**
P
**
of two straight lines
**
l,m
**
is inside the given circle
**
O
**
. 8 circles tangent the lines
**
l,m
**
and circle
**
O
**
, the circles have numbers 1-8 (see the picture). We know the radii of three circles, the centers of which lie on the bisector
**
b
**
of the angle between lines
**
l,m
**
. Find the radius of the fourth circle, whose center lies on the bisector
**
b
**
.

$r_4 = 8, r_6 = 3, r _8 = 2\\ r_2=?$

Appendix. Find the product
$r_3 r_4 r_5 r_6$
as a function from
$R,d$
.
$R$
is the radius circle
**
O
**
,
$d$
- distanse from centrum of circle
**
O
**
to intersection point of two straight lines
**
l,m
**
.

Problem is original.

The answer is 12.

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Well,to solve this mysterious problem,let's first prove a lemma .

Lemma$\overline{PY_{1}}\cdot\overline{PY_{2}}=$ $P$ 's power of $\odot O(<0)$ .ProofFirst,let's show that $\overline{PX_{1}}\cdot\overline{PX_{2}}=$ $P$ 's power of $\odot O$ .It's equivalent to show $X_{1},P,X_{2}$ are collinear.Note that $Q_{1}Z_{1}\perp Z_{1}Z_{2}$ and $P_{2}Z_{2} \perp Z_{1}Z_{2}$ give us $Q_{1}Z_{1}//P_{2}Z_{2}$ $\Longrightarrow$ $\frac{Q_{1}Z_{1}}{P_{2}Z_{2}}=\frac{PQ_{1}}{PP_{2}}$ $\Longrightarrow$ $\frac{Q_{1}X_{1}}{P_{2}X_{2}}=\frac{PQ_{1}}{PP_{2}}$ $\Longrightarrow$ $\frac{Q_{1}X_{1}}{X_{1}O}\cdot\frac{OX_{2}}{X_{2}P_{2}}\cdot\frac{P_{2}P}{PQ_{1}}=1$ .Hence by Menelaus's Theorem,we have $X_{1},P,X_{2}$ are collinear.Now let $U$ denotes the the second intersection of the line $X_{1}PX_{2}$ and $\odot P_{2}$ . A homothetic transformation whose centre and ratio are $P$ and $-\frac{R_{\odot P_{2}}}{R_{\odot Q_{1}}}$ respectively helps us throw $Y_{1}$ , $X_{1}$ to $Y_{2}$ , $U$ ,respectively.Thus, $Y_{1}X_{1}//Y_{2}U$ $\Longrightarrow$ $\angle X_{1}Y_{1}P=\angle UY_{2}P=\angle Y_{2}X_{2}U=\angle Y_{2}X_{2}P$ $\Longrightarrow$ $X_{1},Y_{1},X_{2},Y_{2}$ are concyclic $\Longrightarrow$ $\overline{PY_{1}}\cdot\overline{PY_{2}}=\overline{PX_{1}}\cdot\overline{PX_{2}}=$ $P$ 's power of $\odot O$ . $\Box$

Come backUsing the lamma,we have $\overline{PY_{1}}\cdot\overline{PY_{2}}=\overline{PY_{1}'}\cdot\overline{PY_{2}'}=$ $P$ 's power of $\odot O$ $\Longrightarrow$ $\frac{PY_{1}'}{PY_{1}}=\frac{PY_{2}}{PY_{2}'}$ $\Longrightarrow$ $\frac{R_{\odot{P_{1}}}}{R_{\odot{Q_{1}}}}=\frac{R_{\odot{P_{2}}}}{R_{\odot{Q_{2}}}}$ . So $r_{2}=r_{4}\cdot \frac{r_{6}}{r_{8}}=\boxed{12}$

About the Appendix$r_{3}r_{4}r_{5}r_{6}=(R^{2}-d^{2})^{2}$ (easy to get from our lemma,as you can express $r_{3}r_{5}$ and $r_{4}r_{6}$ as $tan^{2}\theta\cdot (R^{2}-d^{2})$ and $cot^{2}\theta\cdot (R^{2}-d^{2})$ respectively,here $\theta$ is a half of the acute angle that formed by $l$ and $m$ in the picture.)