# Three circles having centres on the three sides of a triangle

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.

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,

$$\frac{2}{r} = \frac{1}{d} + \frac{1}{e} + \frac{1}{f} \tag{\star}$$

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 $$d$$, $$e$$, $$f$$.)

• That was nice.+1 – rah4927 Dec 23 '14 at 18:23

The existing answers have provided the intended solution. On the other hand, there's a catch - the problem has a fatal flaw.

What can we say about the side lengths of the triangle? We have $$b+c$$, $$a+c$$, and $$a+b$$ in proportions $$1:3:2$$, so $$a$$, $$b$$, and $$c$$ are in proportions $$2:0:1$$. That is not a triangle; $$a>b+c$$.

Let the sides be $$a,b,c$$, the radii of the circles opposite them be $$R_A,R_B,R_C$$, and the area of the triangle be $$T$$, mixing notation from the answers of Blue and Jack D'Aurizio. The triangle inequality $$a gives $$\frac1{R_B}+\frac1{R_C}=\frac1{2T}(2a+b+c) < \frac1{2T}(3b+3c) = \frac3{R_A}$$ $$R_A < \frac{3R_BR_C}{R_B+R_C}$$ and similarly for the other two orders; the largest radius must be less than $$\frac32$$ times the harmonic mean of the other two. An acute triangle would be an even stronger restriction. For the given trio of radii $$18,6,9$$, this is not true.

As such, the true answer to the problem:
The given configuration is impossible. There is no such triangle.

Let $$r$$ denote the radius of the inscribed circle of triangle $$ABC$$ and radii of the circles be $$r_a$$, $$r_b$$, $$r_c$$. Also, let the corresponding heights of the triangle $$ABC$$ be $$h_a,h_b$$ and $$h_c$$. Construct $$\triangle A_1B_1C$$ for which the circle $$C$$ is incircle, and $$A_1B_1\parallel AB$$. as shown.

Clearly,

\begin{align} r&= r_c\,\frac{|CH_{c1}|}{|CH_{c}|} = \frac{r_c\,h_c}{h_c+r_c} ,\\ \text{or }\quad \frac 1r&=\frac 1{r_c}+\frac 1{h_c} . \end{align}

Similarly,

\begin{align} \frac 1r&=\frac 1{r_a}+\frac 1{h_a} ,\quad \frac 1r=\frac 1{r_b}+\frac 1{h_b} ,\\ \text{hence }\quad \frac 3r&= \frac 1{r_a}+\frac 1{r_b}+\frac 1{r_c} + \frac 1{h_a}+\frac 1{h_b}+\frac 1{h_c} \tag{*}\label{*} . \end{align}

Combining \eqref{*} with a well-known relation \begin{align} \frac 1r&=\frac 1{h_a}+\frac 1{h_b}+\frac 1{h_c} , \end{align}

we have

\begin{align} \frac 2r&=\frac 1{r_a}+\frac 1{r_b}+\frac 1{r_c} . \end{align}

Given $$r_a=18$$, $$r_b=6$$, $$r_c=9$$, we must have $$r=6$$, but...

Given that $$r$$, we can find all the heights of $$\triangle ABC$$.

\begin{align} h_a&=\frac{r_a r}{r_a-r} ,\quad h_b=\frac{r_b r}{r_b-r} ,\quad h_c=\frac{r_c r}{r_c-r} . \end{align}

Since $$r=r_b=6$$, the height $$h_b$$ must be infinite. As far as two circles with the same radius are touching two lines internally, these lines must be parallel in this case.

Hence, the question is a trap, and as @jmerry stated, the correct answer would be: there is no such triangle.

Obviously, the centers of the three circles are the feet $L_A,L_B,L_C$ of the angle bisectors. Let $I$ be the incenter. An homothethy gives that the radius of the circle on the $BC$-side is $\frac{AL_A}{AI}$ times the inradius $r$. Van Obel's theorem and the bisector theorem give: $$\frac{AL_A}{AI}=\frac{a+b+c}{b+c},$$ hence we know the ratios between the side lengths of $ABC$.

Moreover, if $R_A,R_B,R_C$ are the three radii of the circles, by the previous relation we get: $$\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}=\frac{2}{r}.$$