum_{k=0}^{994} \omega^k=0 \iff \sum_{k=0}^{994}d_k \omega^k=0$

Now we split the sum into 5 smaller sums:
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}d_{5k} \omega^{5k} + \sum_{k=0}^{198}d_{5k+1} \omega^{5k+1} + \sum_{k=0}^{198}d_{5k+2} \omega^{5k+2} + \sum_{k=0}^{198}d_{5k+3} \omega^{5k+3} + \sum_{k=0}^{198}d_{5k+4} \omega^{5k+4}$

Here is the trick: we define that $d_{k+995j}=d_{k}$ (just remember that $d_k$ s are the sides of 995-got) and because $\gcd(5,199)=1$ , we can rewrite our group of sums to this:
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}d_{5k} \omega^{5k} + \sum_{k=0}^{198}d_{5k+199} \omega^{5k+199} + \sum_{k=0}^{198}d_{5k+2\cdot 199} \omega^{5k+2 \cdot 199} + \sum_{k=0}^{198}d_{5k+3 \cdot 199} \omega^{5k+3 \cdot 199} + \sum_{k=0}^{198}d_{5k+4 \cdot 199} \omega^{5k+4 \cdot 199}$

So:
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k} \omega^{5k} + d_{5k+199} \omega^{5k+199} + d_{5k+2\cdot 199} \omega^{5k+2 \cdot 199} + d_{5k+3 \cdot 199} \omega^{5k+3 \cdot 199} +d_{5k+4 \cdot 199} \omega^{5k+4 \cdot 199})$
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}+ d_{5k+199} \omega^{199} + d_{5k+2\cdot 199} \omega^{2 \cdot 199} + d_{5k+3 \cdot 199} \omega^{3 \cdot 199} +d_{5k+4 \cdot 199} \omega^{4 \cdot 199}) \omega^{5k}$

Now if we choose :
$d_{5k} \to \{1,6,11,...,198\cdot 5+1\}$
$d_{5k+199}=d_{5k}+1$
$d_{5k+2\cdot 199}=d_{5k}+2$
$d_{5k+3\cdot 199}=d_{5k}+3$
$d_{5k+4\cdot 199}=d_{5k}+4$

So:
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}+ (d_{5k}+1) \omega^{199} + (d_{5k}+2) \omega^{2 \cdot 199} + (d_{5k}+3) \omega^{3 \cdot 199} +(d_{5k}+4) \omega^{4 \cdot 199}) \omega^{5k}$
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}(1+\omega^{199} + \omega^{2 \cdot 199} + \omega^{3 \cdot 199} + \omega^{4 \cdot 199})+(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})) \omega^{5k}$
$\sum_{k=0}^{994}d_k \omega^k=\sum_{k=0}^{198}(d_{5k}\cdot 0+(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})) \omega^{5k}$
$\sum_{k=0}^{994}d_k \omega^k=(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})\sum_{k=0}^{198} \omega^{5k}$
$\sum_{k=0}^{994}d_k \omega^k=(\omega^{199}+2\omega^{2\cdot 199}+3 \omega^{3 \cdot 199}+4\omega^{4 \cdot 199})\cdot 0$
$\sum_{k=0}^{994}d_k \omega^k= 0$ Q.E.D

This post has been edited 1 time. Last edited by andib2n, Jan 11, 2022, 2:37 PM
Reason: typo in equation

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#9 h Oct 22, 2023, 2:41 AM

Y by

Solved with ARCH hints and Evan's stream

We use complex vectors to represent the sides of the 1990-gon. Rather than utilizing the 1990th roots of unity, since 1990 is 2 mod 4, we can add up the vectors of opposite sides in jumps of 2 and use the 995th roots of unity.

Let $w$ be a 995th root of unity. Pairing consecutive perfect squares $(2k-1)^2$ and $(2k)^2$ as opposite sides, we require
$\sum_{k=1}^{995} a_n \cdot w^k = 0$
where the $a_n$ are of the form $(2m)^2 - (2m-1)^2 = 4m-1$ . We finish using a clever arrangement of $a_n$ . $\blacksquare$

andib2n wrote:

I like the first part of this answer:

v_Enhance wrote:

Solution from Twitch Solves ISL:
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 $(3, 7, 11, \dots, 3981)$ .

But I can't follow the 2nd part, especially how we can get
$\omega^5 z = 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801}$
from
$z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}$

The expression for $\omega^5 z$ follows from $20\omega^{5} + 20\omega^{204} + 20\omega^{403} + 20\omega^{602} + 20\omega^{801}$ being $20\omega^5$ times the sum of the fifth roots of unity, or 0. The remaining expressions follow analogously.

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#10 h Dec 3, 2023, 8:29 PM

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In order to construct the wanted polygon, we'll deal with complex vectors for each of the sides of the polygon, where each pair of opposite sides is of the form $(4i^2,(2i-1)^2)$ .

Let $z$ be a $995-$ th root of unity and let $d_i = 4i^2 = 4i^2 - (2i-1)^2 = 4i-1$ . We want to find a permutation of the $s_i$ 's for which the following holds :
$\sum_{i=0}^{994} s_i \cdot z^i = 0$
Since $995 = 5 \times 199$ , then $z^{199}$ is a $5-$ th root of unity, thus $20 z^{5i} (z^0 + z^{199}+z^{398}+z^{597}+z^{796} )=0$ for all $0\geq i \geq 198$ $...(**)$ .

Now by recursively combining property $(**)$ with the fact that $t = 3z^0 + 7z^{199} + 11z^{398} + 15z^{597} + 19z^{796}$ , we can compute all of the $z^{5i}t$ to prove that the desired sum is no more than
$t\times \left(\sum_{i=0}^{198}z^{5i}\right) = 0$

$\mathbb{Q.E.D.}$

This post has been edited 1 time. Last edited by Cusofay, Dec 3, 2023, 8:30 PM
Reason: .

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#11 h Aug 17, 2024, 6:47 PM • 1 Y

Y by dolphinday

Suppose that $\omega$ is a $995$ th root of unity, and place the sides $(2k-1)^2$ and $(2k)^2$ opposite of each other, so that we only need to find a permutation $(c_0, c_1, \dots,c_{994})$ of $(3,7,\dots, 3975)$ such that

$\sum_{i=0}^{994} c_i \omega^i = 0.$
However, the following construction suffices. $\square$

$\begin{align*} &\phantom{+} 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796} \\ &+ 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801} \\ &+ 43 \omega^{10} + 47 \omega^{209} + 51\omega^{408} + 55 \omega^{607} + 59 \omega^{806} \\ &\vdots \end{align*}$

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#12 h Mar 26, 2025, 2:36 PM

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Solution

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#13 h Mar 27, 2025, 1:58 PM

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Does this hold for n-gons for n = 0 mod 4, if not, why?

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