4. Given a triangle $A B C$, where side $B C$ is twice the length of side $A B$, and a bisector $B M$ is drawn. a) Find the ratio $R_{1} / R_{2}$, where $R_{1}, R_{2}$ are the radii of the circumcircles of triangles $A B M$ and $C B M$ respectively. b) Prove that $3 / 4 < r_{1} / r_{2} < 1$, where $r_{1}, r_{2}$ are the radii of the incircles of triangles $A B M$ and $C B M$ respectively.
Answer. a) $R_{1} / R_{2}=1 / 2$. Hint. a) Let $\angle B M A=\alpha$, then $\angle B M C=\pi-\alpha$. Then in triangles $A B M$ and $C B M$ we have the relations $A B=2 R_{1} \sin \alpha, B C=2 R_{2} \sin (\pi-\alpha)=2 R_{2} \sin \alpha$. Hence $R_{1} / R_{2}=A B / B C=1 / 2$. (Note that in this solution, the fact that $V$ is not used.

b) The area of triangle $C B M$ is twice the area of triangle $A B M$ (this follows, for example, from the area formula using two sides at vertex $B$ and the same angle, equal to half of angle $B$). We have $2 S_{1}=P_{1} r_{1}=(A B+A M+B M) \cdot r_{1}$ and $2 S_{2}=P_{2} r_{2}=(2 A B+2 A M+B M) \cdot r_{2}$ (here $P_{1}, P_{2}$ are the perimeters of triangles $A B M$ and $C B M$, and in calculating the perimeter $P_{2}$ we used the equality $C M=2 A M$, which follows from the property of the angle bisector: $C M / A M=C B / A B$). Let $d=B M, t=A B+A M$. Obviously (from the triangle inequality), $d< t+ d$, i.e., $d<t$.