Given a triangle with vertices $V_1$, $V_2$ and $V_3$, what is the point inside the triangle that is farthest away from any vertex (i.e. the point which maximizes the minimum distance to each vertex).
My first thought is that it is the circumcenter for acute triangles and the midpoint of the longest edge for obtuse triangles, but I can't seem to prove this.
Best Answer
When triangle $\triangle ABC$ is acute, the circumcenter $O$ is the point that maximize the minimal distance.
Let $R = |OA| = |OB| = |OC|$ be the circumradius.
Let $A', B', C'$ be the mid-points of sides $BC$, $CA$ and $AB$ respectively.
Draw six lines segments from $O$ to $A,B,C, A',B',C'$. They will divide triangle $ABC$ into six right angled triangles $\triangle OAC'$, $\triangle OC'B$, $\triangle OBA'$, $\triangle OA'C$, $\triangle OCB'$ and $\triangle OB'A$.
Consider the two triangle $\triangle OAC'$ and $\triangle OB'A$ which are sharing the segment $OA$.
Since $\angle AC'O = \angle OB'A = 90^\circ$, both triangles lie completely inside the circle with $OA$ as diameter (the green circle in above diagram). Since this circle is lying within the circle with $A$ as center passing through $O$ (the blue circle), every point $P$ belongs to triangle $\triangle OAC'$ and $\triangle OB'A$ satisfies $$|PA| \le |OA| = R \quad\implies\quad \min(|PA|, |PB|, |PC|) \le R$$
Similar thing happen to the two pairs of triangles $\triangle OC'B, \triangle OBA'$ and $\triangle OA'C, \triangle OCB'$.
From this, we can conclude
$$\min(|PA|,|PB|,|PC|) \le R\quad\text{ for all points } P \text{ in triangle }ABC$$
When $P = O$, above inequality becomes an equality. This implies
$$\max_{P \in \triangle ABC}\left(\min(|PA|,|PB|,|PC|)\right) = R$$
and $O$ is the point that maximize the minimal distance.
Update
Let us switch to case where triangle $ABC$ is obtuse.
WOLOG, we will assume $a = |BC| > b = |CA| \ge c = |AB|$.
Construct the perpendicular bisector of side $CA$, let it intersect $BC$ at $B''$.
Construct the perpendicular bisector of side $AB$, let it intersect $BC$ at $C''$.
These two perpendicular bisectors split $\triangle ABC$ into 3 regions, two triangles $\triangle BC''B$ (the red one), $\triangle CB'B''$ (the green one) and a pentagon $AC'C''B''B'$ (the blue one).
For any $P$ inside $\triangle ABC$, it is easy to see:
$$ \begin{array}{rcl} P \in \triangle BC''B & \implies & |PB| \le |PA|,|PC|\\ P \in AC'C''B''B' & \implies & |PA| \le |PB|, |PC|\\ P \in \triangle CB'B'' & \implies & |PC| \le |PA|, |PB \end{array} $$
Since the distance functions $|PA|$, $|PB|$, $|PC|$ are convex, their maximum will be achieved at the vertices.
Since both $\triangle BC''C'$ and $\triangle CB'B''$ are right angled triangles, we have $|BC''| > |BC'|$ and $|CB''| > |CB'|$. Together with the obvious equalities, $|AC'| = |BC'|$, $|AC''| = |BC''|$, $|AB'| = |CB'|$, $|AB''| = |AB''|$, there are only two possiblities. The maximum minimum distance is either
where $\beta = \angle C'BC''$ and $\gamma = B''CB'$.
Using sine rule, we have
$$|CB''| : |BC''| = \frac{b}{2\sin\gamma} : \frac{c}{2\sin\beta} = \frac{b}{c} : \frac{c}{b}$$
Since $b \ge c$, we have $|CB''| \ge |BC''|$. This confirms Joe Knapp's conjecture. The maxmium minimum distance is achieved at $B''$, the intersection of the perpendicular bisector of the second longest side with the longest side.
Using the relation $c = 2R\sin\gamma$, one find the maximum minimal distance equals to $\displaystyle\;\frac{bR}{c}\;$ in this case.