Belarusian Mathematical Olympiad

Let $A_1{A}_2=B_1{B}_2=2x$, $\angle A=\alpha$, $\angle B=\beta$, $AB=c$, $AC=b$, $BC=a$. Then we have (see Fig. 1) $$AC=A{B}_2+C{B}_1⟺b=c-B{B}_2+a-B{B}_1=c+a-2x\mathrm{ctg}\beta -\frac{2x}{\sin \beta }⟺$$ $$a+c-b=2x\left(\mathrm{ctg}\beta +\frac{1}{\sin \beta }\right)=2x\mathrm{ctg}\frac{\beta }{2}.$$



Note that (see Fig. 2) $\frac{a+c-b}{2}=r\mathrm{ctg}\frac{\beta }{2}$, where $r$ is the inradius of the triangle $ABC$. Therefore, the condition $AC=A{B}_2+C{B}_1$ is equivalent to $A_1{A}_2=B_1{B}_2=2r$. Similarly, the conditions $BC=B{A}_2+C{A}_1$ and $A_1{A}_2=B_1{B}_2=2r$ are also equivalent.