In the triangle $ABC$, the projection of the center of the circumscribed circle onto the line of the median from vertex $A$ is $P$. What relationship holds between the distances $PA$, $PB$, and $PC$?
I. solution. Let the midpoint of side $BC$ be denoted by $F$. As long as $O, F$, and $P$ are distinct (and also distinct from $A$), from the (right-angled) triangles $OPA$ and $OPF$,

$$
P A^{2}=O A^{2}-O P^{2}=O B^{2}-\left(O F^{2}-P F^{2}\right)=B F^{2}+P F^{2}=\frac{B C^{2}}{4}+P F^{2}
$$

since $O A=O B$. Notice that here $P F$ is the median from $P$ in triangle $PBC$, so by a known relationship,

$$
P F^{2}=\frac{1}{4}\left(2 P B^{2}+2 P C^{2}-B C^{2}\right)
$$

Substituting this in,

$$
P A^{2}=\frac{1}{2}\left(P B^{2}+P C^{2}\right)
$$

In words: $P A$ is the square root of the average of the squares of $P B$ and $P C$.
![](https://cdn.mathpix.com/cropped/2024_05_02_2f70d9c8f27bc7e8bcbag-1.jpg?height=692&width=1500&top_left_y=732&top_left_x=301)

The found relationship is also valid in the cases excluded so far, but it is trivial.

$P$ coincides with $O$ if the line $AF$ passes through $O$, and in this case the triangle is isosceles: $A B=A C$, and the three lengths under consideration are equal. And this is also true if $F$ coincides with $O$ right away, that is, if there is a right angle at $A$ in the initial triangle. (It is impossible for $F$ and $P$ to coincide at a point different from $O$.)

Remark. The expression used for the median can be derived from the fact that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of its diagonals; this, in turn, follows from applying the Pythagorean theorem to the diagonals as hypotenuses, taking the legs in the direction of one pair of sides and the corresponding heights. (It can also be derived using the Law of Cosines, but that is a roundabout way.)

II. solution. Place a Cartesian coordinate system on our figure, with the origin at the center $O$ of the circle, and the y-axis parallel to the median $AF$. This means that $A$ and $F$ have the same x-coordinate, denoted by $p$, so the coordinates of $P$ are $(p, 0)$, and the x-coordinates of $B$ and $C$ are of the form $p-q$ and $p+q$, respectively, and the radius of our circle can be taken as the unit of length, so their y-coordinates can also be expressed. (We only need their squares.)

$$
\begin{aligned}
& P A^{2}=1-p^{2} \\
& P B^{2}=q^{2}+\left(1-(p-q)^{2}\right)=1+2 p q-p^{2} \\
& P C^{2}=q^{2}+\left(1-(p+q)^{2}\right)=1-2 p q-p^{2}
\end{aligned}
$$

Notice that from this,

$$
2 p q=P B^{2}-P A^{2}=P A^{2}-P C^{2}
$$

so

$$
P B^{2}+P C^{2}=2 \cdot P A^{2}
$$

Rátz Ákos (Győr, Révai M. Gimn., IV. o. t.)