The area of an isosceles triangle inscribed in a given circle is $a^{2}$, which is known. Determine this triangle and establish for which values of $a^{2}$ there is a solution and how many there are.
Consider the height of an isosceles triangle as the independent variable, denoted by $x$. If $R$ is the radius of the circle, $0 \leqq x \leqq 2 R$. The half of the base of the triangle is the geometric mean of $x$ and $(2 R-x)$, and thus the area of the triangle is

$$
y=x \sqrt{x(2 R-x)}=\sqrt{x^{3}(2 R-x)}
$$

It is evident that $y$ is real only if $0 \leqq x \leqq 2 R$. Therefore, it is sufficient to examine the variation of the function

$$
f(x)=x^{3}(2 R-x)=-x^{4}+2 R x^{3}
$$

within this interval.

![](https://cdn.mathpix.com/cropped/2024_05_02_22ccdfa44ae1158c89e5g-1.jpg?height=540&width=379&top_left_y=565&top_left_x=862)

$f(x)=0$ (and $y=0$) if $x=0$ and if $x=2 R$. Within this interval, $f(x)$ is continuous and positive everywhere; hence, it must have at least one maximum. Since

$$
f^{\prime}(x)=-4 x^{3}+6 R x^{2}=2 x^{2}(3 R-2 x)=0
$$

if $x=0$ and if $x=\frac{3 R}{2}$, the function $f(x)$ (and $y$) has a maximum at $x=\frac{3 R}{2}$. At this point, $f^{\prime}(x)$ transitions from positive to negative values, and the variation of $f(x)$ is characterized by the following table:

| $x$ | 0 |  | $\frac{3}{2} R$ |  | $2 R$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f^{\prime}(x)$ | 0 | + | 0 | - | - |
|  | 0 |  |  |  |

Accordingly, the maximum value of $y$ and $a^{2}$ is: $\sqrt{\frac{1}{2}\left(\frac{3}{2}\right)^{3} R^{4}}=\frac{3 \sqrt{3}}{4} R^{2}$. This is the area of the equilateral triangle inscribed in the circle.

$y$ takes on every value between 0 and $\frac{3 \sqrt{3}}{4} R^{2}$ twice: once during the increase and once during the decrease. This implies that if $a^{2} \leq \frac{3 \sqrt{3}}{4} R^{2}$, the equation

$$
x \sqrt{x(2 R-x)}=a^{2} \quad \text { or } \quad x^{3}(2 R-x)=a^{4}
$$

has two real, positive solutions.

Supplement. The function $f(x)=x^{3}(2 R-x)$ has an inflection point at $x=0$. It is evident that, since $a^{4}>0$, the equation $f(x)=a^{4}$ cannot have a real solution for which $x>2 R$.

The equation

$$
x^{3}(2 R-x)=a^{4} \quad \text { or } \quad g(x) \equiv x^{4}-2 R x^{3}+a^{4}=0
$$

has at most two positive roots according to Descartes' rule of signs. One of these roots lies between 0 and $\frac{3}{2} R$, and the other between $\frac{3}{2} R$ and $2 R$, if $a^{2}>0, \quad g\left(\frac{3}{2} R\right)=\left(\frac{3}{2} R\right)^{4}-2\left(\frac{3}{2}\right)^{3} R^{4}+a^{4}=a^{4}-\frac{1}{2}\left(\frac{3}{2}\right)^{3} R^{4}<0
$$