Is there an irregular triangle from which a similar triangle can be constructed using its medians as sides?
Solution. Let the sides of the triangle be $a \leq b \leq c$, and the corresponding medians be $s_{a}, s_{b}, s_{c}$. According to the well-known formula for the squares of the medians,

$$
4 s_{a}^{2}=2 c^{2}+2 b^{2}-a^{2}, \quad 4 s_{b}^{2}=2 c^{2}+2 a^{2}-b^{2} \quad \text { and } \quad 4 s_{c}^{2}=2 a^{2}+2 b^{2}-c^{2}
$$

(these can be easily derived from the fact that in any parallelogram, the sum of the squares of the sides equals the sum of the squares of the diagonals, see Figure 1). It is clear that $s_{a} \geq s_{b} \geq s_{c}$. Since two triangles are similar if and only if the ratios of the squares of their corresponding sides are equal, the necessary and sufficient condition for the requirement to be satisfied is

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

![](https://cdn.mathpix.com/cropped/2024_05_02_f6a042a0421b0e12ea8dg-1.jpg?height=480&width=385&top_left_y=641&top_left_x=859)

Figure 1

From these equations, by rearranging, we get that the

$$
2 b^{2} c^{2}+2 b^{4}-a^{2} b^{2}=2 c^{4}+2 a^{2} c^{2}-b^{2} c^{2} \text { and the } 2 a^{2} c^{2}+2 a^{4}-a^{2} b^{2}=2 a^{2} b^{2}+2 b^{4}-b^{2} c^{2}
$$

equalities must hold simultaneously. Adding the two equations and rearranging, we get

$$
0=a^{4}+2 b^{2} c^{2}-c^{4}-2 a^{2} b^{2}=\left(a^{2}-c^{2}\right)\left(a^{2}+c^{2}-2 b^{2}\right)
$$

If $a^{2}=c^{2}$, then $a=c$, and substituting this back into the first equation, we get $b=c$, so in this case the triangle is equilateral. If $a^{2}+c^{2}-2 b^{2}=0$, or $a^{2}+c^{2}=2 b^{2}$ (which is also true for an equilateral triangle), then both equalities in (1) hold, because

$$
\frac{2 c^{2}+\left(a^{2}+c^{2}\right)-a^{2}}{c^{2}}=\frac{2 c^{2}+2 a^{2}-\frac{a^{2}+c^{2}}{2}}{\frac{a^{2}+c^{2}}{2}}=\frac{2 a^{2}+\left(a^{2}+c^{2}\right)-c^{2}}{a^{2}}=3
$$

Therefore, a triangle can be constructed from its medians that is similar to the original triangle if and only if the sum of the squares of two of its sides equals twice the square of the third side. This condition is satisfied, for example, in the right triangle with sides $2, 2 \sqrt{2}$, and $2 \sqrt{3}$ (the medians of which are $\sqrt{3}, \sqrt{6}$, and 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_f6a042a0421b0e12ea8dg-1.jpg?height=449&width=506&top_left_y=2036&top_left_x=796)