Here is a solution: Any point $P$ on $OI$ can be represented as $\lambda O+(1-\lambda)I$ for some real $\lambda$ (we can regard this as a relation between the coordinates of $O,I,P$ on the axis $OI$). For any $P$ let $D(P)=D,E(P)=E,F(P)=F$. We then have $D(P)=\lambda D(O)+(1-\lambda) D(I)$ (this relation takes place on $BC$, where we have chosen origin $B$) and the like (i.e. analogous relations for $E(P),F(P)$; call these relations $(*)$), and since $AF(X)+BD(X)+CE(X)=\frac{\ell}2$ for $X\in\{O,I\}$, it means that it also works for any $P\in OI$ because of $(*)$. We must also prove the converse. For any $P$ satisfying the condition there's a $P'\in OI$ satisfying the condition, where $P'=PD(P)\cap OI$ (maybe $PD(P)$ doesn't intersect $OI$ inside $ABC$, but one of $PD(P),PE(P),PF(P)$ does, so we assume WLOG it's the first one). We have $CE(P)+AF(P)=CE(P')+AF(P')$ and $f(P)=f(P')$, where $f(X)=CE(X)^2-(b-CE(X))^2+AF(X)^2-(c-AF(X))^2$, and I guess these two equations give $CE(P)=CE(P'),AF(P)=AF(P')\Rightarrow P=P'$. [Moderator edit: You can also find 99% of the solution of this problem in another MathLinks thread (the rest is trivial).]