NOTE: I would appreciate it if you provided a hint and not the whole solution.
BdMO 2014 Nationals:
In acute angled triangle ABC, considering a portion of side BC as diameter a
circle is drawn whose radius is 18 units and it touches AB and AC side. Similarly,
considering a portion of sides AC and AB as diameters, two other circles are
drawn whose radii are 6 and 9 units respectively. What is the radius of the incircle
of ∆ABC?
The main problem with this problem is that the figure is incredibly messy.So I drew one circle,and tried getting information from it.I noticed that if $O$ is a centre of one of the above (semi)circles,and if $O$ is located on $AB$, then $CO$ is an angle bisector.Therefore,if we connect all the vertices with the centres,their intersection would be the incentre.Then we need to drop a perpendicular from this point to get the inradius.But that hardly helps us.I have found some similar triangles in the figure.But that does not help me at all.
I have also tried backtracking.I tried to put myself in the problem-maker's shoes and tried to imagine what I would do if I had to create a problem like this.Unfortunately,I couldn't do it..
Any insightful comment,hint will be very much appreciated.
Best Answer
Here's a (not-to-scale) picture of the situation:
Necessarily, each circle center ($D$, $E$, or $F$) is the point where an angle bisector meets an opposite edge; moreover, the points of tangency of a circle with the adjacent edges (for instance, $D^\prime$ and $D^{\prime\prime}$) are simply the feet of perpendiculars from the center to those edges.
We'll write $a$, $b$, $c$ for the lengths of sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, and $d$, $e$, $f$ for the radii of $\bigcirc D$, $\bigcirc E$, $\bigcirc F$. Now, observing that each angle bisector cuts the triangle with into sub-triangles with convenient "bases" and "heights", we can compute the area, $T$, of $\triangle ABC$ in three ways:
$$T \;=\; \frac12 d\;(b+c) \;=\; \frac12 e\;(c+a) \;=\; \frac12 f\;(a+b) \tag{1}$$
Of course, writing $r$ for the inradius of $\triangle ABC$, we have a well-known fourth formula for area: $$T \;=\; \frac12 r\;(a+b+c) \tag{2}$$
We can easily eliminate $a$, $b$, $c$ from the above. For instance, $$\begin{align} b+c = \frac{2T}{d}\quad c+a=\frac{2T}{e}\quad a+b = \frac{2T}{f} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{1}{d}+\frac{1}{e}+\frac{1}{f}\;\right) \tag{3} \\ a+b+c = \frac{2T}{r} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{2}{r}\;\right) \tag{4} \end{align}$$ so that, as @Jack notes,
Addendum (four years later!). As @jmerry has observed, the specific configuration in the problem statement is invalid. If we solve $(1)$ for $a$, $b$, $c$, we find $$a = \left(-\frac1d + \frac1e + \frac1f \right)T \qquad b = \left(\frac1d - \frac1e + \frac1f \right)T \qquad c = \left(\frac1d + \frac1e - \frac1f \right)T$$ With $d=18$, $e=6$, $f=9$, these become $a=2T/9$, $b=0$, $c=T/9$, which do not correspond to a valid triangle ... not even a validly-degenerate one. (It's a good thing I didn't claim my picture was to scale.) A valid triangle requires that the three aspects of the Triangle Inequality hold $$a \leq b+c \qquad b \leq c+a \qquad c \leq a+b$$ which, in turn, require $$\frac3d \geq \frac1e + \frac1f \qquad \frac3e \geq \frac1f+\frac1d \qquad \frac3f \geq \frac1d+\frac1e$$ (The first of these is violated by the given values of $d$, $e$, $f$.)