Given the equilateral triangle $ABC$. For any point $P$ in its plane, we associate the number $AP + BP - CP$ formed from the distances measured from the vertices. Where does this function take its minimum value?
The value of the given expression at the vertices of the triangle $ABC$ is $0, 0, 2AC$ respectively. From now on, let's assume that $P$ is a point different from the vertices of the triangle. Rotate the segment $BP$ around $A$ so that $B$ lands on $C$, and let the new position of $P$ be $P'$. Thus, $BP = CP'$. The angle of rotation is $60^\circ$, so $\triangle APP'$ is an equilateral triangle, and thus $AP = P'P$. Therefore, the value of our function can be expressed using the segments between the points $C, P, P'$: $AP + BP - CP = P'P + CP' - CP$.

If these three points are not collinear, then by the triangle inequality, the value of the function is positive. The value of the function is also positive if the points are collinear but $P'$ is not on the segment $CP$, because in this case either the segment $P'P$ or the segment $CP'$ contains the segment $CP$. If, however, $P'$ is on the segment $CP$, then the value of the function is 0. The minimum of the function (considering the values it takes at the vertices as well) is therefore 0, and it takes this value when $P$ coincides with vertex $A$ or $B$, or when $P'$ is on the segment $CP$ (Figure 1).
![](https://cdn.mathpix.com/cropped/2024_05_02_28ba8411c1b3bb6a408ag-1.jpg?height=432&width=814&top_left_y=564&top_left_x=640)

In the latter case, rotating $A$ around $P$ by $60^\circ$ results in the point $P'$ on the segment $PC$, so $PA < PC$. The half-line $PA$ is rotated to the half-line $PC$ by the same magnitude and direction as the rotation that takes the half-line $BA$ to $BC$, since both rotations are $60^\circ$ and in the opposite direction of the previously applied rotation. Therefore, $B$ and $P$ are on the same $120^\circ$ arc of the circumcircle of $\triangle ABC$, or $P$ is on the arc of the circumcircle $k$ of $\triangle ABC$ that contains $B$ and does not contain $C$. Since $PA < PC$, only the part of this arc that does not contain $C$ is relevant.

If $P$ is on the arc of $k$ that does not contain $C$, then rotating $A$ around $P$ by $60^\circ$ results in the point $P$ on the segment $PC$, which can also be obtained by rotating $P$ around $A$ by $60^\circ$. In this case, the value of the given function is indeed 0. We have seen that the value of the function is also 0 if $P$ coincides with the endpoints of the mentioned arc, so the given function takes its minimum value on the closed arc of $k$ that does not contain $C$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_28ba8411c1b3bb6a408ag-1.jpg?height=472&width=390&top_left_y=1469&top_left_x=846)