Radius of other circle=?

Question

Let $S$ be a circle with centre $O$. A chord $AB$, not a diameter, divides $S$ into two regions $R_1$ and $R_2$ such that $O$ belongs to $R_2$. Let $S_1$ be a circle with centre in $R_1$, touching $AB$ at $X$ and $S$ internally. Let $S_2$ be a circle with centre in $R_2$, touching $AB$ at $Y$, the circle $S$ internally and passing through the centre of $S$. The point $X$ lies on the diameter passing through the centre of $S_2$ and $\angle YXO=30^\circ$. If the radius of $S_2$ is 100 then what is the radius of $S_1$?

(Image of problem text)

I have tried this for over an hour now but I can't get the right answer, which is 60.

After some construction and taking the sine of given angle I got $XY=100\sqrt3$ but radius of circle is still out of reach.

Answer 1

Since $S_2$ touches $S$ internally and passes through the center of $S$, its radius has to be half that of $S$. So since $S_2$ has radius $100$, $S$ has radius $200$.

If a circle touches a line, the radius at the point of contact has to be perpendicular to the line. This tells you that $X$, $Y$ and the center of $S_2$ form a $(30°,60°,90°)$ triangle, with the $60°$ at the center of $S_2$. Therefore that center forms an equilateral triangle with $O$ and $Y$, with an edge length of $100$.

Now assume some coordinates. I'll put $O=(0,0)$ and assume that $AB$ is horizontal, as in the figure in the question. The altitude of an equilateral triangle is $\frac{\sqrt3}2$ its edge length. So in this case $Y=(50\sqrt3,-50)$, with the altitude as $x$ coordinate and half the edge length as $y$ coordinate downwards. Since $XYO$ is a $(30°,30°,120°)$ isosceles triangle, by symmetry you have $X=(-50\sqrt3,-50)$. (This confirms the $\lvert XY\rvert=100\sqrt3$ you found for yourself.)

If you write $C=(-50\sqrt3,-50-r)$ for the center of $S_1$, a circle of radius $r$ around that center will touch $AB$. So now you have to make it touch $S$. To achieve that, you want $200=\lvert OC\rvert+r$ or

\begin{align*} 200-r &= \sqrt{(50\sqrt3)^2+(50+r)^2} \\ (200-r)^2 &= (50\sqrt3)^2+(50+r)^2 \\ 40{,}000 - 400r + r^2 &= 7{,}500 + 2{,}500 + 100r + r^2 \\ 30{,}000 &= 500r \\ r &= 60 \end{align*}

Answer 2

Divide the given radius of $S_2$ by $100$. If $O_2$ is the center of $S_2$ and $M$ the midpoint of $XY$ then $$|O_2Y|=|O_2O|=|OY|=|OX|=1,\quad |OM|={1\over2},\quad |XM|=|MY|={\sqrt{3}\over2}\ .$$ It follows that the unknown radius $r$ satisfies $$2-\sqrt{{3\over4}+\left(r+{1\over2}\right)^2}=r\ ,$$ which then leads to $r={3\over5}$.

Answer 3

Let me suggest a somewhat more synthetic solution. Let $T_1$ be the point of tangency of circles $S_1$ and $S$. Then, by the basic property of a point of tangency, points $O, \, O_1$ and $T_1$ are collinear, where $O_1$ is the center of $S_1$. Let $M$ be the midpoint of $AB$. Draw diameter line $MO$ and let it intersect circle $S$ at point $P$ on the side of circle $S_2$, i.e. $P$ is such that $O$ is inside the segment $MO$. Now, points $T_1, \, X$ and $P$ are actually collinear. Notice also that $O_1X$ is parallel to $OM$ because both of them are perpendicular to $AB$ and so $O_X$ is parallel to $OP$. Since $T_1, \,\, X, \,\, P$ are collinear, $T_1, \,\, O_1, \,\, O$ are also collinear and $O_1X$ is parallel to $OP$, the following relation holds (you can also think that triangles $T_1O_1X$ and $T_1OP$ are similar) $$\frac{T_1X}{T_1P} = \frac{T_1O_1}{T_1O} = \frac{O_1X}{OP}.$$ Since $T_1O_1 = r = O_1X$ and $T_1O = 200 = OP$ the latter equations turn into $$\frac{T_1X}{T_1P} = \frac{r}{200}$$ or if we express $r$ and notice that $T_1P = T_1X + XP$: $$r = \frac{200 \, \, T_1X}{T_1X + XP}$$ Let $L_1$ and $L_2$ be the intersection points of line $XO$ with circle $S$, $L_1$ on the side of $S_1$ and $L_2$ on the side of $L_2$. By the property of a point inside a circle, $$T_1X \cdot XP = L_1X \cdot XL_2 = 100 \cdot 300 = 30000$$ so $$T_1X = \frac{30000}{XP}$$ and thus $$r = \frac{200 \,\cdot \, 30000}{30000 + XP^2}$$ On the other hand, by Pythagoras' identity applied to the right angled triangle $XMP$ $$XP = \sqrt{XM^2 + MP^2} = \sqrt{XM^2 + (MO + OP)^2} = \sqrt{(50 \sqrt{3})^2 + (50+200)^2} = 100 \sqrt{7}.$$ Thus $$r = \frac{200 \,\cdot \, 30000}{30000 + XP^2} = \frac{200 \,\cdot \, 30000}{30000 + (100\sqrt{7})^2} = 60.$$