proposed problems

Consider all triangles ABC with area equal to 1 that are inscribed in a circle with radius 1.
a) Suppose that the minimum sum of the distances from a point inside or on the triangle to the vertices of the triangle is as small as possible. Prove that the triangle is rectangular and has two sides of equal length.
b) Suppose that the minimum sum of the squares of the distances from a point inside or on the triangle to the vertices of the triangle is as small as possible. Prove that the triangle is rectangular and has two sides of equal length.
c) Suppose that the minimum sum of the distances from a point inside or on the triangle to the sides of the triangle is as small as possible. Prove that the triangle is rectangular and has two sides of equal length.
d) Suppose that the minimum sum of the squares of the distances from a point inside or on the triangle to the sides of the triangle is as small as possible. Prove that the triangle has two equal sides, but is not rectangular.

Solution:

Let a,b,c be the lengths of the sides supported by A,B,C resp.
For t between 0 and pi/4 (inclusive), let B=(-cos(t),-sin(t)), C=(cos(t),-sin(t)), A=(sqrt(1-(1/cos(t)-sin(t))2),1/cos(t)-sin(t)). None of these triangles is obtuse. Any two non-congruent triangles that we consider are among those with these vertices. For t=0 we have b=c, for t=pi/4 we have a=b, and somewhere in between we have sin(2t)*(1+cos(2t))=1 and a=c.
Just two non-congruent triangles among them have two equal sides. One is rectangular (the one with t=0 or t=pi/4) and one is not (the one with a=c).
The latter triangle has two angles equal to pi/4+(1/2)*arcsin(cubroot(19+3*sqrt(33))/3+ cubroot(19-3*sqrt(33))/3-2/3), that is about 73.5324396 degrees.

a) It is wellknown, that the sum of the distances from a point P to the vertices of the triangle is minimal in the point of Torricelli. It is also known that this minimal sum is equal to sqrt((a2+b2+c2)/2+ 2*sqrt(3)*D), where D=1 is the area of the triangle. So we seek a triangle for which f:=a2+b2+c2 is minimal.
We calculate f=8*cos2(t)+4*tan(t), and investigate the sign of the derivative f '.
Then we see that f is minimal for t=0 or t=pi/4.

b) It is wellknown that the sum of the squares of the distances from a point P to the vertices of the triangle is minimal in the triangle centroid. It is also known that this minimal sum is equal to (a2+b2+c2)/3.
So we seek a triangle for which f:=a2+b2+c2 is minimal.
Again, we see that f is minimal for t=0 or t=pi/4.

c) It is wellknown that the sum of the distances to the sides is minimal in a vertex with minimal distance to the side supported by that vertex. Since the area of the triangle is equal to 1, this minimal sum is equal to 2/x, where x is the maximum of a,b,c. Since the triangle is inscribed in a triangle with radius 1, the maximum of a,b,c is 2. If the maximum length of a side is 2, then the triangle is rectangular and has two equal sides.

d) It is wellknown that the sum of the squares of the distances from a point P to the sides of the triangle is minimal in the point of Lemoine. It is also known that this minimal sum is equal to 4D2/(a2+b2+c2), where D=1 is the area of the triangle.
So we seek a triangle for which f:=a2+b2+c2 is maximal.
We calculate f=8*cos2(t)+4*tan(t), and investigate the sign of the derivative f '.
The sign goes from positive to negative in only one point t between 0 and pi/4, and there we have sin(2t)*(1+cos(2t))=1. This latter relation implies equality of the sides a and c supported by A and C (resp.).