Given three circles of radius $R$, which touch only at one point and there centers belong to $OD$. Find the length of $AB$.

Question

Given three circles of radius $R$, which touch only at one point and there centers belong to $OD$. The line $OC$ is tangent to $(M, R)$ and intersects $(K,R)$ at $O$. Find the length of $AB$ in terms of $R$.

I attempted to solve it as follows:

$OC^2=(OD-2R)^2\cdot OD^2$

$MO^2+MC^2=OC^2$

$OA\cdot (OA+AB)=(OD-4R)^2+(OD-2R)^2$

I then attempted to combine the equations above, but they didn't give anything useful. Could you please explain to me how to solve the question?

Answer 1

Drop $LP \perp AB$ with $P$ on $AB$. Then $\triangle OPL \sim \triangle OCM $ with similarity ratio $3/5$. Why?

Then $LPA$ is a $3-4-5$ right triangle whence $AB=8R/5$.

(Intermediate steps are left for the reader.)

Answer 2

Hint: Draw perpendicular from $L$ and $M$ to $OC$. Say perp from $L$ meets $OC$ at $H$. Then $AH = HB$.

Now observe that $\triangle OLH \sim \triangle OMC$. Find length of $LH$ using similarity. Then use Pythagoras in $\triangle LHA$.

Find $AH$ and then $AB$.

Answer 3

HINT:

Let the radius be 1, then...