2. Find all positive numbers $k$, for which: Among all triangles $A B C$, in which $|A B|=5 \mathrm{~cm}$ and $|A C|:|B C|=k$, the triangle with the largest area is isosceles.
SOLUTION. For $k=1$, the given characterization is satisfied by any isosceles triangle with the base $AB$ and any height to the side $AB$. Clearly, there is no triangle with the largest area among them.

Assume further that $k>1$ (otherwise, we solve the same problem, but with $A$ and $B$ swapped). On the line $AB$, there exist two distinct points $C_{1}, C_{2}$ such that $\frac{|AC_{1}|}{|BC_{1}|}=\frac{|AC_{2}|}{|BC_{2}|}=k$. All points $C$ in the plane for which $|AC|:|BC|=k$ lie on the Apollonian circle constructed over the diameter $C_{1}C_{2}$ (Fig. 1). It is clear from this that the triangle $ABC$ will have the largest area when the vertex $C$ is at the midpoint of the arc $C_{1}C_{2}$ (in any of the half-planes determined by the line $AB$). Assuming $k>1$, for such a chosen point $C$, it holds that $|AC|>|BC|$ and also

![](https://cdn.mathpix.com/cropped/2024_04_17_7f84c24a01c2b9d734abg-02.jpg?height=284&width=635&top_left_y=888&top_left_x=772)

Fig. 1 $|AC|>|AS|>|AB|$, so the triangle $ABC$ will be isosceles if and only if $|AB|=|BC|$. From this, we derive the equation for the corresponding value of $k$.

For points $C_{1}, C_{2}$, we first have

$$
|AC_{1}|=\frac{k}{k+1}|AB|, \quad |AC_{2}|=\frac{1}{k+1}|AB|
$$

so from the equalities $|AC_{2}|=|AB|+|BC_{2}|$ and $|C_{1}C_{2}|=|BC_{1}|+|BC_{2}|$ we get

$$
|BC_{2}|=\frac{1}{k-1}|AB|, \quad |C_{1}C_{2}|=\frac{2k}{k^2-1}|AB|
$$

and

$$
|BS|=||SC_{1}|-|BC_{1}||=\left|\frac{k}{k^2-1}-\frac{1}{k+1}\right||AB|=\frac{1}{k^2-1}|AB|
$$

$$
|BC|^2=|BS|^2+|CS|^2=|BS|^2+|C_{1}S|^2=\frac{1+k^2}{(k^2-1)^2}|AB|.
$$

From this, we derive the equation

$$
1+k^2=k^4-2k^2+1, \quad \text{or} \quad k^2(k^2-3)=0
$$

which has the only positive solution $k=\sqrt{3}$.

The problem is satisfied by two positive numbers $k, k=\sqrt{3}$ and $k=1/\sqrt{3}$.