We cut out an equilateral triangle from a homogeneous sheet. We hang this triangle at a point $P$ on its perimeter, which is at a distance $x$ from one of its vertices. The median divides the triangle into two parts; in what ratio are the masses of these two parts to each other? When are these two masses equal? When does the ratio of the two masses take on extreme values and what are these extreme values?
$1^{0}$. If we suspend the triangle from any point $P$ on its perimeter, the median through $P$ passes through the centroid $S$ of the triangle; this line $PS$ divides the triangle into two parts, one being $\triangle CPR$, the other being the quadrilateral $APRB$. Since the triangle is made of a homogeneous material, the mass of the two parts is proportional to their areas. Let the area of $\triangle ABC$ be $t$, and the area of $\triangle CPR$ be $t_{1}$; the area of the quadrilateral is $t - t_{1}$. We need to find the value of the expression

$$
y=\frac{t-t_{1}}{t_{1}}=\frac{t}{t_{1}}-1 \ldots
$$

as a function of $AP = x$, where $0 \leqq x \leqq \frac{a}{2}$. (Indeed, we measure the distance of $P$ from the nearest vertex!) Already now,

$$
t_{1}=\frac{1}{2} RC \cdot PP_{1}=\frac{1}{2}\left(RA_{1}+A_{1}C\right) \cdot PP_{1}=\frac{1}{2}\left(RA_{1}+\frac{a}{2}\right) PP_{1} \ldots
$$

From the right triangle $\triangle P P_{1} C$,

$$
PP_{1}=PC \sin 60^{\circ}=(a-x) \frac{\sqrt{3}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_e63a9f0610ef3c4e02c7g-1.jpg?height=461&width=545&top_left_y=873&top_left_x=779)

Let the variable $\angle ASP = \angle RSA_{1} = \alpha$.

Then $RA_{1} = SA_{1} \tan \alpha = \frac{a}{6} \sqrt{3} \tan \alpha$, because

$$
SA_{1}=\frac{1}{3} AA_{1}=\frac{1}{3} a \frac{\sqrt{3}}{2}=\frac{a}{6} \sqrt{3}
$$

From $\triangle PQS$,

$$
\tan \alpha=\frac{PQ}{QS}
$$

However, $PQ = AP \cdot \sin 30^{\circ} = \frac{x}{2}$ and $QS = AS - AQ = \frac{2}{3} a \frac{\sqrt{3}}{2} - x \frac{\sqrt{3}}{2}$.

Thus,

$$
QS=\frac{\sqrt{3}}{6}(2a-3x), \quad \tan \alpha=\frac{x \sqrt{3}}{2a-3x}
$$

and

$$
RA_{1}=\frac{a}{6} \sqrt{3} \cdot \frac{x \sqrt{3}}{2a-3x}=\frac{ax}{2(2a-3x)}
$$

Furthermore, from 2),

$$
t_{1}=\frac{1}{2}\left[\frac{ax}{2(2a-3x)}+\frac{a}{2}\right](a-x) \frac{\sqrt{3}}{2}=\frac{a \sqrt{3}}{4} \cdot \frac{(a-x)^{2}}{2a-3x} \ldots
$$

Since

$$
t=\frac{a^{2}}{4} \sqrt{3}, \quad \text{so} \quad y=\frac{a(2a-3x)}{(a-x)^{2}}-1 \ldots
$$

$2^{0}$. The areas of the two parts separated by $PR$ are equal if $y=1$, that is, if

$$
\frac{a(2a-3x)}{(a-x)^{2}}-1=1
$$

From this, after rearrangement, we get

$$
x(2x-a)=0 \ldots
$$

that is, $x_{1}=0, x_{2}=\frac{a}{2}$. The areas of the two parts are equal if and only if the point $P$ is at one of the vertices of the triangle or at the midpoint of one of its sides. Both cases mean that the line $PR$ is a median of the triangle, which divides the triangle into two congruent triangles.

3. The value of the function $y$ is the same at the endpoints of the interval where $x$ varies, i.e., at $x=0$ and $x=\frac{a}{2}$, where $y=1$. Since $y$ is a continuous function of $x$, it must have an extremum in the interval $\left[0, \frac{a}{2}\right]$.

Let's examine the derivative of $y$!

$$
\begin{gathered}
y'=\frac{-3a(a-x)^{2}+2(a-x)\left(2a^{2}-3ax\right)}{(a-x)^{4}}=\frac{-3a(a-x)+4a^{2}-6ax}{(a-x)^{3}} \\
y'=\frac{a(a-3x)}{(a-x)^{3}}
\end{gathered}
$$

An extremum can only occur where $y'=0$, i.e., if $x=\frac{a}{3}$.

If $x<\frac{a}{3}$, then $y'>0$; if $x>\frac{a}{3}$, then $y'<0$.

Thus, $y$ has a maximum at $x=\frac{a}{3}$. At this point, $y$ reaches its maximum value, which is

$$
y\left(\frac{a}{3}\right)=\frac{a(2a-3\cdot\frac{a}{3})}{(a-\frac{a}{3})^{2}}-1=\frac{a(2a-a)}{\left(\frac{2a}{3}\right)^{2}}-1=\frac{a^2}{\frac{4a^2}{9}}-1=\frac{9}{4}-1=\frac{5}{4}
$$

Therefore, the maximum value of $y$ is $\frac{5}{4}$, and it occurs at $x=\frac{a}{3}$. For all other values of $x$ in the interval $\left[0, \frac{a}{2}\right]$, $y > 1$, meaning the area of the quadrilateral is always greater than the area of the triangle. (Equality holds only when the quadrilateral becomes a triangle).