I am solving this question shown below:
Two circles embedded between two parallel lines whose distance is $x$ and three other circles with indicated diameters touch both the bigger ones. Find $x$
My try:
I used the following facts:
Fact $1:$ If two circles touch externally, distance between their centers is sum of radii.
Fact $2:$ If two circles touch internally, distance between centers is absolute difference of radii.
Fact $3:$ The distance between parallel tangents is Diameter.
Based on these, I have got this figure below:
$O$ and $M$ are centers of Bigger circles.$P,Q,R$ are centers of smaller ones.
By Cosine rule in $\Delta OPM, \Delta OPQ$, we have:
$$\cos \theta=\frac{\left(\frac{x}{2}+4\right)^2+12^2-\left(\frac{x}{2}-4\right)^2}{2(12)\left(\frac{x}{2}+4\right)}=\frac{\left(\frac{x}{2}+4\right)^2+\left(\frac{x}{2}+6\right)^2-10^2}{2\left(\frac{x}{2}+4\right)\left(\frac{x}{2}+6\right)}$$
This gives $$x=12\sqrt{5}$$
Also by cosine rule in $\Delta OMR, OQR$, we get
$$\cos \alpha=\frac{\left(\frac{x}{2}+\frac{9}{2}\right)^2+144-\left(\frac{x}{2}-\frac{9}{2}\right)^2}{2\left(\frac{x}{2}+\frac{9}{2}\right)(12)}=\frac{\left(\frac{x}{2}+\frac{9}{2}\right)^2+\left(\frac{x}{2}+6\right)^2-\frac{441}{4}}{2\left(\frac{x}{2}+\frac{9}{2}\right)\left(\frac{x}{2}+6\right)}$$
This gives $$x=12\sqrt{7}$$
Where i went wrong?
Best Answer
As mentioned in a comment, OP's error was in assuming that point $M$ was on segment $OQ$, an impossibility due to the vertical asymmetry of the figure.
There's probably a slick solution here involving inversive geometry, but here's one that follows OP's trigonometric lead:
We take the large circles $\bigcirc O$ and $\bigcirc O'$ to have radius $r$, and the smaller circles $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ (with $\bigcirc A$ between $\bigcirc B$ and $\bigcirc C$) to have radii $a$, $b$, $c$. Define $\beta := \angle OAB$, $\gamma := \angle OAC$, $\beta' := \angle O'AB$, $\gamma' := \angle O'AC$.
The Law of Cosines in $\triangle OAB$, $\triangle OAC$, $\triangle O'AB$, $\triangle O'AC$ tells us $$\begin{align} \cos\beta &= \frac{-(r+b)^2+(a+b)^2+(r+a)^2}{2(a+b)(r+a)}=\frac{a^2 + a b + a r - b r}{(a + b) (r+a)} \\[4pt] \cos\gamma&=\frac{a^2 + a c + a r - c r}{(a + c) (r+a)} \\[4pt] \cos\beta' &= \frac{a^2 + a b - a r + b r}{(a + b) (r - a)} \\[4pt] \cos\gamma' &= \frac{a^2 + a c - a r + c r}{(a + c) (r - a)} \end{align}$$ Now, since $\beta+\gamma=\angle BAC=\beta'+\gamma'$, we have $\cos(\beta+\gamma)=\cos(\beta'+\gamma')$, so that $$\cos\beta\cos\gamma-\sin\beta\sin\gamma = \cos\beta'\cos\gamma' - \sin\beta'\sin\gamma'$$ With a couple of rounds of squaring, we can replace sines with cosines, and then substitute the above expressions. (A computer algebra system really helps with this process.) When the dust settles, we find $$r^2 = \frac{a^2 (a b + a c + 2 b c)^2}{4 b c (a^2 - b c)} \qquad\to\qquad r = \frac{2\bar{a}+\bar{b}+\bar{c}}{2\bar{a}\,\sqrt{\bar{b}\bar{c} -\bar{a}^2}}\qquad\left(\bar{x}:=\frac1x\right)$$ For OP's specific case, $a=6$, $b=4$, $c=9/2$, so that $r=29/2$. $\square$
We can also consider the Law of Cosines in $\triangle OAO'$ $$|OO'|^2 = |OA|^2 + |O'A|^2 - 2 |OA||O'A|\cos(\beta'-\beta) $$ to determine that (ignoring an extraneous solution) $$|OO'|^2 = \frac{a^2 (a b + a c + 2 b c) (a b + a c - 2 b c)}{b c (a^2 - b c)}= \frac{(\bar{b}+\bar{c}+2\bar{a})(\bar{b}+\bar{c}-2\bar{a})}{\bar{a}^2(\bar{b}\bar{c} - \bar{a}^2)}$$ This is helpful for drawing a figure. (One can also calculate $\angle AOO'$, $\angle BOA$, $\angle COA$ to determine accurate placement of the smaller circles. Frankly, I just eye-balled that. :)
Note that $$\frac{4r^2}{|OO'|^2} = \frac{\bar{b}+\bar{c}+2\bar{a}}{\bar{b}+\bar{c}-2\bar{a}}=\frac{\mu+\bar{a}}{\mu-\bar{a}} \qquad\left(\mu:=\frac12(\bar{b}+\bar{c})\right)$$ This suggests that there might be something interesting going on in the underlying geometry; perhaps something that helps side-step the rather arduous algebraic manipulations above. Investigation is left as an exercise to the reader.