53. The medians $A D$ and $B E$ of triangle $A B C$ are perpendicular to each other. What can the angle $C$ of this triangle be?
53. First solution. This problem can be easily solved algebraically. Let the sides of triangle $ABC$ be denoted by $a, b, c$, and its (perpendicular by the problem's condition) medians $AD$ and $BE$ by $m_a$ and $m_b$ (Fig. $132, a$). Since

$$
m_a^2 = \frac{1}{4} \left(2b^2 + 2c^2 - a^2\right) \quad \text{and} \quad m_b^2 = \frac{1}{4} \left(2a^2 + 2c^2 - b^2\right)
$$

1) See the footnote ${ }^{2}$ on page 158.

6 d. o. Shklyarsky et al.
then (if $M$ is the point of intersection of the medians)

$$
AM^2 = \left(\frac{2}{3} m_a\right)^2 = \frac{1}{9} \left(2b^2 + 2c^2 - a^2\right)
$$

and

$$
BM^2 = \left(\frac{2}{3} m_b\right)^2 = \frac{1}{9} \left(2a^2 + 2c^2 - b^2\right)
$$

Applying the Pythagorean theorem to the right triangle $ABM$, we get

$$
c^2 = \frac{1}{9} \left(2b^2 + 2c^2 - a^2\right) + \frac{1}{9} \left(2a^2 + 2c^2 - b^2\right) = \frac{a^2 + b^2 + 4c^2}{9}
$$

from which it follows that

$$
c^2 = \frac{1}{5} \left(a^2 + b^2\right)
$$

Now, using the cosine rule,

$$
c^2 = a^2 + b^2 - 2ab \cos C, \quad \text{i.e.,} \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{4}{5} \frac{a^2 + b^2}{2ab}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-163.jpg?height=384&width=1038&top_left_y=708&top_left_x=164)

Fig. 132.

Since $a^2 + b^2 - 2ab = (a - b)^2 \geq 0$ and, consequently, $a^2 + b^2 \geq 2ab$, i.e., $\frac{a^2 + b^2}{2ab} \geq 1$, then

$$
\cos C \geq \frac{4}{5}, \text{ i.e., } C \leq \arccos \frac{4}{5} \approx 36^\circ 52'
$$

On the other hand, for any number $k$, where $1 > k \geq \frac{4}{5}$, one can choose the sides $a$ and $b$ of the triangle such that $\frac{a^2 + b^2}{2ab} = \frac{5}{4} k = \alpha \geq 1$, i.e., $a^2 - 2ab \alpha + b^2 = 0$, or

$$
\left(\frac{a}{b}\right)^2 - 2\alpha \left(\frac{a}{b}\right) + 1 = 0
$$

from which it follows that

$$
\frac{a}{b} = \alpha \pm \sqrt{\alpha^2 - 1}
$$

By constructing a triangle $ABC$ where the ratio of the sides $a : b$ has one of these two values and $\cos C = k$, one can verify that the medians $AD$ and $BE$ of this triangle are mutually perpendicular.

Second solution. This problem can also be solved purely geometrically, without using the formula for the length of the median of a triangle. Let $AD = m$ and $BE$ (perpendicular by the problem's condition) be the medians of triangle $ABC$, and $M$ be their point of intersection, so that $AM : MD = 2 : 1$ (Fig. 132, b). Drop a perpendicular $CP$ from vertex $C$ of the triangle to the line $BE$. Since $AE = EC$, then $CP = AM = \frac{2}{3} m$, i.e., point $C$ lies on the line $l$ parallel to $BE$ and at a distance $\frac{2}{3} m$ from $BE$.

Now suppose that points $A$ and $D$ are known, i.e., that segment $AD$ has a fixed position on the plane (and thus a fixed length $m$); this is permissible because in this problem, we need to determine the measure of angle $C$, and therefore we can consider the triangle $ABC$ "up to similarity," i.e., for example, that its median $AD$ has a given length $m$. If $C$ is an arbitrary point on the line $l$ (perpendicular to $AD$ and at a distance $\frac{4}{3} m$ from point $A$), and $B$ and $E$ are the points of intersection of lines $CD$ and $CA$ with the line $k \perp AD$ (or $k \parallel l$), passing through such a point $M$ on segment $AD$ that $AM : MD = 2 : 1$, then $ABC$ is the (desired) triangle with mutually perpendicular medians $AD$ and $BE$ (why?). Thus, our problem reduces to the following: given segment $AD$ and line $l \perp AD$ (Fig. 132, b), what can the angle $ACD$ be, where $C$ is a point on line $l$, i.e., to problem 68a. By moving point $C$ infinitely far along line $l$, we see that angle $C$ can be arbitrarily small; the largest value, in accordance with the solution to problem 68a, is obtained when $C$ coincides with the point $C^*$ where line $l$ is tangent to the circle $S$ passing through $A$ and $D$. Since the center $O$ of circle $S$ lies on the line $p \parallel k$, perpendicular to $AD$ and passing through the midpoint of $AD$, and thus at a distance $\frac{1}{6} m$ from $k$ and $\frac{5}{6} m$ from $l$, the radius $R$ of circle $S$ is

$$
R = \frac{1}{6} m + \frac{1}{3} m + \frac{1}{3} m = \frac{5}{6} m
$$

Since the inscribed angle $AC^*D$ of circle $S$ subtends the chord $AD = m$, by the (applied to triangle $AD C^*$) sine rule, we have

$$
\sin \angle C^* = \frac{AD}{2R} = \frac{m}{\frac{5}{3} m} = \frac{3}{5}, \quad C^* = \arcsin \frac{3}{5}
$$

Thus, the desired angle $C$ lies within the range $0 < C \leq \arcsin \frac{3}{5} \approx 36^\circ 52'$ (or $0 < C \leq \arccos \frac{4}{5}$, since $\arcsin \frac{3}{5} = \arccos \frac{4}{5}$).