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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.

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Here's a (not-to-scale) picture of the situation:

enter image description here

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

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  • 2
    $\begingroup$ That was nice.+1 $\endgroup$ – rah4927 Dec 23 '14 at 18:23
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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 $a<b+c$ 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.

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enter image description here

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.

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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: $$\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}=\frac{2}{r}.$$

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