Belarusian Mathematical Olympiad

Since $A{B}_1$ and $B_1{A}_1$ are the diameters of the circumcircles of triangles $A{C}_1{B}_1$ and $A_{1}C{B}_1$ respectively, $\angle AY{B}_1=\angle A_{1}Y{B}_1=90^{\circ }$. Hence $Y$ belongs to the line $A{A}_1$ and $A{A}_1\perp B_{1}Y$. Similarly, $X$ belongs to the line $B{B}_1$ and $B{B}_1\perp A_{1}X$.



Let $Z_b$ and $Z_a$ be the points of intersection of the lines $B_{1}Y$ and $A_{1}X$ respectively with the hypotenuse $AB$. Let $CB=a$, $AC=b$, $AB=c$ and $A_1{B}_1=x$. We will find the lengths of the segments $B{Z}_a$ and $A{Z}_b$, and then prove the equality $A{Z}_b+Z_{a}B=AB$ from which follows the statement of the problem. Since $\angle B_{1}X{Z}_a=\angle B_1{C}_1{Z}_a=90^{\circ }$, the quadrilateral $B_1{C}_1{Z}_{a}X$ is cyclic. The quadrilateral $C{A}_{1}X{B}_1$ is cyclic too, therefore $$B{A}_1\cdot CB=BX\cdot B{B}_1=B{Z}_a\cdot B{C}_1.(\ast )$$ The triangle $A_{1}B{D}_1$ is similar to the triangle $ABC$, therefore $$\frac{A_{1}B}{AB}=\frac{D_{1}B}{CB}=\frac{A_1{D}_1}{AC}.$$ Hence $A_{1}B=\frac{cx}{b}$, $D_{1}B=\frac{ax}{b}$ and $B{C}_1=x+\frac{ax}{b}=\frac{(a+b)x}{b}$. From $(\ast )$ it follows that $B{Z}_a=\frac{ac}{a+b}$. Similarly one can prove $A{Z}_b=\frac{bc}{a+b}$, so $A{Z}_b+B{Z}_a=AB$.