X OM - II - Task 2

Under what relationship between the sides of a triangle is it similar to the triangle formed by its medians?
Let's denote the medians drawn to the sides $a$, $b$, $c$ of triangle $ABC$ by $m_a$, $m_b$, $m_c$ respectively. We need to investigate in which triangles the medians are proportional to the sides of the triangle. 

First, it is important to note that the larger of two unequal medians is drawn to the shorter side. To verify this, we can use the known formulas for the length of medians:

\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}, \quad m_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}, \quad m_c = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}
\]

From these formulas, it follows that, for example, if $m_a > m_b$, then $m_a^2 > m_b^2$, and according to the above formula, $b^2 > a^2$, hence $b > a$.

Suppose $a \leq b \leq c$, then $m_c \leq m_b \leq m_a$. The triangle with sides $m_a$, $m_b$, $m_c$ is similar to the triangle with sides $a$, $b$, $c$ if and only if

\[
\frac{m_a}{a} = \frac{m_b}{b} = \frac{m_c}{c}
\]

In problem 8, we proved that the area of the triangle equals $\frac{4}{3}$ of the area of the triangle formed by its medians. If these triangles are similar, the ratio of their areas equals the square of the ratio of corresponding sides, so from (2) we have $m_a^2 = \frac{3}{4} c^2$. Therefore, using the first formula from (1),

\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}
\]

Thus,

\[
\frac{3}{4} c^2 = \frac{1}{4} (2b^2 + 2c^2 - a^2)
\]

Simplifying, we get

\[
3c^2 = 2b^2 + 2c^2 - a^2
\]

\[
c^2 = 2b^2 - a^2
\]

Conversely, if condition (3) is satisfied, then the triangle with sides $a$, $b$, $c$ is similar to the triangle formed by its medians. Indeed, considering the relationship (3) in the formulas (1), we obtain

\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2} = \frac{1}{2} \sqrt{2b^2 + 2(2b^2 - a^2) - a^2} = \frac{1}{2} \sqrt{6b^2 - 3a^2} = \frac{\sqrt{3}}{2} \sqrt{2b^2 - a^2} = \frac{\sqrt{3}}{2} c
\]

\[
m_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2} = \frac{1}{2} \sqrt{2a^2 + 2(2b^2 - a^2) - b^2} = \frac{1}{2} \sqrt{4b^2 - b^2} = \frac{1}{2} \sqrt{3b^2} = \frac{\sqrt{3}}{2} b
\]

\[
m_c = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2} = \frac{1}{2} \sqrt{2a^2 + 2b^2 - (2b^2 - a^2)} = \frac{1}{2} \sqrt{3a^2} = \frac{\sqrt{3}}{2} a
\]

From this, it follows that $m_a$, $m_b$, $m_c$ are proportional to $c$, $b$, $a$.

Thus, we have the following result:
A triangle is similar to the triangle formed by its medians if and only if the square of the middle side of the triangle is equal to the arithmetic mean of the squares of the shortest and longest sides.

Note 1. In the solution, we could have avoided referring to problem 8 and derived condition (3) from formulas (1) and (2). This straightforward calculation is proposed as an exercise.