On the plane of triangle $A B C$, we erect perpendiculars at each vertex, and on these - on the same side of the plane, we measure the segments $A A_{1}=k \cdot B C, B B_{1}=k \cdot C A, C C_{1}=k \cdot A B$. Determine $k$ so that the centroid of triangle $A B C$ is equidistant from the points $A_{1}, B_{1}, C_{1}$. What is this distance:1 1 .[^0]


[^0]:    ${ }^{1}$ See the solution to problem 1712. II. in this issue, on page 148.
Solution. We will use the fact that the square of the median $s_{\mathrm{x}}$ to the side $x$ of a triangle with sides $x, y, z$ is, as we will show below, $-s_{\mathbf{x}}^{2}=\left(2 y^{2}+2 z^{2}-x^{2}\right) / 4$, and also that the distance from the centroid to the common endpoint of sides $y$ and $z$ is $2 s_{\mathrm{x}} / 3$.

Let the centroid of triangle $ABC$ be $S$, and the lengths of its sides be $BC=a, CA=b, AB=c$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_18004bdebf903f59d575g-1.jpg?height=613&width=746&top_left_y=347&top_left_x=668)

Thus, from the requirement $S A_{1}=S B_{1}$, we have $S A^{2}+A A_{1}^{2}=S B^{2}+B B_{1}^{2}$, that is,

$$
\frac{2 b^{2}+2 c^{2}-a^{2}}{9}+k^{2} a^{2}=\frac{2 c^{2}+2 a^{2}-b^{2}}{9}+k^{2} b^{2}
$$

rearranging gives

$$
k^{2}\left(a^{2}-b^{2}\right)=\frac{a^{2}-b^{2}}{3}
$$

for $a \neq b$, only $k=1 / \sqrt{3}$ satisfies this $(k>0)$, and for $a=b$, any positive $k$ is valid.

With the found value of $k$,

$$
S A_{1}^{2}=S B_{1}^{2}=\frac{2}{9}\left(a^{2}+b^{2}+c^{2}\right)
$$

and similarly, $S C_{1}^{2}$ is the same. Therefore, $k=1 / \sqrt{3}$ is the only valid value if the triangle $ABC$ has two different sides, and if the triangle is equilateral, any value of $k$ is valid. The required distance is

$$
\frac{1}{3} \sqrt{2\left(a^{2}+b^{2}+c^{2}\right)}
$$

The formula used can be derived from this: the sum of the squares of the sides of a parallelogram $KLMN$ is equal to the sum of the squares of its diagonals (Figure 2).
![](https://cdn.mathpix.com/cropped/2024_05_02_18004bdebf903f59d575g-1.jpg?height=264&width=1192&top_left_y=1862&top_left_x=455)

Figure 2

Let $\angle L K N \leq 90^{\circ}$ and the projections of $N, M$ on the line $KL$ be $N_{1}, M_{1}$, so $K N_{1}=L M_{1}$ and $M M_{1}=N N_{1}$, thus $K M^{2}+L N^{2}=\left(K L+L M_{1}\right)^{2}+M M_{1}^{2}+\left(K L-K N_{1}\right)^{2}+N N_{1}^{2}=2 K L^{2}+2\left(L M_{1}^{2}+M M_{1}^{2}\right)=K L^{2}+M N^{2}+L M^{2}+N K^{2}$.

Indeed, if the sides are $K L=M N=y, K N=L M=z$, and the intersection of the diagonals is $O$, then $K O=s_{x}=K M / 2$ is the median to the side $x$ of triangle $KLN$, and $4 s_{x}^{2}+x^{2}=2 y^{2}+2 z^{2}$, from which we derived the expression for $s_{x}$ above.