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http://www.mathlinks.ro/Forum/viewtopic.php?p=336213#336213 if $a>b>c$, there are two segments $PQ,P'Q$, with equal lenghts: $P,P' \in a$ $Q,Q' \in b$ $CP=CQ'$ $CP'=CQ$ so, the segments $PQ,P'Q'$ are symmetric through the bisector of $\angle C$

For $D\in \{A,B,C\}$ I note $DP=x$, $DQ=y$ and $\frac 12 xy\sin D=\frac 12 S$, i.e. $xy=k$ (constant). But $PQ^2=x^2+y^2-2xy\cos D$. Therefore, $PQ$ is minimum $\Longleftrightarrow x^2+y^2$ is minimum ($xy$ - constant) $\Longleftrightarrow$ $x=y=\sqrt{\frac{S}{\sin D}}$ a.s.o.

Nice and quick proof! Why should we check three cases? We don't want to find a minimum segment for every angle, but for the entrire triangle. I think that the smallest segment corresponds to the smallest angle of the triangle

YES ! Prove easily that $\min PQ=\min_{D\in \{A,B,C\}} \sqrt{2S\tan \frac D2}=\sqrt{2S\tan \frac{\min \{A,B,C\}}{2}}.$ Remark. In my opinion, although this problem is nicely, the its level is ** (from ******).