area between intersecting circles (Estonia Open 1998 Senior 2.1)

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area between intersecting circles (Estonia Open 1998 Senior 2.1)

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#1 h Mar 27, 2020, 1:50 AM

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Circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively lie on a plane such that that the circle $C_2$ passes through $O_1$ . The ratio of radius of circle $C_1$ to $O_1O_2$ is $\sqrt{2+\sqrt3}$ .
a) Prove that the circles $C_1$ and $C_2$ intersect at two distinct points.
b) Let $A,B$ be these points of intersection. What proportion of the area of circle is $C_1$ is the area of the sector $AO_1B$ ?

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#2 h Mar 27, 2020, 9:33 AM

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Circle $C_{1}$ , midpoint $O_{1}(0,0)$ and radius $r$ .
Circle $C_{2}$ , midpoint $O_{2}(r\sqrt{2-\sqrt{3}},0)$ and radius $r\sqrt{2-\sqrt{3}}$ .

Points $A(\frac{r\sqrt{2+\sqrt{3}}}{2},\frac{r\sqrt{2-\sqrt{3}}}{2})$ and $A(\frac{r\sqrt{2+\sqrt{3}}}{2},-\frac{r\sqrt{2-\sqrt{3}}}{2})$ .

Area circle $C_{1}$ equals $\pi r^{2}$ .
Area $ABO_{1}$ equals $\frac{r^{2}}{4}$ .

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#3 h Mar 27, 2020, 8:25 PM

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$(a)$ Since $r(C_1):O_1O_2>1$ $\Rightarrow$ $C_2$ lies inside $C_1$ . Now observe that the ratio between radius and distance is equivalent to the ratio between radii. Also $r(C_1):r(C_2)<2$ (it's easy to show) $\Rightarrow$ $r(C_1)<2r(C_2)$ which implies that $C_1$ and $C_2$ intersect at two distinct points. $\Box$
$(b)$ Let $\alpha$ be an exterior angle $\angle AO_1B$ . Then the ratio between the areas must be $\frac{2\pi(2+\sqrt 3)}{\alpha}$ (by the sector's formula). By condition: $\angle AO_2O_1=\angle BO_2O_1$ because $AO_2=O_1O_2=BO_2$ and $AO_1=BO_1$ . Apply the cosine law in $\triangle AO_2O_1$ :
$cos(\angle AO_2O_1)=\frac{AO_2^2+O_1O_2^2-AO_1^2}{2AO_2\cdot O_1O_2}=\frac{2r^2-k^2r^2}{2r^2}=\frac{2-k^2}{2}=\frac{2-(2+\sqrt 3)}{2}=\frac{-\sqrt 3}{2}$ .
There are denote $r$ as radius of $C_2$ and $k$ as ratio by condition. Hence $\angle AO_2O_1=\frac{5\pi}{6} \Rightarrow \alpha = \frac{5\pi}{3}$ .
Therefore the ratio between areas equal $\frac{6(2+\sqrt 3)}{5}$ .

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