geometrygre-exam
The following is problem 8 from a GRE exam found here.
The problem states that the two circles with radius $r=3$ intersect each other such that the area of the darkened region is equal to the sum of areas of the dashed regions. Find the area of the darkened region. Thanks.
I'll calculate the result for two points randomly uniformly picked the unit disk; it scales with $R^2$.
Consider $p_1$ at distance $r_1$ from the origin. Where can $p_2$ lie such that the origin is closer to either point than the other point? For the origin to be closer to $p_1$ than $p_2$, we must have $p_2$ outside the circle around $p_1$ with radius $r_1$. For the origin to be closer to $p_2$ than $p_1$, we must have $p_2$ on the origin's side of the line equidistant from the origin and $p_2$.
Thus, given $p_1$ at $(0,r_1)$, the admissible region for $p_2$ is the difference of two circular segments, the segment of the circle around the origin with radius $1$ up to $y=r_1/2$ minus the segment of the circle around $p_1$ with radius $r_1$ up to $y=r_1/2$. The area whose expected value we seek is proportional to the squared distance $d^2$ between $p_1$ and $p_2$. Thus, to calculate the desired expected value, we need the area of a circular segment and the integral over a circular segment of the squared distance to a given point on its axis. In detail:
$$ \begin{align} A(r,a)&:=\int_{-r}^a\mathrm dy\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\mathrm dx\;, \\ B(r,a,b)&:=\int_{-r}^a\mathrm dy\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\mathrm dx\left(x^2+(y-b)^2\right)\;, \\ \langle d^2\rangle&=\frac{\displaystyle\int_0^1\mathrm dr_1r_1\left(B\left(1,\frac{r_1}2,r_1\right)-B\left(r_1,-\frac{r_1}2,0\right)\right)}{\displaystyle\int_0^1\mathrm dr_1r_1\left(A\left(1,\frac{r_1}2\right)-A\left(r_1,-\frac{r_1}2\right)\right)}\;. \end{align} $$
Some help from our electronic friends yields the integrals:
$$ \begin{align} A(r,a)=&a\sqrt{r^2-a^2}+r^2\arctan\frac a{\sqrt{r^2-a^2}}+\frac\pi2r^2\;, \\ B(r,a,b)=&\frac16\left( \left(6b^2a+8b\left(a^2-r^2\right)+r^2a+2a^3\right)\sqrt{r^2-a^2}\right. \\ &\left.+3r^2\left(r^2+2b^2\right)\left(\frac\pi2+\arctan\frac a{\sqrt{r^2-a^2}}\right)\right)\;. \end{align} $$
In the negative terms with $r=r_1$, the arctangent is
$$\arctan\frac{-r_1/2}{\sqrt{r_1^2-(r_1/2)^2}}=-\arctan\frac1{\sqrt3}=-\frac\pi6\;.$$
Then further integral crunching yields
$$ \begin{align} \langle d^2\rangle&=\frac{\displaystyle\frac1{288}(88\pi+81\sqrt3)-\frac1{288}\left(8\pi-3\sqrt3\right)}{\displaystyle\frac1{16}(4\pi+3\sqrt3)-\frac14\left(\frac\pi3-\frac{\sqrt3}4\right)} \\ &= \frac{\displaystyle\frac5{18}\pi+\frac7{24}\sqrt3}{\displaystyle\frac\pi6+\frac{\sqrt3}4} \\ &= \frac16\frac{20\pi+21\sqrt3}{2\pi+3\sqrt3} \\ &\approx1.44034\;, \end{align} $$
which is confirmed by numerical simulations. The average squared distance of two points uniformly randomly picked in the unit disk without conditions is $1$, so the condition that the lune determined by the points contain the origin increases the average squared distance considerably.
The area to be averaged is $2A(d,-d/2)=d^2(4\pi-3\sqrt3)/6$, so the expected area is
$$ \frac{4\pi-3\sqrt3}6\langle d^2\rangle=\frac1{36}\frac{(4\pi-3\sqrt3)(20\pi+21\sqrt3)}{2\pi+3\sqrt3}\approx1.76927\;. $$
The problem with three points seems considerably more involved.
[Edit:]
Some more details on the calculation, to respond to the comments.
The basic idea is that the conditional expectation value of $d^2$ is the integral of $d^2$ over the volume satisfying the condition, divided by the volume satisfying the condition. Because we're dividing two integrals, we don't need to worry about constant factors. If we were instead calculating the probability for the condition to be fulfilled, we'd have to multiply the denominator by $2\pi$ for the missing integration over $\phi_1$ and divide it by $\pi^2$ for normalizing the density, for an overall factor of $2/\pi$, so the probability for the condition to be fulfilled is
$$\frac2\pi\left(\frac\pi6+\frac{\sqrt3}4\right)=\frac13+\frac{\sqrt3}{2\pi}\approx0.609\;.$$
But since we're dividing two integrals, this factor $2/\pi$ cancels. The integration is performed in polar coordinates for $p_1$ and in Cartesian coordinates for $p_2$; the origin for the coordinates is the centre of the unit circle in case of the polar coordinates for $p_1$ and the centre of the respective circle with radius $r$ in case of the Cartesian coordinates for $p_2$, and those are two different circles in the two terms of the difference.
The difference is the difference of the integrals over two different circular segments (as discussed in the comments). Both segments have the line equidistant between $p_1$ and the centre of the unit circle as their linear boundary. The positive term is the integral over the lower segment of the unit circle $C(O,1)$ defined by that line, and the negative term is the integral over the lower segment of the circle $C(p_1,r_1)$ defined by that line, where $O$ is the origin and $C(p,r)$ denotes the circle with origin $p$ and radius $r$.
Since the segments both have the same form and just differ in their centre and radius, the calculation is organized into calculating integrals $A$ and $B$ with general radius $r$ and boundary at $a$ and then substituting specific values in the two terms of the difference. Since Cartesian coordinates relative to the centre of the respective circle are used, the $y$ coordinate $b$ of $p_1$ relative to the respective centre also needs to be taken into account in calculating the squared distance $d^2$ from $p_1$.
So in the integrals $A$ and $B$ over a circular segment ($A$ being the integral of $1$, i.e. the volume, and $B$ being the integral of $d^2$), the parameter $r$ is the radius of the circle, the parameter $a$ is the upper $y$ coordinate (relative to the circle's centre) at which the circular segment ends, and the parameter $b$ is the $y$ coordinate of $p_1$ (relative to the circle's centre).
For the positive term, which is for the circular segment of the unit circle $C(O,1)$, the radius $r$ is $1$, the $y$ coordinate $a$ of the bisector relative to the centre at $O$ is $r_1/2$, and the $y$ coordinate $b$ of $p_1$ relative to the centre at $O$ is $r_1$ (since $p_1$ is at $(0,r_1)$). For the negative term, which is for the circular segment of the circle $C(p_1,r_1)$, the radius $r$ is $r_1$, the $y$ coordinate $a$ of the bisector relative to the centre at $p_1$ is $-r_1/2$, and the $y$ coordinate $b$ of $p_1$ relative to the centre at $p_1$ is $0$.
Since we're integrating over the lower segments of the respective circles, the $y$ coordinate runs from its minimal value $-r$ to its value $a$ at the linear boundary, and the $x$ coordinate takes its entire range of values from $-\sqrt{r^2-y^2}$ to $\sqrt{r^2-y^2}$.
This is all for the integration over $p_2$ given $p_1$ fixed at $(0,r_1)$. Then we also have to integrate over $p_1$. The integration over the polar angle $\phi_1$ of $p_1$ just yields a constant factor $2\pi$ that cancels (see above), since it's only the relative orientation of $p_1$ and $p_2$ that matters, not the absolute orientation of $p_1$. So we can just integrate the result we got for $p_1$ at $(0,r_1$) over $r_1$, taking into account the Jacobian $r_1$ of the polar coordinates.
If you draw the line segment from $C$ to $D$, it divides the shaded region into two pieces called circular segments. One of the circular segments is a segment of the circle with center at $A$, and the other is a segment of the circle with center at $B$.
You can see here -- http://mathworld.wolfram.com/CircularSegment.html -- how to compute the area of a circular segment.
Best Answer
Denote the area of the darkened region as $A$, and denote the area of each of the dashed regions as $B$. The areas of the dashed regions are equal since the two intersecting circles have equal area. So $A+B=\pi\cdot 3^2=9\pi$, since $A$ and $B$ added give the area of the circle with radius $3$. But based on given information, $A=2B$. Can you proceed from there?