A Nice Geometry Problem (1)

Let the center of the unit circle be $C(0,c)$ . Then, we have $\begin{cases} (y-c)^2 + x^2 = 1 & \small \color{#3D99F6} \text{Equation of the unit circle} \\ y = x^2 & \small \color{#3D99F6} \text{Equation of the parabola} \end{cases}$ .

Considering $x \ge 0$ . The points which the circle and parabola intersect satisfy the two equation and:

$\begin{aligned} (y-c)^2 + {\color{#3D99F6} x^2} & = 1 & \small \color{#3D99F6} \text{Since }y = x^2 \\ (y-c)^2 + {\color{#3D99F6} y} & = 1 \\ y^2 - (2c-1)y + c^2 - 1 & = 0 \end{aligned}$

Solving the quadratic equation for $y$ , $y = \dfrac {2c-1 \pm \sqrt{(2c-1)^2-4(c^2-1)}}2 = \dfrac {2c-1 \color{#3D99F6}\pm \sqrt{5-4c}}2$ . For the circle to be tangent to the parabola or only 1 contact point $\color{#3D99F6} 5-4c = 0$ $\implies c = \frac 54$ , $y = \frac {2\cdot \frac 54 - 1 \pm 0}2 = \frac 34$ and $x = \sqrt y = \frac {\sqrt 3}2$ . Implies $AB = \frac {\sqrt 3}2$ and $\angle ACB = \frac \pi 3$ .

Now, the area of the grey area $=$ area of the parabola for $0 \le y \le \frac 34$ $-$ area of $\frac 23 \pi$ circular segment of radius 1.

$\begin{aligned} A_{\text{grey}} & = A_{\text{parabola}} - A_{\text{segment}} \\ & = A_{\text{parabola}} - \left(A_{\text{sector}} - A_{\text{triangle}}\right) \\ & = 2 \int_0^\frac 34 x \ dy - \frac \pi 3 + 2\left(\frac 12 \cdot AB\cdot BC\right) \\ & = 2 \int_0^\frac 34 \sqrt y \ dy - \frac \pi 3 + \frac {\sqrt 3}2\cdot \frac 12 \\ & = 2 \cdot \frac 23 \cdot y^\frac 32 \bigg|_0^\frac 34 - \frac \pi 3 + \frac {\sqrt 3}4 \\ & = \frac {\sqrt 3}2 - \frac \pi 3 + \frac {\sqrt 3}4 \\ & = \frac {3\sqrt 3}4 - \frac \pi 3 \end{aligned}$

Therefore, $A+B=3+4 = \boxed{7}$ .

Let the point $(0,a)$ be the center of the circle. The parabola clearly will intersect the lower half of the circle, which has equation $y = - \sqrt{1 - x^2} + a$ .

At the intersection points, the slopes of the tangent lines to the parabola and circle must be equal; to find the intersection points, we equate their derivatives.

$\begin{aligned} \dfrac{d}{dx} (- \sqrt{1 - x^2} + a) &= \dfrac{d}{dx} (x^2) \\ \\ \dfrac{x}{\sqrt{1 - x^2}} &= 2x \\ \\ \sqrt{1 - x^2} &= \dfrac{1}{2} \qquad \small \text{[since clearly } x \not= 0] \\ \\ x = \pm \dfrac{\sqrt{3}}{2} \qquad \implies \qquad y = x^2 &= \dfrac{3}{4} \qquad \implies \qquad a = y + \sqrt{1 - x^2} = \dfrac{5}{4} \end{aligned}$

The area between the two curves will then be

$\begin{aligned} \int_{\text{-}\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \left(-\sqrt{1 - x^2} + \dfrac{5}{4} - x^2 \right) dx &= 2 \displaystyle\int_{0}^{\frac{\sqrt{3}}{2}} \left(-\sqrt{1 - x^2} + \dfrac{5}{4} - x^2 \right) dx \\ \\ &= 2 \int_{0}^{\frac{\sqrt{3}}{2}} \left( -\sqrt{1 - x^2} \right) dx \ + \ 2 \int_{0}^{\frac{\sqrt{3}}{2}} \left( \dfrac{5}{4} - x^2 \right) dx \\ \\ &= 2 \int_{0}^{\frac{\pi}{3}} \left( -\sqrt{1 - \sin^2 \theta} \right) \cos \theta \ d\theta \ + \ 2 \left[ \dfrac{5x}{4} - \dfrac{x^3}{3} \right]_{0}^{\frac{\sqrt{3}}{2}} \quad \small \text{[In the first integral, substituting } x = \sin \theta, dx = \cos \theta \ d\theta] \\ \\ &= -2 \int_{0}^{\frac{\pi}{3}} \cos^2 \theta \ d\theta \ + \ 2 \left( \dfrac{5 \sqrt{3}}{8} - \dfrac{\sqrt{3}}{8} \right) \\ \\ &= -2 \int_{0}^{\frac{\pi}{3}} \dfrac{1}{2} \left( \cos 2 \theta + 1 \right) \ d\theta \ + \ \sqrt{3} \\ \\ &= - \left[ \dfrac{1}{2} \sin 2 \theta + \theta \right]_{0}^{\frac{\pi}{3}} \ + \ \sqrt{3} \\ \\ &= - \left( \dfrac{\sqrt{3}}{4} + \dfrac{\pi}{3} \right) \ + \ \sqrt{3} \\ \\ &= \dfrac{3\sqrt{3}}{4} - \dfrac{\pi}{3} \end{aligned}$

so $A = 3$ and $B = 4$ giving the answer $A + B = \boxed{7}$