Question

Let n> 2 be an integer and $f:P\to R$ a function defined on the set of points in the plane, with the property that for any regular n-gon $A_1{A}_2...A_n,$ $f\left(A_1\right)+f\left(A_2\right)+...+F\left(A_n\right)=0.$ Prove that f is the zero function.

• A. f is zero function .
• B. f is not a zero function.
• C. f is indefinite function .
• D. f is not a function .

Answer 1

The correct option is A f is zero function .

Let A be an arbitrary point. Consider a regular n-gon $A{A}_1{A}_2...A_{n-1}$.Let k be an integer $0\le k\le n-1.$.A rotation with center A of angle $\frac{2k\pi }{n}$ sends the polygon $A{A}_1{A}_2...A_{n-1}$ to $A_{k0}{A}_{k1}...A_{k,n-1,}$ where $A_{k0}=A$ and $A_{ki}$ is the image of $A_{ki}$ or all $i=1,2,...,n-1$. From the condition of the statement, we have $\sum _{k=0}^{n-1}\sum _{i=0}^{n-1}f\left(A_{ki}\right)=0.$ Observe that in the sum the number f(A) appears n times, therefore $nf\left(A\right)+\sum _{k=0}^{n-1}\sum _{i=0}^{n-1}f\left(A_{ki}\right)=0$.On the other hand, we have $\sum _{k=0}^{n-1}\sum _{i=0}^{n-1}f\left(A_{ki}\right)=\sum _{i=0}^{n-1}\sum _{k=0}^{n-1}f\left(A_{ki}\right)=0,$ since the polygons $A_{0i}{A}_{1i}...A_{n-1,i}$ are all regular n-gons. From the two equalities above we deduce $f\left(A\right)=0$, hence $f$ is the zero function.