A point of inflection of a cubic parabola is the center of a circle; this circle is internally tangent to the parabola at points $A(\sqrt{2}, \sqrt{2})$ and $B(-\sqrt{2},-\sqrt{2})$ and intersects it at two other points. The tangents drawn to the parabola at points $A$ and $B$ are parallel to each other. If the abscissa of one of the intersection points is $x_{1}<0$, determine the magnitude of the area bounded by the two curves between $x_{2}=\sqrt{2}$ and these $x_{1}$ limits!
If the tangents at points $A, B$ of a circle are parallel, then $A B$ is the diameter of the circle and the midpoint of $A B$, i.e., the origin is the center of the circle, whose equation is:

$$
x^{2}+y^{2}=r^{2}
$$

The circle passes through the point $(\sqrt{2}, \sqrt{2})$; thus:

$$
2+2=r^{2}
$$

Accordingly, the circle in question is:

$$
x^{2}+y^{2}=4 \ldots
$$

The equation of the cubic parabola is:

$$
y=a x^{3}+b x^{2}+c x+d \ldots
$$

The abscissa of the inflection point, as seen in problem 1341: $-\frac{b}{3 a}$. Here $-\frac{b}{3 a}=0$, so $b=0$. Since the curve passes through the origin, equation (2) is satisfied by $(0,0)$ and thus $d=0$. Therefore, the equation of the curve is:

$$
y=a x^{3}+c x \text { and } y^{\prime}=3 a x^{2}+c \ldots
$$

The curve passes through the point $(\sqrt{2}, \sqrt{2})$, so

$$
\sqrt{2}=a \cdot 2 \sqrt{2}+c \sqrt{2}, \quad \text { i.e., } \quad 2 a+c=1 \ldots
$$

The curve and the circle have a common tangent at points $A, B$. The slope of the tangent to the circle at point $A$ is -1, because the tangent of the direction of the radius $O A$ is: $\frac{\sqrt{2}}{\sqrt{2}}=1$. Therefore, the derivative in (3) is -1 when $x=\sqrt{2}$, i.e.,

$$
6 a+c=-1 \ldots
$$

From (4) and (5): $4 a=-2, a=-\frac{1}{2}$ and thus $c=2$, so the curve in question is:

$$
y=-\frac{1}{2} x^{3}+2 x \ldots
$$

This curve has a local maximum at $x=+\frac{2}{3} \sqrt{3} \sim 1.16$, and a local minimum at $x=-\frac{2}{3} \sqrt{3} \sim -1.16$.

Now let's find the common points of the circle (1) and the curve (6). We need to solve the system of equations (1) and (6). We eliminate $y$ by substituting the expression from (6) into (1):

$$
x^{2}+\left(-\frac{1}{2} x^{3}+2 x\right)^{2}=4, \text { rearranging: } x^{6}-8 x^{4}+20 x^{2}-16=0 \ldots
$$

This equation has $\sqrt{2}$ and $-\sqrt{2}$ as double roots due to the tangency, so the left side of the equation must be divisible by

$$
(x-\sqrt{2})^{2}(x+\sqrt{2})^{2}=\left(x^{2}-2\right)^{2}
$$

Indeed,

$$
x^{6}-8 x^{4}+20 x^{2}-16=\left(x^{2}-2\right)^{2}\left(x^{2}-4\right)
$$

The intersection points of the parabola and the circle correspond to the roots of the equation $x^{2}-4=0$: $x^{\prime}=-2, x^{\prime \prime}=2$. These points lie on the x-axis.

![](https://cdn.mathpix.com/cropped/2024_05_02_0ce2f1c3ac295a009c8cg-1.jpg?height=593&width=629&top_left_y=2116&top_left_x=737)

The area bounded by the two curves, between $x_{1}=-2$ and $x_{2}=\sqrt{2}$, can be found by subtracting twice the area of the lune-shaped region bounded by the arcs of the two curves between $x_{2}=\sqrt{2}$ and $x^{\prime \prime}=2$ from the area of the circle, and then taking half of the resulting difference. 

The chord $A A_{1}$ in the circle defines a sector $A F A_{1}$, which is cut from a $90^{\circ}$ sector. The area of the $90^{\circ}$ sector is: $\frac{r^{2} \pi}{4}=\pi$. The area of the triangle $A O A_{1}$ is: $\frac{r^{2}}{2}=2$. Thus, the area of the sector is: $t_{1}=\pi-2$ and half of this, the area of the mixed curve $A E F$ (where $\widehat{A F}$ is the arc of the circle): $\frac{\pi}{2}-1$.

From this, we need to subtract the area $t_{2}$ bounded by $A E$, $E F$, and the arc of the parabola $E F$; since $O E=\sqrt{2}$ and $O F=2$,

$$
t_{2}=\int_{\sqrt{2}}^{2}\left(-\frac{1}{2} x^{3}+2 x\right) d x=\left[-\frac{x^{4}}{8}+x^{2}\right]_{\sqrt{2}}^{2}=-\frac{16}{8}+4+\frac{4}{8}-2=\frac{1}{2}
$$

The area of the lune-shaped region is:

$$
t_{3}=\frac{1}{2} t_{1}-t_{2}=\frac{\pi}{2}-1-\frac{1}{2}=\frac{\pi-3}{2}
$$

The area we are looking for is: $\quad t=\frac{1}{2}\left(r^{2} \pi-2 t_{3}\right)=\frac{1}{2}\left(4 \pi-2 \frac{\pi-3}{2}\right)=\frac{3(\pi+1)}{2}$.

$$
t=6.2124 \text { square units. }
$$

Ferenc Kieweg (Kegyesrendi g. VIII. o., Bp.)