A pair of real numbers $(a, b)$ with $a^{2}+b^{2} \leq \frac{1}{4}$ is chosen at random. If $p$ is the probability that the curves with equations $y=a x^{2}+2 b x-a$ and $y=x^{2}$ intersect, then $100 p$ is closest to
(A) 65
(B) 69
(C) 53
(D) 57
(E) 61
The curves with equations $y=a x^{2}+2 b x-a$ and $y=x^{2}$ intersect exactly when the equation

$$
a x^{2}+2 b x-a=x^{2}
$$

has at least one real solution.

This equation has at least one real solution when the quadratic equation

$$
(a-1) x^{2}+2 b x-a=0
$$

has at least one real solution.

(Note that when $a=1$, this equation is actually linear as long as $b \neq 0$ and so will have at least one real solution.)

This quadratic equation has at least one real solution when its discriminant is non-negative; that is, when

$$
(2 b)^{2}-4(a-1)(-a) \geq 0
$$

Manipulating algebraically, we obtain the equivalent inequalities:

$$
\begin{aligned}
4 a^{2}-4 a+4 b^{2} & \geq 0 \\
a^{2}-a+b^{2} & \geq 0 \\
a^{2}-a+\frac{1}{4}+b^{2} & \geq \frac{1}{4} \\
a^{2}-2 a\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}+b^{2} & \geq \frac{1}{4} \\
\left(a-\frac{1}{2}\right)^{2}+b^{2} & \geq\left(\frac{1}{2}\right)^{2}
\end{aligned}
$$

Therefore, we want to determine the probability that a point $(a, b)$ satisfies $\left(a-\frac{1}{2}\right)^{2}+b^{2} \geq\left(\frac{1}{2}\right)^{2}$ given that it satisfies $a^{2}+b^{2} \leq\left(\frac{1}{2}\right)^{2}$.

The equation $\left(x-\frac{1}{2}\right)^{2}+y^{2}=\left(\frac{1}{2}\right)^{2}$ represents a circle with centre $\left(\frac{1}{2}, 0\right)$ and radius $\frac{1}{2}$.

The inequality $\left(x-\frac{1}{2}\right)^{2}+y^{2} \geq\left(\frac{1}{2}\right)^{2}$ represents the region outside of this circle.

The equation $x^{2}+y^{2}=\left(\frac{1}{2}\right)^{2}$ represents a circle with centre $(0,0)$ and radius $\frac{1}{2}$.

The inequality $x^{2}+y^{2} \leq\left(\frac{1}{2}\right)^{2}$ represents the region inside this circle.

Re-phrasing the problem in an equivalent geometric way, we want to determine the probability that a point $(a, b)$ is outside the circle with centre $\left(\frac{1}{2}, 0\right)$ and radius $\frac{1}{2}$ given that it is inside the circle with centre $(0,0)$ and radius $\frac{1}{2}$.

Note that each of these two circles passes through the centre of the other circle.
![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-035.jpg?height=454&width=950&top_left_y=1404&top_left_x=585)

Putting this another way, we want to determine the fraction of the area of the circle centred at the origin that is outside the circle centred at $\left(\frac{1}{2}, 0\right)$. This region is shaded in the first diagram above.

Let $r=\frac{1}{2}$.

The area of each circle is $\pi r^{2}$.

To determine the shaded area, we subtract the unshaded area from the area of the circle. By symmetry, the unshaded area below the $x$-axis will equal the unshaded area above the $x$-axis. In the second diagram above, the origin is labelled $O$, the point where the circle centred at $O$ intersects the $x$-axis is labelled $B$, and the point of intersect in the first quadrant of the two cicles is labelled $A$. The unshaded region above the $x$-axis consists of $\triangle A O B$ plus two curvilinear regions.

Since $O A$ and $O B$ are radii of the circle centred at $O$, then $O A=O B=r$.

Since $A B$ is a radius of the circle centred at $B$, then $A B=r$.

This means that $\triangle O A B$ is equilateral, and so it has three $60^{\circ}$ angles.

When $\triangle O A B$ is combined with the curvilinear region above and to the right of $\triangle O A B$, we thus obtain a sector of the circle with centre $O$ and central angle $60^{\circ}$.

Since $60^{\circ}$ is $\frac{1}{6}$ of $360^{\circ}$, this sector is $\frac{1}{6}$ of the entire circle and so has area $\frac{1}{6} \pi r^{2}$.

Similarly, when $\triangle O A B$ is combined with the curvilinear region above and to the left of $\triangle O A B$, we obtain a sector of the circle with centre $B$ and central angle $60^{\circ}$; this sector has area $\frac{1}{6} \pi r^{2}$. If $K$ is the area of equilateral $\triangle O A B$, the area of each curvilinear region is $\frac{1}{6} \pi r^{2}-K$ and so the unshaded area above the $x$-axis is equal to $2\left(\frac{1}{6} \pi r^{2}-K\right)+K$ which simplifies to $\frac{1}{3} \pi r^{2}-K$. Thus, the total unshaded area inside the circle of area $\pi r^{2}$ is $\frac{2}{3} \pi r^{2}-2 K$, which means that the shaded area inside the circle is $\pi r^{2}-\left(\frac{2}{3} \pi r^{2}-2 K\right)$ which is equal to $\frac{1}{3} \pi r^{2}+2 K$.

Finally, we need to use the area of an equilateral triangle with side length $r$. This area is equal to $\frac{\sqrt{3}}{4} r^{2}$. (To see this, we could use an altitude to divide the equilateral triangle into two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. Using the ratio of side lengths in these special triangles, this altitude has length $\frac{\sqrt{3}}{2} r$ and so the area of the equilateral triangle is $\frac{1}{2} r\left(\frac{\sqrt{3}}{2}\right)$ or $\frac{\sqrt{3}}{4} r^{2}$.)

Therefore, the fraction of the circle that is shaded is

$$
\frac{\frac{1}{3} \pi r^{2}+\frac{\sqrt{3}}{2} r^{2}}{\pi r^{2}}=\frac{\frac{1}{3} \pi+\frac{\sqrt{3}}{2}}{\pi}=\frac{2 \pi+3 \sqrt{3}}{6 \pi} \approx 0.609
$$

This means that the probability, $p$, that a point $(a, b)$ satisfies the given conditions is approximately 0.609 and so $100 p \approx 60.9$.

Of the given choices, this is closest to 61, or (E).

ANswer: (E)