A triangle's three side lines are tangent to four circles whose radii are four consecutive elements of a geometric sequence. What is the largest angle of the triangle?
It is known that for the area $t$ of a triangle, the radii $\varrho, \varrho_{a}, \varrho_{b}$, and $\varrho_{c}$ of the inscribed and excircles, and the semiperimeter $s$, the following relationships hold:

$$
t=\varrho \cdot s=\varrho_{a}(s-a)=\varrho_{b}(s-b)=\varrho_{c}(s-c)
$$

In our problem - since the inscribed circle is the smallest and any two radii have different lengths - we can assume that the sides are in the following order: $a1)$.

According to these,

$$
\begin{gathered}
q=\frac{\varrho_{a}}{\varrho}=\frac{\varrho_{c}}{\varrho_{b}}, \quad \text { i.e., } \quad \frac{s}{s-a}=\frac{s-b}{s-c} \\
s(s-c)=(s-b)(s-a) \\
(a+b+c)(a+b-c)=(c+a-b)(c-(a-b)) \\
(a+b)^{2}-c^{2}=c^{2}-(a-b)^{2} \\
a^{2}+b^{2}=c^{2}
\end{gathered}
$$

Thus, by the converse of the Pythagorean theorem, the triangle can only be a right triangle - if it exists at all - and its largest angle can only be $\gamma=90^{\circ}$.

However, since in this determination (1) we only used two ratios, we do not yet know if there is a right triangle for which the

$$
\frac{\varrho_{b}}{\varrho_{a}}=q
$$

requirement is also satisfied. Clarifying this would not have been possible within the time of the competition (see the following 3rd remark).

Remarks. 1. Some may still know the following formulas, which allow for "easier" (i.e., using logarithm tables) calculation of angles from the sides, compared to the "cumbersome" cosine rule:

$$
\operatorname{tg} \frac{\gamma}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}, \quad \sin \frac{\gamma}{2}=\sqrt{\frac{(s-a)(s-b)}{a b}}, \quad \cos \frac{\gamma}{2}=\sqrt{\frac{s(s-c)}{a b}}
$$

(The similar formulas for the other two angles can be written by cyclic permutation of the sides, while $s$ remains unchanged.) According to the first formula,

$$
\operatorname{tg} \frac{\gamma}{2}=\sqrt{\frac{\varrho \cdot \varrho_{c}}{\varrho_{a} \cdot \varrho_{c}}}=\sqrt{\frac{\varrho^{2} q^{3}}{\varrho^{2} q^{3}}}=1
$$

$\frac{\gamma}{2}=45^{\circ}, \gamma=90^{\circ}$. Here, the existence of the triangle is guaranteed because the formulas only give meaningless results for non-existent triangles.

2. The mentioned formulas - and many similar ones - are from the era when the most advanced tool for computation was the logarithm table. In the following - only in large steps - came the mechanical calculators with gears, first manual, then motorized, and today the electronic calculators. (We omitted the slide rule, which was only good for quick estimates compared to these.) - The formulas are still valid today. - However, the $\operatorname{tg} x / 2$ appears, for example, in the transformation of certain integrals; but it is impossible to list all such small but well-usable tricks in theory.
3. For a right triangle that also satisfies the $\varrho_{b}: \varrho_{a}=q$ requirement, we get a cubic equation for the tangent of half the larger acute angle. Similarly to the above,

$$
\begin{aligned}
& \operatorname{tg} \frac{\beta}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}=\sqrt{\frac{\varrho \cdot \varrho_{b}}{\varrho_{a} \cdot \varrho_{c}}}=\frac{1}{q} \quad \text { and } \\
& \operatorname{tg} \frac{\alpha}{2}=\frac{1}{q^{2}}
\end{aligned}
$$

(i.e., the tangents of the half-angles also form a geometric sequence with 3 elements and the same ratio). From these,

$$
\begin{gathered}
\operatorname{tg}^{2} \frac{\beta}{2}=\frac{1}{q^{2}}=\operatorname{tg} \frac{\alpha}{2}=\operatorname{tg}\left(45^{\circ}-\frac{\beta}{2}\right)= \\
=\frac{1-\operatorname{tg} \frac{\beta}{2}}{1+\operatorname{tg} \frac{\beta}{2}}=\frac{1-\frac{1}{q}}{1+\frac{1}{q}}=\frac{q-1}{q+1} \\
q^{3}-q^{2}-q-1=0
\end{gathered}
$$

(This "beautiful" equation also pops up in the solution of many other simply formulated problems.) Its approximate solution is:

$$
q=1.83929, \quad \alpha=32^{\circ} 56^{\prime} 6^{\prime \prime}, \quad \beta=57^{\circ} 3^{\prime} 54^{\prime \prime}
$$

Second solution. In addition to the notation used in the previous solution, let the vertices of the triangle be $A, B, C$, the center of the inscribed circle be $O$, and the centers of the excircles be $O_{a}, O_{b}, O_{c}$ such that the circle centered at $O_{a}$ touches the extensions of the sides $A B$ and $A C$, and the side $B C$ from the outside, and so on. Thus, the line $O O_{a}$ bisects the angle $B A C$, and the line $O_{b} O_{c}$ bisects the exterior angles at $A$, making it perpendicular to $O A$. $O$ is the orthocenter of the triangle $O_{a} O_{b} O_{c}$ and an interior point of the triangle, so this triangle is acute-angled.

![](https://cdn.mathpix.com/cropped/2024_05_02_823261d193f0d3d28e5bg-2.jpg?height=1093&width=943&top_left_y=993&top_left_x=580)

Let the points of tangency of the circles on the line $A B$ be $E, E_{a}, E_{b}, E_{c}$, respectively. From the similar right triangles $A O E$ and $A O_{a} E_{a}$, as well as $A O_{b} E_{b}$ and $A O_{c} E_{c}$,

$$
\frac{A O_{a}}{A O}=\frac{E_{a} O_{a}}{E O}=\frac{\varrho_{a}}{\varrho}=q=\frac{\varrho_{c}}{\varrho_{b}}=\frac{E_{c} O_{c}}{E_{b} O_{b}}=\frac{A O_{c}}{A O_{b}}
$$

(The elements to the right of $q$ are in the same logical order as the others from left to $q$.) Let's make a small swap in the ratios at the two ends of the sequence:

$$
\frac{A O_{a}}{A O_{c}}=\frac{A O}{A O_{b}}
$$

According to this, the ratios of the legs of the right triangles $A O_{a} O_{c}$ and $A O O_{b}$ are equal. Therefore, these two triangles are similar. For one of their corresponding angles and the vertex angle of the latter,

$$
A O_{a} O_{c} \varangle\left(=O O_{a} B \varangle\right)=A O O_{b} \varangle=O_{a} O B \varangle,
$$

which means that the angles of the right triangle $O O_{a} B$ are equal, so each is $45^{\circ}$.

On the other hand, the Thales circle over the segment $O O_{a}$ passes through $B$ and $C$, and these are on the opposite side of $O O_{a}$, so

$$
\frac{A C B \varangle}{2}=O C B \varangle=O O_{a} B \varangle=45^{\circ}
$$

thus $A C B \varangle=90^{\circ}$, which is then the largest angle of the triangle $A B C$. This gives us the answer to the question of the problem.