Bulgarian Mathematical Competitions

Solution:

We shall use the standard notation for the elements of $\triangle ABC$. We may assume that $b > c$. Denote by $I$ the incenter of $\triangle ABC$. Then the condition $A_1N = r\sqrt{3}$ implies that $INA_1$ is a right-angled triangle with $\text{NIA 1 = 60}$. We shall prove that $AA_1 \perp IN$. If so, then $\text{AA 1N = 60}$. Now, if $M$ is the midpoint of the segment $AA_1$, then $\triangle MNA_1$ is equilateral and therefore $\triangle ANM$ is isosceles with $\text{ANM = MAN = 30}$. Thus $\text{ANC = 90}$.

To prove that $AA_1 \perp IN$, note that this is equivalent to the equality $AI^2 - IA_1^2 = AN^2 - A_1N^2$. The Cosine theorem for $\triangle ANB$ gives
$$AN^2 = c^2 + BN^2 + 2c \cdot BN \cos \beta = c^2 + BN^2 + 2c \cdot BN \frac{a^2 + c^2 - b^2}{2ac}$$
On the other hand, the Menelaus theorem for $\triangle ABC$ and the line $B_1C_1$ implies that
$$\frac{CB_1}{B_1A} \cdot \frac{AC_1}{C_1B} \cdot \frac{BN}{NC} = 1$$
Hence $\frac{p-c}{p-b} \cdot \frac{BN}{BN + a} = 1$, i.e., $BN = \frac{a(p-b)}{b-c}$, where $p$ is the semiperimeter of $\triangle ABC$. Then $A_1N = BN + p - b$ and therefore
$$\begin{aligned} AN^2 - A_1N^2 & = c^2 - (p-b)^2 + BN \frac{a^2 + c^2 - b^2 - a^2 - ac + ab}{a} \\ & = c^2 - (p-b)^2 - \frac{a(p-b)}{b-c} \cdot \frac{2(b-c)(p-a)}{a} \\ & = c^2 - (p-b)^2 - 2(p-a)(p-b) = (p-a)^2 \\ & = AI^2 - IC_1^2 = AI^2 - IA_1^2 \end{aligned}$$
which completes the proof.