# Nearest point of the vertices of a triangle

I would like to know if the point closest to the three vertices of an equilateral triangle is the centre of its circumcircle, and if so, how to prove it.

By closest point, I mean that any other point has at least one distance to one of the vertices of the triangle greater than the radius of the circumcircle, i.e, if a point is not the centre of circumcircle, I want to show that the distance of that point to one of the vertices is greater than the distance of the centre of the circumcircle to the vertices.

• Define "closest to the three vertices". You mean the minimum of the sum of the those three distances (the distances from the special point to each vertex)? Apr 22, 2022 at 8:16
• By closest point, I mean that any other point has at least one distance to one of the vertices of the triangle greater than the radius of the circumcircle, i.e, if a point is not the centre of circumcircle, I want to show that the distance of that point to one of the vertices is greater than the distance of the centre of the circumcircle to the vertices. Apr 22, 2022 at 8:19
• By symmetry, the closest point is on the three medians and so is the center.
– lhf
Apr 22, 2022 at 10:08

Draw a circle from each vertex which involves the circumcentre. As any two circles intersect each other exactly twice, and the circumcentre is therefore three of these intersections, the mutual intersection of the three circles is exactly the circumcentre.

Any other point therefore lies outside a circle.

• How does this show that the circumcenter is the "closest point" to the vertices? The things you have written also hold for obtuse triangles, in which case the circumcentre is not the "closest point" Apr 22, 2022 at 8:49
• One need additionally to prove that circumcenter is the only common point of these three disks (here disk is union of boundary and interior of a circle) for equilateral triangle. This can be shown using tangent lines to circles through circumcenter. Apr 22, 2022 at 9:36

Let $$P$$ be our candidate for the "closest point", $$A$$, $$B$$, $$C$$ the vertices of the triangle and $$M$$ the circumcenter of the triangle.

Let $$d_A = \|A-P\|,$$ $$d_B = \|B-P\|,$$ $$d_C = \|C-P\|,$$ $$d_M = \|M-P\|.$$

Let $$R = \|A-M\| = \|B-M\| = \|C-M\|.$$

With some tedious calculations, we find $$\frac{1}{3} \left(d_A^2+d_B^2+d_C^2\right) = d_M^2 + R^2$$ (holds in equilateral triangles, only). This obviously means that the largest of the three values $$d_A^2,$$ $$d_B^2,$$ $$d_C^2$$ is greater than or equal to $$d_M^2+R^2.$$ If they all were smaller than $$d_M^2+R^2,$$ the equation could not hold. Therefore, $$\max(d_A,d_B,d_C)\geq R,$$ independently of the location of $$P$$. But we already know one configuration with $$d_A=d_B=d_C=R,$$ specifically the one in which $$P=M.$$ As we have shown that we cannot do better than this, this is the optimal solution.

We even know that this solution is unique. The equation above shows us that $$d_A=d_B=d_C=R$$ cannot hold if $$P\neq M,$$ because $$P\neq M$$ implies $$d_M^2\neq 0,$$ which in turn implies that one of $$d_A^2,$$ $$d_B^2,$$ $$d_C^2$$ must be greater than (not equal to) $$R^2.$$

So according to your definition the closest point should have distasnce from each vertice at most the radius the circumcircle. If we draw the 3 circles with centers each vertice and radius $$\frac{a}{\sqrt{3}}$$ (suppose $$a$$ is the lenght of one edge of the equilateral triangle) these 3 circles meet at one exactly one point, the circumcenter. That proves your closest point of definition.