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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.

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    $\begingroup$ I think you are right for acute triangles , but wrong for obtuse triangles but also no proof yet $\endgroup$ – Willemien Aug 17 '17 at 19:02
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When triangle $\triangle ABC$ is acute, the circumcenter $O$ is the point that maximize the minimal distance.

An acute triangle

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.

An obtuse triangle

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

  • achieved at $B''$ with value $|AB''| = |CB''| = \frac{b}{2\sin\gamma}$
  • or achieved at $C''$ with value $|AC''| = |BC''| = \frac{c}{2\sin\beta}$.

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.

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  • $\begingroup$ Nice! Any ideas of an approach for the obtuse case? I think @Joe Knapp's conjecture about the point being the intersection of the longest side with the perpendicular bisector of the second longest side may be correct. $\endgroup$ – pmat Aug 17 '17 at 22:59
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    $\begingroup$ @pmat Joe Knapp's conjecture should be correct. The two perpendicular bisectors split the triangle into 3 regions ( two triangles and a pentagon ). one region for each vertex closest to the point $P$. Since the distance function is a convex function on the region it controls, the maximum of minimum distance has to be achieved at a vertex of one of these 3 shapes. It is clear we can rule out all but 2 vertices, i.e. the intersection of the two bisectors with the longest side. If one can verify the intersection with the bisector of the second longest side has larger distance, then we are done. $\endgroup$ – achille hui Aug 18 '17 at 0:06
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One way to determine the point graphically would be to expand circles of equal radius from each vertex until they box in a small area/point.

Here's an example of an obtuse triangle that violates your rule:

enter image description here

The circumcenter formed by the midlines of each side is outside the triangle and the big red point is the furthest point, well away from the midpoint of the longest side. It's interesting that the point is on the midline from the second-longest side, so maybe that's the rule: for obtuse triangles, the desired point is the intersection of the longest side with the midline from the second longest side.

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    $\begingroup$ And if the obtuse triangle is isosceles then there are two such points on the longest side $\endgroup$ – Henry Aug 18 '17 at 1:07
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For an obtuse triangle the midpoint of the longest side is a false solution.

Say you have an isosceles triangle with an obtuse apex angle. The midpoint of the base is so close to the apex that you can slide along the base to increase that distance until you are equally far from the apex and the base vertex you moved towards. That is a maximal distance point to all vertices. Note that it is the intersection of the base with the perpendicular bisector of one of the other sides.

Therein lies a clue for scalene obtuse triangles. With all three sides having different lengths choose the longest side, then construct the perpendicular bisector of the second longest side, and where they intersect is the point you want. See @Joe knapp's answer.

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