A geometry and algebra problem — RMO training

Several Olympiad problems deal with functions defined on certain sets of points. These problems are interesting in that they combine both geometrical and algebraic ideas.


Let n>2 be an integer and f:P \rightarrow \Re a function defined on the set of points in the plane, with the property that for any regular n-gon A_{1}A_{2} \ldots A_{n}

f(A_{1})+f(A_{2})+ \ldots + f(A_{n})=0.

Prove that f is the zero function.


Core Concept: 

In Euclidean geometry, the only motions permissible are rigid motions — translations, rotations, and reflections.


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 \frac{2\pi k }{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_{kl})=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.

More later,

Nalin Pithwa



Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s