Consider a circle $O_1$ with radius $R$ and a point $A$ outside the circle. It is known that $\angle BAC=60^\circ$, where $AB$ and $AC$ are tangent to $O_1$. We construct infinitely many circles $O_i$ $(i=1,2,\dots\>)$ such that for $i>1$, $O_i$ is tangent to $O_{i-1}$ and $O_{i+1}$, that they share the same tangent lines $AB$ and $AC$ with respect to $A$, and that none of the $O_i$ are larger than $O_1$. Find the total area of these circles. I know this problem was easy, but it still appeared in the TST, and so I posted it. It was kind of a disappointment for me.
Problem
Source: Taiwan 1st TST, 1st independent study, question 1
Tags: geometry, trigonometry, ratio, geometric sequence, geometry proposed
11.04.2009 04:26
Generalization: $\Gamma_i$ is a circle of radius $r_i$ inscribed in an angle of measure $2\alpha$ such that each $\Gamma_i$ is externally tangent to $\Gamma_{i+1}$ and $r_{i+1} < r_i.$ Then the sum of the areas of the circles $\Gamma_i$ equals the area of a circle of radius $r =\frac {_1}{^2} r_0 (\sqrt{ \sin \alpha} + \sqrt{\text{csc} \alpha}).$ Let $a,b$ the sides of the subject angle with vertex $P$ and denote $X_i$ the tangency point of $\Gamma_i(O_i,r_i)$ with line $a.$ From the similar $\triangle POX_0 \sim \triangle PO_1X_1$ we get $\frac{O_1X_1}{O_0X_0}=\frac{r_1}{r_0}=\frac{PO_1}{PO_0}=\frac{PO_0-O_0O_1}{PO_0}=\frac{PO_0-r_0-r_1}{PO_0}$ Plugging $PO_0=r_0 \cdot \csc \alpha$ yields $\frac{r_1}{r_0}=\frac{\csc \alpha-1}{\csc \alpha+1}=\frac{1-\sin \alpha}{1+\sin \alpha}$ Radii $r_0,r_1,r_2,r_3...$ form a decreasing geometric progression with ratio $k=\frac{1-\sin \alpha}{1+\sin \alpha}$ $\Longrightarrow$ Areas $S_i$ of $\Gamma_i$ form another decreasing geometric progession with ratio $k^2.$ $\sum_{i=0}^{\infty}S_i= \frac{S_0}{1-k^2} = \frac{\pi {r_0}^2}{1-\left ( \frac{1-\sin \alpha}{1+\sin \alpha} \right)^2}= \left ( \frac{1+ \sin \alpha}{2 \sqrt{\sin \alpha}} \right)^2 \cdot \pi {r_0}^2$ Let $\varrho$ be the radius of the circle equivalent to the chain of circles $\Gamma_i.$ Then $\pi \varrho^2=\left ( \frac{1+ \sin \alpha}{2 \sqrt{\sin \alpha}} \right)^2 \cdot \pi {r_0}^2 \ \Longrightarrow \ \varrho=\frac{_1}{^2}r_0 ( \sqrt{\sin \alpha}+\sqrt{\csc \alpha}).$