circleseuclidean-geometrygeometrypolygons
Find the value of $R/r$:
I go with co-ordinate geometry, considering the centre of the circles is at the origin, then the equation of the circle becomes as
$$ x^2 + y^2 = R^2 $$
$$ x^2 + y^2 = r^2 $$
After this I not able to solve this.
2nd Attempt :-
here is my nice answer:
let $C$ be the center of enclosed circle then its radius $r$ is given by set formula by HC Rajpoot
$$r=\frac{abc}{2\sqrt{abc(a+b+c)}+ab+bc+ca}$$
where, $a=1, b=2, c=3$ are radii of three externally touching circles then
$$r=\frac{1\times2\times 3}{2\sqrt{1\times 2\times 3(1+2+3)}+1\times 2+2\times 3+3\times 1}=\frac{6}{23}$$
now, drop a perpendicular $CN$ of length $y$ from center $C$ to the x-axis to get a right triangle $C_1NC$ in which hypotenuse
$C_1C=1+\frac6{23}=\frac{29}{23}$. apply pythagorean
$$C_1N=\sqrt{(C_1C)^2-(CN)^2}=\sqrt{\left(\frac{29}{23}\right)^2-y^2}\ \ \ \ ..........(1)$$
similarly, in right triangle $CNC_2$ in which hypotenuse
$C_2C=2+\frac6{23}=\frac{52}{23}$. apply pythagorean
$$NC_2=\sqrt{(C_2C)^2-(CN)^2}=\sqrt{\left(\frac{52}{23}\right)^2-y^2}\ \ \ \ ........(2)$$
since, circle $C_1$ & $C_2$ are touching hence, $$C_1N+NC_2=C_1C_2=1+2=3$$
$$\sqrt{\left(\frac{29}{23}\right)^2-y^2}+\sqrt{\left(\frac{52}{23}\right)^2-y^2}=3$$
$$\sqrt{\frac{841}{529}-y^2}=3-\sqrt{\frac{2704}{529}-y^2}$$ taking squares on both sides, i get
$$\sqrt{\frac{2704}{529}-y^2}=\frac{1104}{529}$$
again i take square,
$$y^2=\frac{2704}{529}-\left(\frac{1104}{529}\right)^2=\frac{400}{529}$$
$$y=\frac{20}{23}$$
above is the correct value of ordinate of center $C$ of the enclosed circle
$\boldsymbol{r\lt1}$
Using similar triangles, we get the $x$-coordinate of $Q$ to be $r+(1-r)=1$.
The sum of the orange and lavender segments times $r$ should be the length of the lavender segment; that is, $$ \frac{\color{#C00000}{r}}{\color{#00A000}{1}}(\color{#FF8000}{r}+\color{#8080FF}{x})=\color{#8080FF}{x} $$ solving gives $$ x=\frac{r^2}{1-r} $$
$\boldsymbol{r\gt1}$
Using similar triangles, we get the $x$-coordinate of $Q$ to be $r-(r-1)=1$.
The sum of the lavender and orange segments should be $r$ times the length of the lavender segment; that is, $$ \color{#8080FF}{x}+\color{#FF8000}{r}=\frac{\color{#C00000}{r}}{\color{#00A000}{1}}\color{#8080FF}{x} $$ solving gives $$ x=\frac{r}{r-1} $$
If the center of $C_1$ is on the left of $(0,0)$, then the $x$-coordinate of $Q$ will be $-1$.
Best Answer
The hint:
Let $ABCDE$ be our regular pentagon, $O$ be a center of the circles and $T$ be a touching point to $AB$.
Thus, $OT=r$, $OA=R$ and $\measuredangle ATO=90^{\circ}.$
Hence, $$\measuredangle AOT=\frac{360^{\circ}}{5\cdot2}=36^{\circ}$$ and $$AT=R\sin36^{\circ}=r\tan36^{\circ}.$$ Can you end it now?