Problem 851

Let $S$ be the point on $AC$, on the same side of $A$ as $C$, such that $AS = \sqrt{2}AB/2$. This point satisfies $AS^2 = \frac{AB^2}{2} = AB \cdot AM$, which is the power of $A$ with respect to the circumscribed circle of $BMP$; therefore, if this circumscribed circle passes through $S$ then it is tangent to $AB$. We will show that this is true, by proving that $\angle MSB = \angle MPB$. Let $N$ be the

midpoint of $AC$. Since $AS = \frac{\sqrt{2}AB}{2} = \frac{AC+BC}{2} = AN + MN$, we have $NS = NM$. From this we obtain $\angle NMS = \angle NSM$.

Note that triangles $ASB$ and $AMS$ are similar, because $\frac{AB}{AS} = \sqrt{2} = \frac{AS}{AM}$. From this it follows that $\angle AMS = \angle ASB$. Then $\angle ABC = \angle AMN = \angle AMS - \angle NMS = \angle ASB - \angle NSM = \angle MSB$.

On the other hand, since $M$ is the midpoint of the hypotenuse of the right triangle $APB$, we have $MP = MB$. This means that $\angle MPB = \angle MBP = \angle ABC = \angle MSB$, as we wanted to prove.