In a triangle with an obtuse angle, the median from the vertex of the obtuse angle forms a right angle with one of the sides of the angle.

a) What relationship exists between the measures of the three sides?

b) What relationship exists between the measures of the three medians?






(Translation provided as requested, maintaining the original text's line breaks and format.)
1. Let the obtuse angle vertex of the triangle $ABC$ be $C$, and let the median $CC_0$ from $C$ be perpendicular to $CB$ ($C_0$ is the midpoint of $AB$).

![](https://cdn.mathpix.com/cropped/2024_05_02_6d02a099b98cb42ba0aeg-1.jpg?height=526&width=746&top_left_y=252&top_left_x=668)

Thus, in the right triangle $C_0BC$, with the usual notation for sides and medians (Figure 1),

$$
CC_0^2 = s_c^2 = \left(\frac{c}{2}\right)^2 - a^2
$$

Reflecting the triangle $ABC$ over $C_0$, $A$ and $B$ swap places, and the image of $C$ is $C'$. Thus, $C'A = CB$, and from the right triangle $CC_1A$,

$$
CC'^2 = 4s_c^2 = b^2 - a^2
$$

From these two relationships, eliminating $s_c$ yields a simple relationship between the three sides:

$$
c^2 = 3a^2 + b^2, \quad \text{or} \quad 3a^2 + b^2 - c^2 = 0.
$$

2. Let the midpoint of $BC$ be $A_0$, and the centroid of triangle $ABC$ be $S$. Thus, $SA_0 = AA_0 / 3 = s_a / 3$, and $SB = 2s_b / 3$, $SC = 2s_c / 3$. From the right triangles $SCA_0$ and $SCB$,

$$
\begin{aligned}
& CA_0^2 = \frac{a^2}{4} = \frac{s_a^2}{9} - \frac{4s_c^2}{9} \\
& CB^2 = a^2 = \frac{4s_b^2}{9} - \frac{4s_c^2}{9}
\end{aligned}
$$

Eliminating these, we obtain a relationship between the medians:

$$
3s_c^2 = s_a^2 - s_b^2, \quad \text{or} \quad s_a^2 - s_b^2 - 3s_c^2 = 0
$$

The found relationship (1) and the relationship (2) between the medians distinguish the roles of the sides and medians in different coefficients. The same can be said about the medians in relationship (2).

Remarks. 1. According to our proof, (1) and (2) are necessary conditions for $SC$ and $CB$ to be perpendicular. It is easy to show, based on the known relationships between the sides and medians - one of which we insert here:

$$
4s_c^2 = 2a^2 + 2b^2 - c^2, \quad 9c^2 = 8s_a^2 + 8s_b^2 - 4s_c^2
$$

that these conditions are also sufficient for the desired perpendicularity. - Since the problem was originally a competition problem and the competition time was limited, the problem did not ask for the converse.

2. The similarity of the found relationships (1) and (2) (which is even more evident if (2) is written as

$$
s_a^2 = 3s_c^2 + s_b^2
$$

suggests that the triangle formed by the medians as sides also has the same property as the original triangle. Indeed, if the midpoints of segments $C'S$ and $AS$ are denoted by $D$ and $E$ respectively, in the triangle $ADS$,

$$
DS = \frac{2}{3}s_c, \quad AD(=BS) = \frac{2}{3}s_b, \quad AS = \frac{2}{3}s_a
$$

thus this triangle is similar to the triangle formed by the medians as sides, with an obtuse angle at $D$, and the median $DE$ is perpendicular to the side $DS$. (This also provides another proof for (2).)[^0]

![](https://cdn.mathpix.com/cropped/2024_05_02_6d02a099b98cb42ba0aeg-2.jpg?height=526&width=765&top_left_y=127&top_left_x=672)


[^0]: ${ }^{1} \mathrm{The}$ first was recently used in problem 1772, K. M. L. 44 (1972) 10 - 11. The second is derived by similarly writing $4s_a^2$ and $4s_b^2$ and solving the system of three equations for $a^2$, $b^2$, and $c^2$ as unknowns.