Problem 11.2. The incircle of $\triangle A B C$ has radius $r$ and is tangent to the sides $A B, B C$ and $C A$ at points $C_{1}, A_{1}$ and $B_{1}$, respectively. If $N=B C \cap B_{1} C_{1}$ and $A A_{1}=2 A_{1} N=2 r \sqrt{3}$, find $\Varangle A N C$.

Sava Grozdev, Svetlozar Doychev
11.2. We shall use the standard notation for the elements of $\triangle A B C$. We may assume that $b>c$. Denote by $I$ the incenter of $\triangle A B C$. Then the condition $A_{1} N=r \sqrt{3}$ implies that $I N A_{1}$ is a right-angled triangle with $\Varangle N I A_{1}=60^{\circ}$. We shall prove that $A A_{1} \perp I N$. If so, then $\Varangle A A_{1} N=60^{\circ}$. Now, if $M$ is the midpoint of the segment $A A_{1}$, then $\triangle M N A_{1}$ is equilateral and therefore $\triangle A N M$ is isosceles with $\Varangle A N M=\Varangle M A N=30^{\circ}$. Thus $\Varangle A N C=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-068.jpg?height=524&width=812&top_left_y=724&top_left_x=517)

To prove that $A A_{1} \perp I N$, note that this is equivalent to the equality $A I^{2}-I A_{1}^{2}=A N^{2}-A_{1} N^{2}$. The Cosine theorem for $\triangle A N B$ gives

$$
A N^{2}=c^{2}+B N^{2}+2 c \cdot B N \cos \beta=c^{2}+B N^{2}+2 c . B N \frac{a^{2}+c^{2}-b^{2}}{2 a c}
$$

On the other hand, the Menelaus theorem for $\triangle A B C$ and the line $B_{1} C_{1}$ implies that

$$
\frac{C B_{1}}{B_{1} A} \cdot \frac{A C_{1}}{C_{1} B} \cdot \frac{B N}{N C}=1
$$

Hence $\frac{p-c}{p-b} \cdot \frac{B N}{B N+a}=1$, i.e., $B N=\frac{a(p-b)}{b-c}$, where $p$ is the semiperimeter of $\triangle A B C$. Then $A_{1} N=B N+p-b$ and therefore

$$
\begin{aligned}
A N^{2}-A_{1} N^{2} & =c^{2}-(p-b)^{2}+B N \frac{a^{2}+c^{2}-b^{2}-a^{2}-a c+a b}{a} \\
& =c^{2}-(p-b)^{2}-\frac{a(p-b)}{b-c} \cdot \frac{2(b-c)(p-a)}{a} \\
& =c^{2}-(p-b)^{2}-2(p-a)(p-b)=(p-a)^{2} \\
& =A I^{2}-I C_{1}^{2}=A I^{2}-I A_{1}^{2}
\end{aligned}
$$

which completes the proof.

Remark. The fact that $A A_{1} \perp I N$ holds true for any $\triangle A B C$. It can be proved as above or using complex numbers or inversion. Note also that if $P$ is
the second common point of the line $A A_{1}$ and the incircle of $\triangle A B C$, then the line $N P$ is tangent to this circle. The data of our problem give $P \equiv M$.