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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)

Figure

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.

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    $\begingroup$ This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – user296602 Aug 31 '16 at 6:17
  • $\begingroup$ Please improve your title to be more specific. See A good title $\endgroup$ – 6005 Aug 31 '16 at 6:22
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    $\begingroup$ @6005, T. Bongers, A---B, ok now? $\endgroup$ – Display name Aug 31 '16 at 6:33
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    $\begingroup$ No they touch at one point @A---B $\endgroup$ – Display name Aug 31 '16 at 6:35
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    $\begingroup$ @A---B not given in question, though can be drawn. $\endgroup$ – Display name Aug 31 '16 at 7:05
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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*}

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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 $r={3\over5}$.

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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 $$r = \frac{200 \,\cdot \, 30000}{30000 + XP^2} = \frac{200 \,\cdot \, 30000}{30000 + (100\sqrt{7})^2} = 60.$$

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