What's the measure of the radius of the biggest circle below?

For reference: In figure A is the tangency point:: LE=2(TE), $\overset{\LARGE{\frown}}{AN}=60^o$

$\frac{(TE)^2}{R-r}= 10$

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Calculate R.(Answer:80)

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$\triangle AON: \angle AON = 60^o, AO=r, ON=r \implies\\ AN =r \therefore \triangle AON(equilateral)$

$\triangle ATL(right): (LN+r)^2=(3x)^2+(r+OT)^2 \implies\\ LN^2+2LNr+r^2 = 9x^2+r^2+2rOT+OT^2 \rightarrow\\ LN^2+2LNr=9x^2+2rOT+OT^2$

....??? I"freeze" here...

Can I say that A, O. T and O1 are collinear?

Solution 1:

First note that $A, O $ and $O_1$ are collinear and as $A, O$ and $T$ are collinear, all four points $A, O, T$ and $O_1$ must be collinear. As $\angle O_1AL = 60^\circ$ and $O_1A = O_1L = R$, $\triangle O_1AL$ is equilateral with side length $R$.

Now there are multiple ways to get to the answer. One approach would be,

$LT = \frac{R \sqrt 3}{2} \implies TE = \frac{R}{2 \sqrt 3}$

$OT = \frac{R}{2} - r$

$OT^2 + TE^2 = OE^2 \implies \left(\frac{R}{2} - r\right)^2 + \left(\frac{R}{2 \sqrt3}\right)^2 = r^2$

And we obtain $R = 3r$

Now we also know that $TE^2 = \frac{R^2}{12} = 10(R-r) = \frac{20 R}{3}$.

$\therefore R = 80$

Solution 2:

Hint 1: Try proving angle $∠NOT=120°$ by getting the angles of quadrilateral $NOTL$ using triangles $△ANO$ and $△ALT$. This also proves that points $A$, $O$, $T$ are collinear.
Hint 2: Since the circles are internally tangent, points $A$, $O$ and $O_1$ must also be collinear. Combined with Hint 1, this answers your question about the collinearity of points $A$, $O$, $T$ and $O_1$ : yes, they're collinear.
Hint 3: Try constructing lines $AO_1$ and $LO_1$. Is triangle $ALO_1$ equilateral?
Hint 4: After connecting lines $NE$, $OE$ and constructing the line perpendicular to $AL$ in $△ALO_1$ from angle $∠AO_1L$ (the perpendicular line is also collinear with point E), some new 30-60-90 and 60-60-60 triangles are formed, which can then be used to finish solving the problem.