Let's determine the point on the arc $AB$ of a circle for which the sum of the distances from points $A$ and $B$ is maximized.
Let $C$ be any point on the arc $AB = i$. Given that we are to examine the sum of the distances from $C$ to $A$ and $B$, let us rotate $B$ around $C$ onto the extension of the segment $AC$, and let the new position of $B$ be $B'$.

![](https://cdn.mathpix.com/cropped/2024_05_02_ba9d3d589005d242e45ag-1.jpg?height=529&width=542&top_left_y=267&top_left_x=781)

Then the triangle $CBB'$ is isosceles, and the angles at the base $BB'$, according to the external angle theorem, are half as large as the angle $ACB$, which we know is a constant $\gamma$ for any position of $C$, since the rays $CA$ and $CB$ always intercept the arc $i_1$ complementary to $i$ in the circle containing $i$. Therefore, $\angle AB'B = \gamma / 2$, which is constant.

Since $B'$ is always on the same side of the line $AB$ as $C$ and the arc, the geometric locus of $B'$ is the circular arc of the angle $\gamma / 2$ subtended by the segment $AB$ on the side containing $i$, and $C$ can be recovered for each position of $B'$ by intersecting $AB'$ with $i$. The center of the circular arc is the point on the perpendicular bisector $m$ of the segment $AB$ that subtends the angle $2 \cdot (\gamma / 2) = \gamma$ with $AB$, which is the common point $D$ of $i$ and $m$.

The sum $AC + CB = AB'$, being the chord of the circular arc, is maximized when it is a diameter, i.e., when $AB'$ passes through $D$, and in this case, the sought position of $C$ is precisely $D$. Thus, our result can be simply stated: the sought point is the midpoint of the given arc $AB$.

Ferenc Németh (Tata, Eötvös J., II. o. t.)

Remarks. 1. Some noticed that the key to the solution could be read in the same issue of our journal (1975, No. 3) in problem 1352, which was the problem set for this exercise.

2. However, many submitted more complicated solutions based on trigonometric calculations. However, the use of more powerful tools is only justified if they lead to a simpler, shorter solution.