n-gon function

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n-gon function

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Source: Romanian IMO Team Selection Test TST 1996, problem 1

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#1 h Sep 13, 2005, 10:11 AM • 3 Y

Y by Adventure10, centslordm, Mango247

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function such that for every regular $n$ -gon $A_1A_2 \ldots A_n$ we have $f(A_1)+f(A_2)+\cdots +f(A_n)=0$ . Prove that $f(x)=0$ for all reals $x$ .

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#2 h Oct 1, 2005, 1:42 PM • 2 Y

Y by Adventure10, Mango247

ehsan2004 wrote:

Prove that $f(x)=0$ for all reals $x$ .

You wanted to say " $f(A)=0$ for all points $A$ ", right?

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#3 h Oct 1, 2005, 1:51 PM • 2 Y

Y by Adventure10, Mango247

perfect_radio wrote:

ehsan2004 wrote:

Prove that $f(x)=0$ for all reals $x$ .

You wanted to say " $f(A)=0$ for all points $A$ ", right?

excuse me, my meant was $f(x)\equiv 0$

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#4 h Oct 1, 2005, 1:57 PM • 2 Y

Y by Adventure10, Mango247

Take $A \neq B$ . Let $\ell$ be the perpendicular bisector of $AB$ . Construct a rhombus $ACBD$ , with $C,D \in \ell$ and $\measuredangle DAC = \measuredangle DBC = \dfrac{\pi}{3}$ . This yields $f(A)+f(C)+f(D)=0=f(B)+f(C)+f(D)$ , so $f(A)=f(B)=t$ , $\forall A \neq B$ .

Therefore $nt=0$ , so $t=0$ .

Have I done something wrong? It looks too good to be true

(because I used the condition given only for $n=3$ )

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enescu

#5 h Oct 3, 2005, 3:36 PM • 2 Y

Y by Adventure10, Mango247

Actually, $n$ is fixed in the original statement.

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#6 h Oct 3, 2005, 6:10 PM • 2 Y

Y by Adventure10, Mango247

enescu wrote:

Actually, $n$ is fixed in the original statement.

Oops... sorry

. do you know the solution for $n \geq 4$ ?

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enescu

#7 h Oct 3, 2005, 6:27 PM • 7 Y

Y by Batominovski, wateringanddrowned, Adventure10, Upwgs_2008, Mango247, and 2 other users

Yes. Let $A$ be an arbitrary point. Consider a regular $n-$ gon $AA_{1}A_{2}\ldots A_{n-1}.$ Let $k$ be an integer, $0\leq k\leq n-1.$ A rotation with center $A$ of angle $\dfrac{2k\pi}{n}$ sends the polygon $AA_{1}A_{2}\ldots A_{n-1}$ to $A_{k0}A_{k1}\ldots A_{k,n-1},$ where $A_{k0}=A$ and $A_{ki}$ is the image of $A_{i}$ , for all $i=1,2,\ldots,n-1.$

From the condition of the statement, we have
$\sum_{k=0}^{n-1} \sum_{i=0}^{n-1}{f(A_{ki})}=0.$
Observe that in the sum the number $f(A)$ appears $n$ times, therefore
$nf(A)+\sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=0.$
On the other hand, we have
$\sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=\sum_{i=1}^{n-1} \sum_{k=0}^{n-1}{f(A_{ki})}=0,$
since the polygons $A_{0i}A_{1i}\ldots A_{n-1,i}$ are all regular $n-$ gons. From the two equalities above we deduce $f(A)=0,$ hence $f$ is the zero function.

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#8 h Jun 30, 2021, 8:39 AM • 1 Y

Y by MS_asdfgzxcvb

We may assume $n$ is even, since for $n$ odd, the sum of the vertices in any $2n$ -gon is zero.

Now let $A_1\cdots A_n$ be a regular $n$ -gon. For each $i$ and $j$ , let $M_{ij}$ be the midpoint of $\overline{A_iA_j}$ (so in particular, $M_iM_i=A_i$ ), and let $O$ be the center of the $n$ -gon.

We know since $M_{i1}M_{i2}\cdots M_{in}$ and $M_{1,1+i}M_{2,2+i}\cdots M_{n,n+i}$ are regular $n$ -gons that $\begin{align*} 0=\sum_i\sum_jf(M_{ij}) =n\cdot f(O)+\sum_j\sum_{\substack{i<n\\ i\ne n/2}}f(M_{j,j+i}) &=n\cdot f(O) \end{align*}$

This post has been edited 1 time. Last edited by TheUltimate123, Jun 30, 2021, 8:40 AM

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#9 h Aug 24, 2021, 7:01 PM

Y by

ehsan2004 wrote:

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function such that for every regular $n$ -gon $A_1A_2 \ldots A_n$ we have $f(A_1)+f(A_2)+\cdots +f(A_n)=0$ . Prove that $f(x)=0$ for all reals $x$ .

The claim for just $n=4$ :
https://aops.com/community/p1703551

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#10 h Wednesday at 10:09 AM

Y by

For $n$ even, draw a lot of $n$ -gons with diametres the segments through the respective vertices and the centre of some $n$ -gon, and the calculation works out to give $f(\text{ centre }) = 0$ , so we're done.
For $n$ odd, notice that the statement then holds for $2n$ as well, so we're done.

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