Let's find among the isosceles triangles with a unit base length the one in which the diameter of the inscribed circle is the largest. Let's also construct this triangle.

Among the isosceles triangles with a unit base length, we need to find the one where the diameter of the inscribed circle is the largest. Let's construct this triangle as well.
I. Solution. a) Express the diameter $2 \varrho$ of the inscribed circle as a function of the variable $x$ using the known relationship $\varrho=t / s$ ( $m$ is the height along the axis, Fig. 1):

$$
2 \varrho=\frac{2 t}{s}=\frac{4 t}{2 s}=\frac{2 x m}{2+x}=\frac{x}{2+x} \sqrt{4-x^{2}}=x \sqrt{\frac{2-x}{2+x}} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_59df0309a51532f57f58g-1.jpg?height=691&width=339&top_left_y=446&top_left_x=882)

Fig. 1

Naturally, we only interpret the function for $x>0$, so $2 \varrho>0$, thus the maximum, if it exists, occurs at the same $x$ value as the maximum of its square, the

$$
y=(2 \varrho)^{2}=\frac{x^{2}(2-x)}{2+x}=\frac{2 x^{2}-x^{3}}{x+2}
$$

function, which we will find.

The extremum can only occur where the derivative vanishes:

$$
\begin{gathered}
y^{\prime}=\frac{\left(4 x-3 x^{2}\right)(x+2)-\left(2 x^{2}-x^{3}\right) \cdot 1}{(x+2)^{2}}= \\
=\frac{-2 x\left(x^{2}+2 x-4\right)}{(x+2)^{2}}=\frac{-2 x(x+\sqrt{5}+1)}{(x+2)^{2}}(x+1-\sqrt{5})=0
\end{gathered}
$$

which, due to the constraint $x>0$, only occurs at

$$
x_{0}=\sqrt{5}-1 \quad(\approx 1.236)
$$

since in the last form, the variable factors in the leading fraction are positive, and the fraction is always negative.

There is a maximum at this point because $y^{\prime}>0$ if $x<x_{0}$, and $y^{\prime}<0$ if $x>x_{0}$.

Alternatively, we can use the trigonometric form. Let $z_{0}=\arcsin \left(\frac{1}{2}\right)$, then $\cos z_{0}=\sqrt{\frac{\sqrt{5}+1}{4}}$

$2 z_{0}=51^{\circ} 50^{\prime}$, since the leading fraction is always positive. The cosine function is monotonically decreasing in the neighborhood of $z_{0}$, so the [ ]-enclosed factor is also, and $\varrho^{\prime}>0$ if $z<z_{0}$, and $\varrho^{\prime}<0$ if $z>z_{0}$, thus there is a maximum. The base length is also given by $B C=2 \cos 2 z_{0}=4 \cos ^{2} z_{0}-2=\sqrt{5}-1$ from this.

Bálint Chikán (Eger, Gárdonyi G. Gymnasium, 3rd grade)