Two inscribed circles are tangential to a chord with diameter line at $30^\circ$ with chord. Find radius ratio of two circles

Question

I found the problem here (I can't see deleted posts) but the post got downvoted and deleted soon, but I felt so inspired to find the solution that can't let the problem rot by itself, so, rather re-pharased, the problem is:

Let $\omega$ be a circle with center at $O$, $\omega_1$ be the circle touching $\omega$ internally at $B$, passing through $O$ and touching a chord $\ell$ in $A$ such, that $\angle BDA=30^\circ=\frac{\pi}{6}$ where $D=BO\cap\ell$. And let $\omega_2$ be the circle, touching $\omega$ internally and touching $\ell$ in $D$. Find the radii ratio of $\omega_1,\,\omega_2$.

Answer 1

tl;dr, but long things short: looking up how to inscribe a circle between an arc and the chord made things very easy (considering only similar triangles and sine and cosine of $30^\circ$).

Rather than considering properties given, I'd construct $\omega_1,\ell,\omega_2$ such that they have the properties.
Given an arbitrary point $B$ on $\omega$, how do we construct $\omega_1$ to touch $\omega$ and to pass through $\omega$'s center? Basing on that the radii $FB,OB$ should be parallel on lie on the same line as both $FB,OB$ are perpendicular to the common tangent line at $B$, so $F\in OB,FB=\frac{1}{2}OB$. Given $B$, such $\omega_1$ is unique.
Now we construct $\ell$: consider $\triangle FAD,\,\angle FAD=90^\circ$ so $\angle DFA=60^\circ$ and such $A$ is unique up to symmetry about $BO$.
Now we have to construct $\omega_2$: taking an arbitrary point $E$ on continuation of line $AB$ beyond $\omega$ we can construct line $EI$, touching $\omega$ and bisector of $\angle DEI$ will contain center of $\omega_2$, then perpendicular bisector will contain $E$ as being the bisector of $\angle DEI$, thus $E$ is not needed. The center of $\omega_2$ will be intersection of that bisector with the perpendicular to $AD$ containing $D$ (as $\omega_2$ touches $AD$ in $D$), but how to make $\omega_2$ touch $\omega$?
And then I thought: why do I have to do that? The result of inscribing a circle between an arc and it's chord should be known and I can just look it up (which can't be at a contest, obviously). And indeed (google translate of http://www.problems.ru/view_problem_details_new.php?id=111699),

Lemma 1: Chord $AB$ splits circle S into two arcs. The circle $S_1$ touches the chord $AB$ at the point $M$ and one of the arcs at the point $N$. Prove that a) the line $MN$ passes through the middle $P$ of the second arc; b) the length of the tangent $PQ$ to the circle $S1$ is equal to $PA$.
Proof: Let $O$ and $O_1$ be the centers of circles $S$ and $S_1$, respectively. Since $OP = ON$ and $O_1M = O_1N$ (radii of the same circle), the triangles $OPN$ and $O_1MN$ are isosceles, and $OPN$ is their common angle at the bases. Therefore, the points $N$, $M$, and $P$ lie on the same line.
Another way. We consider homothety centered at the tangent point $N$ of the circle, taking the circle $S_1$ to the circle $S$. The tangent $AB$ to the circle $S_1$ goes to the tangent $l$ to the circle $S$ parallel to it, the tangent parallel to the chord $AB$ divides the arc $AB$ in half. Then the point $M$ goes to the middle $P$ of the arc $AB$ that does not contain the point $N$. Therefore, the line $MN$ passes through the middle $P$ of this arc.
b) We extend the radius $OP$ of the circle $S$ to the intersection with the chord $AB$ at the point $K$. Then $K$ is the midpoint of the chord $AB$. Applying the tangent and secant theorem, the product of segments of intersecting chords, and the Pythagorean theorem, we obtain $$\begin{align*} PQ^2 &= PM\cdot PN = PM\cdot (PM + MN) = PM^2 + PM\cdot MN =\\ &= (PK^2 + KM^2) + AM\cdot MB = (PK^2 + KM^2) + (AK + KM)\cdot(BK-KM) =\\ &= (PK^2 + KM^2) + (AK + KM)\cdot(AK-KM) = (PK^2 + KM^2) + (AK^2-KM^2) =\\ &= PK^2 + AK^2 = AP^2. \end{align*}$$ Therefore, $PQ = AP$.
Another way. We extend $PO$ to the intersection with the circle S at the point $L$. Right triangles $PKM$ and $PNL$ are similar, therefore $=$, whence $PM \cdot PN = PK \cdot PL$. In addition, $AK$ is the height of the right triangle $APL$ drawn from the top of the right angle $PAL$. Hence, $$PQ^2 = PM \cdot PN = PK \cdot PL = PA^2.$$

and this was the key point.

For constructing $I$ we take $J$ as the middle of the arc for the chord $\ell$, containing $B$: we take $I$ as intersection of $\omega$ and the perpendicular to $AD$, passing through $O$ and intersecting $AD$ at $C$: as $\triangle CDO\sim\triangle ADF$ ($FA\perp AD$ as radius and the touching line of $\omega_1$), hence $\frac{DA}{DC}=\frac{DF}{DO}$, but $DF=\frac{FA}{\sin 30^\circ}=\frac{FO}{\sin 30^\circ}=2FO$ hence $OD=FO$ and $DC=CA$, anyway a line passing through center and perpendicular to a chord contains diameter thus meeteing the arcs in their middle by symmetry. And we take $I=JD\cap\omega$.
Now by constructing $GH$ as the bisector perpendicular of $ID$, meeting $ID$ at $H$ and meeting the line, perpendicular to $AD$ passing through $D$ at $G$ we consider $\triangle JCD\sim\triangle DHG$ as $\angle CJD=\angle GDH$ and $\angle H=\angle C=90^\circ$.
The rest is pretty straightforward: as we get $OC=OI+IC,CD,JI=JD+DI=JD+2DH$ in terms of $OF=r$ and $DG=x$ and the triangle similarity and additionally $JI=4r\cos\angle CJD$ in terms of $r$ we can derive the equation, bounding $x$ and $r$ and solwe it.

Answer 2

A solution with details and justifications:

Let $r = BO$ be the radius of the big circle $\omega$. Furthermore, let $O_1$ be the center of the circle $\omega_1$ and let $r_1$ be its radius. Let $O_2$ be the center of the circle $\omega_2$ and let $r_2$ be its radius. By assumption, since $\omega_1$ is internally tangent to $\omega$ at the point $B$, the three points $O, \, O_1, \, B$ are collinear. Furthermore, the center $O$ of $\omega$ lies on $\omega_1$ and since $O_1 \in BO$, the segment $BO$ is a diameter of $\omega_1$, so $r = 2\, r_1$ and $O_1$ is the midpoint of $BO$, i.e. $BO_1 = OO_1 = r_1 = \frac{r}{2}$.

Since $AD$ is tangent to $\omega_1$ at point $A$, we know that $\angle \, DAO_1 = 90^{\circ}$. Combined with the assumption that $\angle \, ADO_1 = 30^{\circ}$, we conclude that $\angle \, AO_1O = \angle \, AO_1D = 60^{\circ}$ and $DO_1 = 2\, AO_1 = 2\, r_1 = r$. Since $AO_1 = r_1 = OO_1$, triangle $AOO_1$ is isosceles with $\angle \, AO_1D = 60^{\circ}$, so it is equilateral and thus $OO_1 = AO_1 = AO = r_1$. However, $OO_1 = r_1$ so $$DO = DO_1 - OO_1 = 2r_1 - r_1 = r_1 = OO_1 = AO_1 = AO$$ Let $M$ be the midpoint of the segment $AD$. Since $DO = AO$, triangle $ADO$ is isosceles and $OM$ is its median, so it is in fact the orthogonal bisector of $AD$. Moreover, $\angle\, OAD = \angle \, ADO = 30^{\circ}$, so $OM = \frac{1}{2} \, DO = \frac{1}{2} \, AO = \frac{r_1}{2}$ and $DM = AM = \frac{\sqrt{3}}{2}\, AO = \frac{\sqrt{3}}{2}\, r_1$.

Denote by $F$ the point of tangency between the circles $\omega$ and $\omega_2$. Extend the line $OM$ until it intersects the big circle $\omega$ at the points $K$ and $L$, where $L$ and $F$ are on the same side of the line $l$, and $K$ and $F$ are on different sides of $l$. Then $KL$ is a diameter of $\omega$ (because $O \in KL$) and $KL \, \perp \, AD$.

Lemma: The three points $F, \, D$ and $K$ are collinear.

Proof of Lemma: One can use some version of Thales' intercept theorem or homothety. For example: since $F$ is the point of tangency of $\omega$ and $\omega_1$, the three points $F,\, O_2$ and $O$ are collinear. Therefore, there exists a homothety with center $F$ that maps point $O_2$ to point $O$. Then, the scaling factor of this homothety is $\frac{FO}{FO_2} = \frac{r}{r_2}$. By applying the said homothety, the circle $\omega_2$ is mapped to a new circle, whose center is the image of the center $O_2$ of $\omega_2$, which by construction is the point $O$. However the radius of the new circle should be equal to $\frac{r}{r_2} \cdot r_2 = r$ and since there exists exactly one circle with center $O$ and radius $r$, and that circle is $\omega$, we conclude that $\omega_2$ is mapped to $\omega$ under the constructed homothey. Because $\omega_2$ is tangent to $AD$ at $D$ and $O_2$ is the center of $\omega_2$, we observe that $O_2D \, \perp \, AD$. Then, the line $O_2D$ is mapped by the homothety to a line through point $O$ parallel to $O_2D$ and therefore perpendicular to $AD$. However, by construction, the unique line through point $O$ perpendicular to $AD$ is the line $OM$, which has been extended to the diameter $KL$, so the line $O_2D$ is mapped by the homothety to the line $KL$ and in particular the ray $O_2D$ is mapped to the ray $OK$. Then, since the homothety maps the circle $\omega_1$ to the circle $\omega$ and the ray $O_2D$ to the ray $OK$, the intersection point $D = \omega_1 \cap O_2D$ is mapped to the corresponding intersection point $K = \omega \cap OK$ and therefore the points $F, \, D$ and $K$ are collinear. $\,\,\,\,\, \square$

As already established, $KM \, \perp \, AD$ so $\angle \, KMD = 90^{\circ}$. By Pythagoras' theorem $$KD = \sqrt{KM^2 + DM^2} = \sqrt{(KO + OM)^2 + DM^2} = \sqrt{(BO + OM)^2 + DM^2}= \sqrt{\left(2\,r_1 + \frac{r_1}{2}\right)^2 + \frac{3}{4}\, r_1^2} = \sqrt{7}\, r_1$$
If we denote by $C$ the diametrically opposite point of $B$ on $\omega$, then the center $O$ and the point $D$ lie on the diameter $BC$ and furthermore $$BD = BO_1 + OO_1 + DO = 3\, r_1 \,\,\,\text{ and }\,\,\, CD = BC - BD = 2\, BO - 3\, r_1 = 4\, r_1 - 3\, r_1 = r_1$$. By one of these power-of-point theorems $$BD \cdot CD = FD \cdot KD$$ which translates to the equation $$3\, r_1 \cdot r_1 = FD \cdot \sqrt{7}\, r_1$$ and if we solve it for $FD$, $$FD = \frac{3}{\sqrt{7}}\, r_1$$ By the homothety defined in the Lemma above, or if you prefer by Thales' intercept theorem for the points $D \in FK$ and $O_2 \in FO$ $$\frac{FD}{FK} = \frac{FD}{FD + KD} = \frac{FO_2}{FO} = \frac{r_2}{r} = \frac{r_2}{2\, r_1}$$ which can be reformulated into the equation $$\frac{r_2}{r_1} = 2 \, \frac{FD}{FD + KD} = 2\, \frac{\frac{3}{\sqrt{7}}\, r_1}{ \frac{3}{\sqrt{7}}\, r_1 + \sqrt{7}\, r_1} = \frac{3}{5} $$

Answer 3

Let $r_1$ and $r_2$ be the radii of the circles $\omega_1$ and $\omega_2$, respectively. From the right triangle $ADO_1$, we get $OD = r_1\csc 30^\circ - r_1 = r_1$. Then, apply the cosine rule to the triangle $ODO_2$

$$OO_2^2 = (2r_1-r_2)^2 = r_1^2+r_2^2-2r_1r_2\cos(30^\circ + 90^\circ)$$

which leads to the ratio

$$\frac{r_1}{r_2} = \frac53$$