Solution by dr Kortram for the christmas problem.

For the figures see HERE .

Problem 1: A triangle ABC with area 1 has a circumscribed circle with radius 1. Locate a point P on or inside the circumscribed circle such that PA+PB+PC is minimal, and determine this minimum value.

Problem 2: A triangle ABC with area 1 has a circumscribed circle with radius 1. Locate a point P on or inside the circumscribed circle such that PA+PB+PC is maximal, and determine this maximum value.

For preparation, I first investigate which triangles are possible. One appears at first sight: the rectangular one with hypotenusa equal to 2 and two sides of equal length (fig 1).
Proposition: all other possible triangles are acute.
Proof: (see fig 2) Take one of the vertices, for instance C, and let this one be at the top. The base AB has a length which is smaller than the diameter of the circle, ie AB is smaller than 2. So the height is greater than 1. Then the arc which angle C supports is smaller than pi, therefore angle C is smaller than 90 degrees.

I can describe the coordinates of the vertices of the possible triangles as follows:
Let the circumscribed circle have center (0,0). There is a t in the interval [0,pi/2) (even in a smaller interval [0,a]) such that (see fig 3)
A=(-cos(t),-sin(t)), B=(cos(t),-sin(t)).
For C there remain only at most two possibilities. Since base*height=2, we readily find that
C=(...,1/cos(t)-sin(t)), and because the distance from C to (0,0) is 1 we have
C=( plus or minus squareroot(1-(1/cos(t)-sin(t))2), 1/cos(t)-sin(t)).

I will use the following theorem of Ptolemy from classical geometry (see figure 4):
If a quadrangle ABCD is inscribed in a circle then AC.BD = AB.CD + AD.BC.
If ABCD is not inscribed in a circle, then AC.BD is smaller than AB.CD + AD.BC.

The most important ingredient for the solution of problem 1 is the following proposition: (see fig 5):

Given a triangle ABC, I construct on the sides equilateral triangles ACB' and CBA' like in the figure. Then BB' = AA', and the intersection point T of AA' and BB' has the following property: for every point P which doesn't coincide with T, PA+PB+PC is greater than TA+TB+TC.

Proof (fig 5): under rotation with center C over 60 degrees, triangle ACA' is transformed into triangle B'CB. Hence you see that AA' and BB' have equal lengths and that the smallest angle between them is 60 degrees.
Since the angles ATB' and ACB' are both 60 degrees, quadrangle AB'CT is inscibed in a circle. Because AC=CB'=B'A, the theorem of Ptolemy simply says: TB'=TA+TC; and for points P that are not lying on the circle through A,C and B', PB' is smaller than PA+PC. If P does lie on that circle, then PB'=PA+PC.
Now TA+TB+TC=TB'+TB=B'B and
BB' is smaller than or equal to PB'+PB, which is smaller than or equal to PA+PC+PB.
The first equality holds only if P lies on BB', the second one only if P lies on the circle through A,C,B'.
So TA+TC+TB is smaller than PA+PC+PB for all P that don't coincide with T.

Of course, from the fact that TA+TB+TC is minimal, it follows that T lies on the straight line CC' (C' is such that triangle BAC' is equilateral).

Solution of problem 1 (figure 6):

For the rectangular triangle ABC with two equal sides we have CC'=1+squareroot(3).
For an acute triangle we have
(CC')2 = 1-(1/cos(t)-sin(t))2 + (1/cos(t)+sqrt(3)cos(t))2 = 4cos2(t) + 2tan(t) + 2sqrt(3)
= (1+sqrt(3))2 + 2tan(t)(1-sin(2t)), and this is greater than (1+sqrt(3))2.
For a minimum, we have to locate A,B and C such that they have coordinates (-1,0),(1,0) and (0,1). We let P be the point (0.wortel(3)/3). The minimal sum is then 1+sqrt(3).


For problem 2, I do first some preparation.
If triangle ABC is acute, and H' is the point of reflection of the orthocenter H of triangle ABC under reflection in side AB, then triangle AH'B has the same circumscribed circle as triangle ABC (see fig 7).

Proof: angle H'AB = angle HAB = 90 degrees - beta = angle H'CB, so quadrangle H'ACB is inscribed in a circle.

Every point P inside triangle ABC lies in at least one of the triangles AHB,BHC and CHA, and a reflection in an appropriate side maps P to a point P' inside the circumscribed circle of triangle ABC. Remark that PA+PB=P'A+P'B and that PC is smaller than or equal to P'C (fig 8).
For the solution of problem 2 we only have to look at points on the sides or outside the triangle, and in figure 9 we see that we can restrict ourselves to points on the circumference of the circle (Q lies farther away from each vertex than P).

Choose P on the circle. There are t and x with
0 smaller than or equal to t, t smaller than or equal to x, x smaller than or equal to pi/2,
such that (fig 10)

A=(-cos(t),-sin(t)), B=(cos(t),-sin(t)), P=(cos(x),-sin(x)).
PA+PB = sqrt((cos(x)+cos(t))2+(sin(x)-sin(t))2) + sqrt((cos(x)-cos(t))2+(sin(x)-sin(t))2) =
sqrt(2+2cos(x+t))+sqrt(2-2cos(x-t)) = 2cos((x+t)/2)+2sin((x-t)/2) = 4sin(pi/4-t/2)cos(pi/4-x/2).
This is maximal if x=pi/2, so PA+PB is smaller than or equal to 4sin(pi/4-t/2).
When we choose the triangle such that t=0, then we see that
PA+PB is smaller than or equal to 2sqrt(2), and since PC is smaller than or equal to 2 (diameter of the circle) we find that
PA+PB+PC is smaller than or equal to 2sqrt(2)+2.
We do have this value in the rectangular triangle with two equal sides.
A=(-1,0),B=(1,0),C=(0,1); choose P=(0,-1).