Problem 38. In an isosceles triangle $\mathrm{ABC}$, the lateral side is equal to $b$. The projection of the base onto the lateral side is equal to $m$. Find the radius of the circle circumscribed around triangle $A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-032.jpg?height=640&width=514&top_left_y=1476&top_left_x=405)
Solution. From formula II $a^{2}=2 b m$, hence $a=\sqrt{2 b m}$. From the right triangle $A D C: h^{2}=b^{2}-\frac{a^{2}}{4}$, or $h^{2}=b^{2}-\frac{b m}{2}=\frac{b(2 b-m)}{2}$, or

$$
h=\sqrt{\frac{b(2 b-m)}{2}}
$$

$S_{A B C}=\frac{1}{2} a h=\frac{1}{2} \sqrt{2 b m} \cdot \sqrt{\frac{b(2 b-m)}{2}}$, or:

$$
4 S_{A B C}=2 b \sqrt{m(2 b-m)}
$$

And then $R=\frac{a b c}{4 S_{A B C}}=\frac{b^{2} \cdot \sqrt{2 b m}}{2 b \cdot \sqrt{m(2 b-m)}} ; R=\frac{b}{2} \sqrt{\frac{2 b}{2 b-m}}$.