# A circle of radius $r$ is inscribed into a triangle.

A circle of radius $$r$$ is inscribed into a triangle $$ABC$$. Tangent lines to this circle parallel to the sides of the triangle cut out three smaller triangles, $$\triangle A_cB_cC$$, $$\triangle A_bBC_b$$, $$\triangle AB_aC_a$$. The radii of the circles inscribed in these smaller triangles are equal to $$1$$, $$2$$ and $$3$$, respectively. Find $$r$$.

I have no idea how to start...

• I solved your problem. If you want to see my solution, show please your attempts. Commented Jul 6, 2019 at 19:06
• A drawing could help to clear things out. Commented Jul 6, 2019 at 19:25
• In order to get responses that suit your needs, please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck. Commented Jul 6, 2019 at 20:23
• He says that he does not know how to start. DonAntonio suggested a useful approach - @sailormars2016 draw a diagram and the radiuses to the tangent lines.
– Moti
Commented Jul 7, 2019 at 1:10

Consider triangle ABC, A as top vertex, B the right on and C the left. let;s denote the heights as $$h_a$$ , $$h_b$$ and $$h_c$$, and the radii of circles $$r_a=1$$, $$r_b=2$$ and $$r_c=3$$ and unknown radius as $$r$$ . The triangles created by tangents to circles and parallel to bases are similar to main triangle, let't denote the heights of these triangles as $$h'_a$$. $$h'_b$$ and $$h'_c$$; we may write:

$$\frac{r_a}{r}=\frac{h'_a}{h_a}$$

$$h_a=2r+h'_a$$ .

Therefore:

$$h_a=2r+\frac{r_ah_a}{r}$$

Which gives:

$$2r^2-rh_a +h_a=0$$

Similarly we get:

$$2r^2-rh_b +2h_b=0$$

$$2r^2-rh_c +3h_c=0$$

Now we this statement: If three perpendiculars from a point inside a triangle are droped on the sides (here the radii of circle r) we have:

$$\frac{r}{h_a} +\frac{r}{h_b}+\frac{r}{h_c}=1$$

Now we have a system of four equations for four unknown $$h_a,. h_b,.h_c$$ and $$r$$. Solving this system will give you r. Wolfram alpha gives $$r=6, h_a=14.5, h_b=18, h_c=24$$. If we use generalized Descartes theorem and assume the sides of triangle circles with radius infinity, where $$k_s=\frac{1}{∞}=0$$ will be the curvature of sides we have:

$$(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+0+\frac{1}{r})^2=2(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+0+\frac{1}{r^2})$$

It finally results in:

$$23 r^2+132r-36=0$$

Which gives $$r=6$$

Let $$|CE|=H_c$$ and $$|CD|=h_c$$ be the altitudes of similar triangles $$\triangle ABC$$ and $$\triangle A_cB_cC$$. Then \begin{align} \frac{|CD|}{r_c} &= \frac{|CE|}{r} \tag{1}\label{1} ,\\ \frac{H_c-2r}{r_c} &= \frac{H_c}{r} \tag{2}\label{2} ,\\ H_c &= \frac{2r^2}{r-r_c} \tag{3}\label{3} . \end{align}

Similarly, two other altitudes of $$\triangle ABC$$ in terms of $$r,r_a,r_b$$ are

\begin{align} H_a &= \frac{2r^2}{r-r_a} \tag{4}\label{4} ,\\ H_b &= \frac{2r^2}{r-r_b} \tag{5}\label{5} , \end{align}

and we can apply a well-known relation

\begin{align} \frac1r&= \frac1{H_a}+\frac1{H_b}+\frac1{H_c} \tag{6}\label{6} \end{align}

to find out that $$r$$ in terms of $$r_a,r_b,r_c$$ is just \begin{align} r&=r_a+r_b+r_c \tag{7}\label{7} . \end{align}

The original question would be solved by now, but we can do more than that: we can completely solve the $$\triangle ABC$$.

Using known Heron-like formula for the area, we have

\begin{align} S&= \frac1{\sqrt{ {(\tfrac1{H_a}+\tfrac1{H_b}+\tfrac1{H_c})} {(-\tfrac1{H_a}+\tfrac1{H_b}+\tfrac1{H_c})} {(\tfrac1{H_a}-\tfrac1{H_b}+\tfrac1{H_c})} {(\tfrac1{H_a}+\tfrac1{H_b}-\tfrac1{H_c})} }} \\ &=\frac{r^{7/2}}{\sqrt{r_a r_b r_c}} \tag{8}\label{8} . \end{align}

Next, we can find the semiperimeter $$\rho$$ and circumradius $$R$$ of $$\triangle ABC$$:

\begin{align} \rho&=\frac Sr =\frac{r^{5/2}}{\sqrt{r_a r_b r_c}} \tag{9}\label{9} ,\\ R&= \frac{2\,S^2}{H_a H_b H_c} =\tfrac14\,\frac{r(r-r_a)(r-r_b)(r-r_c)}{r_a r_b r_c} \tag{10}\label{10} . \end{align}

Now we are ready to find the three side lengths of $$\triangle ABC$$ as the roots of cubic equation in terms of $$\rho,r,R$$:

\begin{align} x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,\rho\,r\,R&=0 \tag{11}\label{11} . \end{align}

In particular, for $$r_a=1,\ r_b=2,\ r_c=3$$ we have

\begin{align} r&=6 ,\quad S=216 ,\quad \rho=36 ,\quad R=15 \tag{12}\label{12} , \end{align}

\eqref{11} becomes

\begin{align} x^3-72\,x^2+1692\,x-12960&=0 \tag{13}\label{13} \end{align}

with three roots $$\{18,\, 24,\, 30\}$$, that is, the sought triangle is the famous $$3-4-5$$ right-angled triangle, scaled by $$6$$.

Note that the side lengths are inversely proportional to corresponding radii of incircles.

For another example, the picture illustrates a solution for $$r_a=7,\ r_b=5,\ r_c=3$$. In this case we have $$r=15$$ and the side lengths are

\begin{align} a&=\tfrac{120\sqrt7}7 ,\quad b=\tfrac{150\sqrt7}7 ,\quad c=\tfrac{180\sqrt7}7 \tag{14}\label{14} . \end{align}

Edit

In fact, solution of the cubic equation \eqref{11} is unnecessary: since the area and the altitudes are known, the side lengths can be found explicitly as

\begin{align} a&=r\,(r-r_a)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}} \tag{15}\label{15} ,\\ b&=r\,(r-r_b)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}} \tag{16}\label{16} ,\\ c&=r\,(r-r_c)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}} \tag{17}\label{17} . \end{align}