Alt-Center

The circle with its four sections is re-drawn above, with the two secants aligned with the horizontal and vertical directions, with the longer secant being horizontal, and the shorter secant being vertical. We will take the origin of the coordinate system to be at the center of the circle. Then the intersection point of the two secants is $(a, b)$ . From the length conditions we can deduce the values of $a$ and $b$ , we have

$\sqrt{1.5} = 2 \sqrt{1 - a^2} \Rightarrow a = - \sqrt{ \dfrac{5}{8} }$

and

$1.5 \sqrt{1.5} = 2 \sqrt{1 - b^2} \Rightarrow b = - \sqrt{ \dfrac{5}{32} }$

Next, we identify the four points of intersection between the secants and the circle with the radian angle each makes with the positive x-axis as $t_1, t_2, t_3, t_4$ . Since the given circle is a unit circle, we have

$\sin t_1 = \sin t_3 = b$ and $\cos t_2 = \cos t_4 = a$

It follows that $\cos t_1 = \sqrt{1 - b^2}$ and $\cos t_3 = - \sqrt{1 - b^2 }$ and $\sin t_2 = \sqrt{1 - a^2}$ and $\sin t_4 = -\sqrt{ 1 - a^2}$

Next, to find the blue area, we form the 2D vector from the intersection point $(a, b)$ to a point on the circumference of the circle.

$r(t) = (x(t), y(t) ) = (\cos t , \sin t) - (a, b)$

Differentiating with respect to $t$ ,

$r'(t) = ( x'(t) , y'(t) ) = ( - \sin t , \cos t)$

Now the area is given by this formula which follows from Green's Theorem.

$\text{Blue Area} = \dfrac{1}{2} ( \displaystyle \int_{t_1}^{t_2} x(t) y'(t) - y(t) x'(t) dt + \int_{t_3}^{t_4} x(t) y'(t) - y(t) x'(t) dt )$

Plugging in the expression for $r(t)$ and $r'(t)$ , and simplifying, the above expression reduces to

$\text{Blue Area} = \dfrac{1}{2} ( (t_2 - t_1) + (t_4 - t_3) - a (\sin t_2 - \sin t_1+\sin t_4-\sin t_3 ) + b (\cos t_2 - \cos t_1 + \cos t_4 - \cos t_3) )$

We note that $t_1 + t_3 = \pi$ and $t_2 + t_4 = 2 \pi$ . Therefore, the expression becomes,

$\text{Blue Area} = \dfrac{1}{2} ( \pi - a ( \sqrt{1 - a^2} - b - \sqrt{1 - a^2} - b ) + b (a - \sqrt{1 - b^2} + a + \sqrt{1 - b^2} ) )$

Simplifying, we get

$\text{Blue Area} = \dfrac{1}{2} ( \pi + 4 a b ) = \dfrac{1}{2} ( \pi + 4 \sqrt{ \dfrac{5}{8} } \sqrt{ \dfrac{5}{32} } )$

$= \dfrac{1}{2} ( \pi + \dfrac{5}{4} ) = \dfrac{ 4 \pi + 5 }{8}$

It follows that the gray area is $\pi (1)^2 - \dfrac{ 4 \pi + 5}{8} = \dfrac{ 4 \pi - 5 }{8 }$

Hence the ratio of the areas is $\dfrac{ 4 \pi + 5 }{4 \pi - 5 }$ ; therefore, $A = 4$ and $B = 5$ , making the answer $4 + 5= 9$ .

Let's consider the circle as $x^2+y^2=1$ , so that $y(x)=\sqrt{1-x^2}$ . Points coordinates are*:

$\displaystyle A\left(\frac{1}{2}\sqrt{\frac{3}{2}},\sqrt{\frac{5}{8}}\right)$

$\displaystyle B\left(\sqrt{\frac{5}{32}},\frac{3}{4}\sqrt{\frac{3}{2}}\right)$

$\displaystyle C\left(-\frac{1}{2}\sqrt{\frac{3}{2}},\sqrt{\frac{5}{8}}\right)$

$\displaystyle D\left(\sqrt{\frac{5}{32}},-\frac{3}{4}\sqrt{\frac{3}{2}}\right)$

We can evaluate the area of $1$ as:

$\displaystyle Area(1)=\int_{-\frac{1}{2}\sqrt{\frac{3}{2}}}^{\sqrt{\frac{5}{32}}} \sqrt{1-x^2}-\sqrt{\frac{5}{8}}$ .

Now, to calculate the area of $2$ , let's rotate anticlockwise** of $\frac{\pi}{2}$ radiants the entire circle. The situation is described in the second picture. $\displaystyle A2\left(-\sqrt{\frac{5}{8}},\frac{1}{2}\sqrt{\frac{3}{2}}\right)$

$\displaystyle B2\left(-\frac{3}{4}\sqrt{\frac{3}{2}},\sqrt{\frac{5}{32}}\right)$

$\displaystyle D2\left(\frac{3}{4}\sqrt{\frac{3}{2}},\sqrt{\frac{5}{32}}\right)$

The area of $2$ can be calculated

$\displaystyle Area(2)=\int_{-\sqrt{\frac{5}{8}}}^{\frac{3}{4}\sqrt{\frac{3}{2}}} \sqrt{1-x^2}-\sqrt{\frac{5}{32}}dx$ .

Let's define $W$ the white area and $B$ the blue area. We have:

$\displaystyle W=Area(1)+Area(2)=\int_{-\frac{1}{2}\sqrt{\frac{3}{2}}}^{\sqrt{\frac{5}{32}}} \sqrt{1-x^2}-\sqrt{\frac{5}{8}} dx+\int_{-\sqrt{\frac{5}{8}}}^{\frac{3}{4}\sqrt{\frac{3}{2}}} \sqrt{1-x^2}-\sqrt{\frac{5}{32}} dx=\frac{1}{8}(4\pi+5)$ Since the radius $R=1$ ,

$\displaystyle B=\pi-W=\pi-\frac{1}{8}(4\pi+5)=\frac{1}{8}(4\pi-5)$

The ratio is $\frac{4\pi+5}{4\pi-5}$ , so $A=4$ , $B=5$ . Eventually, $A+B=9$ .

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*: $A_x=\frac{1}{2}\sqrt{\frac{3}{2}}$ , so that $A_y=y(A_x)=\sqrt{1-\frac{3}{8}}=\sqrt{\frac{5}{8}}$ . $B_y=\frac{3}{4}\sqrt{\frac{3}{2}}=\sqrt{1-{B_x}^2}$ , so that $B_x=\sqrt{\frac{5}{32}}$

**: To rotate the points I multilpied the rotation matrix $\left( \begin{array}{c}\\ \sin{\theta} & \cos{\theta} \\ \cos{\theta} & -\sin{\theta} \end {array} \right)$ , $\theta=\frac{\pi}{2}$ by the respective points.