I was trying to directly prove the Kakutani's Fixed Point Theorem thorough Sperner's Lemma (I still am) when the following question came to me:
Let $\triangle$ be an equilateral triangle and $P$ be it's center point. Is there a way to partition $\triangle$ into three congruent regions such that "if a convex set $S$ inside $\triangle$ intercepts all three regions, then $S$ must contain $P$"?
At first I thought, if $\triangle = \triangle ABC$, about using the regions $\triangle ABP, \triangle BCP, \triangle CAP$, but that doesn't quite work, as shown...
No, it is not possible, under a much weaker assumption: it's impossible even if all we know is that all three regions have positive area.
Let $D,E,F$ be arbitrary points from three different regions, chosen so that no three of $P, D, E, F$ are collinear. (This is actually the only place where we need assumptions about the regions; this does not work if the union of two regions is entirely contained in a line through $P$.) Point $P$ must be contained in $\triangle DEF$, or else $\triangle DEF$ is a counterexample convex set.
Let $PD$ intersect $EF$ at $X$, $PE$ intersect $DF$ at $Y$, and $PF$ intersect $DE$ at $Z$. Then:
In particular, the interior of quadrilateral $DYPZ$ (the intersection of $\triangle DEY$ and $\triangle DFZ$) must be entirely from the same region as $D$.
Similarly, the interior of quadrilateral $EXPZ$ must be entirely from the same region as $E$, and the interior of quadrilateral $FXPY$ must be entirely from the same region as $F$.
For all $\epsilon>0$, the set of all points within distance $\epsilon$ of segment $XY$ intersects all three quadrilaterals (and therefore all three regions), and we can choose $\epsilon$ small enough that it does not contain $P$.