Let $A B C$ be a given triangle inscribed in a circle $k$, $P$ be any point in the plane of $A B C$, and let $m$ denote the minimum of the sum $A P+B P+C P$. For which of the triangles inscribed in the circle $k$ will $m$ be maximal?
It is known that if the angles of a triangle are less than $120^{\circ}$, then the point $P$ mentioned in the problem is the so-called isogonal point, from which each side of the triangle is seen at an angle of $120^{\circ}$. If, however, the triangle has an angle of $120^{\circ}$ or larger, for example at vertex $A$, then $P$ coincides with $A$ and $m=AB+AC$. The proof of this can be found on page 240 of István Reiman's book "A geometriai és határterületei" (Gondolat Publishing, 1986). First, let's find the maximum of $m$ in the case where $\angle BAC \geq 120^{\circ}$. In this case, $m=AB+AC$, which can be written using the usual notation as:

$$
m=2 \cdot R \cdot \sin \beta + 2 \cdot R \cdot \sin \gamma = 2 \cdot R \cdot (\sin \beta + \sin \gamma)
$$

Using the fact that $\sin \beta + \sin \gamma = 2 \cdot \sin \frac{\beta + \gamma}{2} \cdot \cos \frac{\beta - \gamma}{2}$, and that $\frac{\beta + \gamma}{2} \leq 30^{\circ}$, we get that $m \leq 2 \cdot R \cdot 2 \cdot \frac{1}{2} \cdot \cos \frac{\beta - \gamma}{2} \leq 2 \cdot R$.

Next, we assume that the triangle $ABC$ has an isogonal point, which we denote by $P$. Rotate the triangle $BPA$ along with the circle $k$ by $60^{\circ}$ around vertex $B$ (see the figure). Since the sides are seen from point $P$ at an angle of $120^{\circ}$ and the rotation angle is $60^{\circ}$, $\angle CPP' = \angle PPA' = 180^{\circ}$, so the points $C, P, P', A'$ lie on a straight line. Due to the rotation, $BP = BP' = PP'$ and $AP = A'P'$, so $AP + BP + CP = A'P' + PP' + CP = CA'$. The circle with a diameter of $3R$ drawn around the midpoint of segment $BD$ covers the figure (indeed, $k'$ passes through the center of $k$), so $CA' \leq 3R$. It is obvious that if $ABC$ is an equilateral triangle, then $m = CA' = 3R$.

Therefore, among the triangles inscribed in circle $k$, $m$ is the largest for the equilateral triangle, and the maximum is $3R$.

Solution by Zoltán Vörös

Remark. Several solvers noticed that the problem is related to problem F. 3086. It can be assumed that the problem setter, upon further reflection on this problem, came across this elegant solution.

![](https://cdn.mathpix.com/cropped/2024_05_02_4b40aeb9510815061695g-1.jpg?height=789&width=984&top_left_y=1064&top_left_x=581)