Let $f:[0,1]\to\mathbb R^k$ be a continuous function with $f(1) = \overrightarrow 0$.
Is it true that there always exist $k$ points $0 \le a\_1 \le a\_2 \le \ldots \le a\_k \le 1$ such that $\sum\_{i=1}^k (-1)^{k+1} f(a\_k) = f(a\_1) - f(a\_2) + f(a\_3) - \ldots = \frac{f(0)}{2}$?


When $k=1$ we have to find one point $a\_1 \in [0,1]$ with $f(a\_1) = \frac{f(0)+f(1)}{2}$, which is the intermediate value theorem.


When $k=2$ the situation is more complicated.
We believe the statement is still true.
However, we noted that if we are not finding $\frac{f(0)}{2}$ but, say, $.49 f(0)$ instead, there exist counter-examples.
A counter-example looks like a sine function $f(t) = (t, \sin 100t)$, rotated in the 2-d plane a little bit.
(This is not the example, but you can imagine that changing the constants $.49$ and $100$ a bit makes it work.)


For bigger $k$ we have no idea.


Are there any similar results known before?



A generalization of intermediate value theorem on R^k
The statement is true. It is almost precisely Lemma 2 in the paper D.Burago, "Periodic metrics", Adv. Soviet Math. 9, (1992), 205-210. The proof is short but not easy to invent. The paper can be read on Google Books [here](http://books.google.com/books?id=8f8tky1MoisC&pg=PA207). 


Notes on the text: the intervals in the formulation of Lemma 2 are in fact disjoint, the term "antipodal map" means "odd map" (i.e. $\varphi(-x)=-\varphi(x)$).


For $k=2$, this fact is classic and one can replace 0.5 by any inverse integer ($1/3, 1/4,\dots$) but for every other value there is a counter-example. 


