31**. In triangle $ABC$, it is given: $\widehat{A}=60^{\circ}, \widehat{B}=45^{\circ}$. Circles are constructed on the medians $[A M]$ and $[B N]$ as diameters. In what ratio does the radical axis of these circles divide the midline of the triangle, parallel to side $[A B]$?
31. Let the circle constructed on the diameter $[A M]$ intersect the side $[B C]$ at point $P$, and the circle constructed on

![](https://cdn.mathpix.com/cropped/2024_05_21_5cdadc99c18e78f1502ag-103.jpg?height=517&width=571&top_left_y=247&top_left_x=343)

Fig. 53 diameter $[B N]$, intersect the side $[A C]$ at point $Q$ (Fig. 53). Since angles $A P M$ and $B Q N$ subtend the diameters $[A M]$ and $[B N]$, $[A P]$ and $[B Q]$ are altitudes of triangle $A B C$. The power of point $C$ relative to the first circle is $p_{1}^{2}=|C M||C P|=$ $=\frac{a}{2} \cdot b \sin \gamma$, where $\gamma=\widehat{A C} B$. The power of the same point relative to the second circle is $p_{2}^{2}=|C N| \cdot|C Q|=$ $=\frac{b}{2} a \sin \gamma$.

Thus, $p_{1}^{2}=p_{2}^{2}=\frac{a b}{2} \sin \gamma ;$ hence, $C$ lies on the radical axis of the given circles. But the radical axis is perpendicular to the line of centers ($O_{1} O_{2}$), which is parallel to the base (explain). Therefore, the radical axis coincides with the altitude of the triangle $\mathrm{CH}$. This altitude divides the base of the triangle into segments:

$$
\left.|H A|=h \operatorname{ctg} 60^{\circ}=\frac{h}{\sqrt{3}} ; \quad|H B|=: h \operatorname{ctg} 45^{\circ}=h \quad \text { (here } \quad h=|H C|\right) .
$$

Thus, $|H B|:|H A|=\sqrt{3}$. The altitude divides the midline in the same ratio.