Let $a, b, c$ represent the lengths of the sides of triangle $ABC$. Draw the tangents to the incircle of triangle $ABC$ that are parallel to the sides. These tangents cut off three new triangles from triangle $ABC$; in each of these new triangles, we again draw the incircle. Calculate the sum of the areas of the four incircles.
I. solution. It is known that the radius $\varrho$ of the inscribed circle $k$ of a triangle is equal to the $t / s$ ratio, where $t$ is the area and $s$ is half the perimeter.

![](https://cdn.mathpix.com/cropped/2024_05_02_3cfbc16666d76a4445bdg-1.jpg?height=488&width=507&top_left_y=257&top_left_x=798)

The triangle $H_{1}$ cut off by the tangent parallel to side $a$ is similar to the original triangle $H$, so the radius $\varrho_{1}$ of the inscribed circle $k_{1}$ is the same fraction of $\varrho$ as any linear dimension of $H_{1}$ is of the corresponding dimension of $H$. It is convenient to take the ratio of the altitudes $m_{a 1}$ and $m_{a}$ perpendicular to $a$, because $m_{a 1}=m_{a}-2 \varrho$. On the other hand, $m_{a}=2 t / a$, so

$$
m_{a 1}=\frac{2 t-2 a \varrho}{a}, \quad \text { and } \quad \varrho_{1}=\frac{t-a \varrho}{t} \cdot \varrho=\varrho-\frac{\varrho^{2}}{t} a
$$

By substituting $b$ and then $c$ for $a$, we get the radii $\varrho_{2}$ and $\varrho_{3}$ of the circles inscribed in the triangles cut off by the tangents parallel to sides $b$ and $c$, respectively. Thus, the sum of the areas of the four circles, after factoring out, expanding the squares, and combining similar terms, and finally expressing everything in terms of the sides,

$$
\begin{gathered}
\pi\left(\varrho^{2}+\varrho_{1}^{2}+\varrho_{2}^{2}+\varrho_{3}^{2}\right)=\pi\left[4 \varrho^{2}-\frac{2 \varrho^{3}}{t}(a+b+c)+\frac{\varrho^{4}}{t^{2}}\left(a^{2}+b^{2}+c^{2}\right)\right]= \\
=\frac{\pi \varrho^{4}}{t^{2}}\left(a^{2}+b^{2}+c^{2}\right)=\frac{\pi t^{2}}{s^{4}}\left(a^{2}+b^{2}+c^{2}\right)= \\
=\frac{\pi(-a+b+c)(a-b+c)(a+b-c)\left(a^{2}+b^{2}+c^{2}\right)}{(a+b+c)^{3}}
\end{gathered}
$$

(since the sum of the first two terms in the square brackets is 0).

II. solution. Using the same notation as above, we determine the ratio $\varrho_{1} : \varrho$ from the ratio of the perimeters of $H_{1}$ and $H$. Let the new vertices of $H_{1}$ be $B_{1}, C_{1}$ (the latter on $A C$) and let $k$ touch the lines $A B, B_{1} C_{1}, C A, B C$ at points $C^{\prime}, A^{\prime \prime}$, $B^{\prime}, A^{\prime}$, respectively. Then, based on the equality of the tangent segments drawn from points $B_{1}, C_{1}, B, C$ and $A$ to $k$, the perimeter of $H_{1}$ is

$$
\begin{gathered}
A B_{1}+B_{1} C_{1}+C_{1} A=\left(A B_{1}+B_{1} A^{\prime \prime}\right)+\left(A^{\prime \prime} C_{1}+C_{1} A\right)=\left(A B_{1}+B_{1} C^{\prime}\right)+ \\
+\left(B^{\prime} C_{1}+C_{1} A\right)=A C^{\prime}+B^{\prime} A=A B-B C^{\prime}+A C-B^{\prime} C= \\
=A B+A C-\left(B A^{\prime}+A^{\prime} C\right)=c+b-a=2(s-a)
\end{gathered}
$$

Thus, $\varrho_{1}=\varrho(s-a) / s$, and with similar rearrangement steps as above,

$$
\begin{aligned}
& \pi\left(\varrho^{2}+\varrho_{1}^{2}+\varrho_{2}^{2}+\varrho_{3}^{2}\right)=\frac{\pi \varrho^{2}}{s^{2}}\left[s^{2}+(s-a)^{2}+(s-b)^{2}+(s-c)^{2}\right]= \\
& =\frac{\pi t^{2}}{s^{4}}\left[4 s^{2}-2 s(a+b+c)+\left(a^{2}+b^{2}+c^{2}\right)\right]=\frac{\pi t^{2}}{s^{4}}\left(a^{2}+b^{2}+c^{2}\right)
\end{aligned}
$$

Sarolta Nagy (Budapest, Hámán K. g. IV. o. t.)

Remark. The linear magnification ratio $\lambda$ between $H_{1}$ and $H$ can also be obtained from the fact that $k$ is the ex-circle of $H_{1}$. The sides of $H_{1}$ are $a_{1}=\lambda a, \lambda b, \lambda c$, half the perimeter is $s_{1}=\lambda s$, and the area is $t_{1}=\lambda^{2} t$, so applying the known expression $\varrho_{a}=t /(s-a)$ for the radius of the ex-circle in a meaningful way,

$$
t / s=\varrho=t_{1} /\left(s_{1}-a_{1}\right)=\lambda t /(s-a), \quad \text { and thus } \quad \lambda=(s-a) / s
$$[^0]


[^0]:    ${ }^{1}$ See, for example, Kürschák J.-Hajós Gy.-Neukomm Gy.-Surányi J.: Matematikai Versenytételek I, 3rd edition, Tankönyvkiadó, Budapest, 1965, p. 36.