Problem 1305

## Solution.

In triangle $ABC$, we denote the lengths of the sides by $a, b, c$, the semi-perimeter by $s$, the radius of the inscribed circle by $r$, the radius of the circumscribed circle by $R$, and the area by $P$. For the triangle $A'B'C'$, we will use the same letters with a prime. We will first show that triangles $ABC$ and $A'B'C'$ have the same radii of their inscribed circles.

If we denote the common value of the three fractions in the problem by $\lambda$, solving the system

$$|AP| + |BP| = \lambda c, \quad |BP| + |CP| = \lambda a, \quad |CP| + |AP| = \lambda b$$

we get

$$|AP| = \lambda(s-a), \quad |BP| = \lambda(s-b), \quad |CP| = \lambda(s-c)$$

Multiplying these and using Heron's formula and the formula $P = rs$ gives

$$|AP||BP||CP| = \lambda^3 (s-a)(s-b)(s-c) = \lambda^3 \frac{P^2}{s} = \lambda^3 Pr.$$

Furthermore, triangles $PBC$ and $PC'B'$ are similar because they have three equal angles by the inscribed angle theorem over $\overline{BC'}$ and $\overline{B'C}$. From this, it follows that

$$\frac{a'}{a} = \frac{|B'C'|}{|BC|} = \frac{|C'P|}{|BP|} = \frac{|B'P|}{|CP|}$$

Multiplying this and two analogous relations, we get

$$\frac{a' b' c'}{abc} = \frac{|A'P||B'P||C'P|}{|AP||BP||CP|}$$

On the other hand, from the formula $P = \frac{abc}{4R}$ and the fact that triangles $ABC$ and $A'B'C'$ share the same circumscribed circle, it follows that

$$\frac{a' b' c'}{abc} = \frac{P'}{P}$$

From (5), we also get

$$\frac{|B'P| + |C'P|}{|B'C'|} = \frac{|BP| + |CP|}{|BC|} = \lambda$$

and similarly,

$$\frac{|A'P| + |B'P|}{|A'B'|} = \frac{|C'P| + |A'P|}{|C'A'|} = \lambda$$

We conclude that point $P$ has the same property with respect to triangle $A'B'C'$ as it does with respect to the original triangle, and even with the same ratio $\lambda$. Therefore, an equality analogous to (4) follows directly:

$$|A'P||B'P||C'P| = \lambda^3 P' r'$$

Combining relations (4), (6), (7), and (8) gives

$$\frac{P'}{P} = \frac{P' r'}{P r}$$

from which we finally conclude $r' = r$.

Let $I$ be the center of the inscribed circle of triangle $ABC$ and $D$ its tangency point with side $\overline{BC}$. Points $I'$ and $D'$ are defined analogously with respect to triangle $A'B'C'$. It is a known fact that

$$|BD| = s - b, \quad |CD| = s - c$$

which, in combination with (3), gives

$$\frac{|BP|}{|CP|} = \frac{|BD|}{|CD|}$$

By the angle bisector theorem, we know that $PD$ is the angle bisector of $\angle BPC$. Using relations for triangle $A'B'C'$ analogous to (3), it follows that $PD'$ is the angle bisector of $\angle B'PC'$, from which it follows that points $D, P,$ and $D'$ lie on the same line. Furthermore,

$$\begin{aligned} \angle C'D'P & = 180^\circ - \angle D'PC' - \angle PC'D' = 180^\circ - \frac{1}{2} \angle B'PC' - \angle CCB' \\ & = 180^\circ - \frac{1}{2} \angle BPC - \angle CBB' = 180^\circ - \angle BPD - \angle DBP = \angle PDB. \end{aligned}$$

Since $I'D' \perp B'C'$ and $ID \perp BC$, we have $\angle DD'I' = \angle IDD'$. If $I' \neq I$, we conclude that $DD'II'$ or $DD'I'I$ is an isosceles trapezoid, so the line $II'$ is parallel to the line $PD$. Similarly, it would follow that $II'$ is parallel to the other two lines passing through $P$ and one tangency point of the inscribed circle with a side of triangle $ABC$. This leads to a contradiction with the fact that at least two of these three lines through $P$ are distinct, thus proving $I' = I$. Therefore, the inscribed circles of triangles $ABC$ and $A'B'C'$ coincide.