<s>[INST] $2$ circles internally tangent to a third one with given values of radii and length of 2 chords. Find the value of a certain product of lengths The line AB goes through P. The radius of the red circle is 81, the large black circle is 99, and the blue circle 18. The line XY is tangent to the red and blue circles.
Find NX*NA.
I made an attempt via finding many side lengths (Found that the distance between O_2 and X is 54, and that ABN and YXN are similar. However, this didn't lead to anywhere. I also found the ratio between AX and XN is 9/2.
I've also found that sin(ANB) is 2/3.
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Edit: Adding a solution using similarity of triangles and Pythagoras theorem
$\small O_2X^2 = XT^2 + O_2T^2 = XT^2 +  O_1O_2^2 - (O_1X - XT)^2= 18^2 + 81^2 - 63^2$
$\implies \small O_2X = 54$.
As $O_1O = O_2P = 18, O_1P = OO_2 = 81, $ then we must have $ \small \ O_1P \parallel OB$.
Now you can show $ \small \angle AO_1P = \angle OO_1O_2$.
If $\small AU \perp O_1P \text { and } O_2W \perp AO, \triangle O_1 O_2 W \cong \triangle O_1 A U$.
$\small AU^2 = AO_1^2 - O_1U^2 = 81^2 - 9^2, AP^2 = AU^2 + PU^2 = 81^2 - 9^2 + (81-9)^2$
$ \displaystyle \small AU = 36 \sqrt{5}, AP = 108, AV = \frac{11}{9} AU = 44 \sqrt5, OV = 11$
$\small AO_2^2 = AV^2 + VO_2^2 \implies AO_2 = 54 \sqrt5$
Now note that $ \displaystyle \small \angle O_2AX = \frac{1}{2} \angle O_2O_1X \implies \triangle O_2AS \sim \triangle O_2O_1M$
That leads to $ \small O_2S = 18 \sqrt5$ and applying Pythagoras theorem gives you $\small AS$ and $\small SX$. Adding them,
$\small AX = 36 (\sqrt{10}+1)$
As $ \small \ O_1P \parallel OB, \angle AXP = \angle ANB \implies PX \parallel BN$
$ \therefore \displaystyle \small \frac{AN}{AX} = \frac{AB}{AP} = \frac{OB}{O_1P} = \frac{99}{81} = \frac{11}{9} \implies 
AN = \frac{11}{9} \cdot AX, XN = \frac{2}{9} \cdot AX$
$ \displaystyle \small NX \cdot NA = \frac{2}{9} \cdot \frac{11}{9} \cdot AX^2 = 352 (11 + 2 \sqrt{10})$
Solution using trigonometry:
For calculation of $ \small AP$ and $ \small O_2X$ and establishing $ \small PX \parallel BN$, please see previous solution.
If $ \displaystyle \small \angle PAO_2 = \alpha$, $\sin \alpha = \sin (\frac{1}{2} \angle PO_1O_2) = \frac{9}{81} = \frac{1}{9}$
If $ \displaystyle \small \angle O_2 A X = \beta$, $\sin \beta = \sin (\frac{1}{2} \angle O_2 O_1 X) = \frac{27}{81} = \frac{1}{3}$
If $ \small \angle AXP = \gamma \text { and } R \text { is circumradius of } \triangle APX, R = 81$.
$ \displaystyle \small 2R \sin \gamma = AP \implies \sin \gamma = \frac{2}{3}$
$ \small AX = 2 R \sin \angle APX = 2 R \sin (180^0 - (\alpha + \beta) - \gamma) \small = 162 \sin (\alpha+\beta+\gamma)$
$ \displaystyle \small \sin (\alpha + \beta) = \frac{1}{9} \cdot \frac{2\sqrt2}{3} + \frac{1}{3} \cdot \frac{4\sqrt5}{9} = \frac{4\sqrt5+2\sqrt2}{27}$
$ \displaystyle \small \cos(\alpha + \beta) = \frac{8\sqrt{10}-1}{27}, \cos \gamma = \frac{\sqrt5}{3}$
$ \displaystyle \small \sin(\alpha+\beta+\gamma) = \frac{2 (\sqrt{10}+1)}{9}$
So $ \small AX = 36 (\sqrt{10}+1)$
From here, follow the same steps as in previous solution.
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