Dear All My Friends, We solve this problem by using cosin and angle bisector theorems: Suppose $ a$, $ b$, $ c$ are side lengths of $ ABC$. We calculate $ AB_{2}$ and $ AC_{2}$. They are: $ AB_{2}=\frac{b \cdot c}{(a-b+c)}$ and $ AC_{2}=\frac{b \cdot c}{(a+b-c)}$ Two given triangle areas are equal if and only if: $ b \cdot c=\frac{b\cdot c}{(a+b-c)}\cdot \frac{b\cdot c}{(a-b+c)}$ From this: $ b^{2}-b\cdot c+c^{2}-a^{2}=0$ Moreover by cosin theorem: $ a^{2}=b^{2}+c^{2}-2\cdot b\cdot c\cdot cos(A)$ Therefore: $ cos(A)=\frac{1}{2}$ and we are done. Calculation: $ AC_{2}=\frac{b \cdot c}{a+b-c}$ Denote $ AC_{2}=x$, $ r$ is radius of incircle, $ h_{c}$ is length of altitude from $ C$, $ N$ is touch point of incircle with $ AC$ We use areas of some triangles. Denote $ S(ABC)$ as area of $ \triangle ABC$ $ S(AB_{1}C_{2})=S(AB_{1}I)+S(AC_{2}I)=S(AB_{1}I)+S(AB_{1}I)\cdot \frac{AC_{2}}{AB_{1}}$ $ S(AB_{1}C_{2})=S(AB_{1}I)+S(AB_{1}I)\cdot \frac{2 \cdot x}{b}=S(AB_{1}I)\cdot \frac{b+2 \cdot x}{b}$ $ x\cdot \frac{h_{c}}{4}=S(AB_{1}I)\cdot \frac{b+2 \cdot x}{b}$ $ x\cdot \frac{2\cdot S(ABC)/c}{4}=\frac{r\cdot b}{4}\cdot \frac{b+2\cdot x}{b}$ $ x\cdot\frac{S(ABC)}{2\cdot c}=\frac{2\cdot S(ABC)/(a+b+c)}{4}\cdot (b+2\cdot x)$ $ \frac{x}{c}=\frac{(b+2\cdot x)}{(a+b+c)}$ $ x\cdot (a+b+c)=c\cdot (b+2\cdot x)$ $ AC_{2}=x=\frac{b\cdot c}{a+b-c}$ Best regards, Bui Quang Tuan

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Dear All My Friends, I suggest another problem. It is exactly as original but only one slightly changed: we replace Incenter $ I$ with Circumcenter $ O$. How about the solution now? Best regards, Bui Quang Tuan