3. Inside a right triangle $ABC$ with hypotenuse $AC$, a point $M$ is taken such that the areas of triangles $ABM$ and $BCM$ are one-third and one-fourth of the area of triangle $ABC$, respectively. Find $BM$, if $AM=60$ and $CM=70$. If the answer is not an integer, round it to the nearest integer.
Solution. Denoting $A B=c, B C=a$, we get

$$
\left\{\begin{array}{l}
\left(c-\frac{c}{4}\right)^{2}+\left(\frac{a}{3}\right)^{2}=60^{2} \\
\left(\frac{c}{4}\right)^{2}+\left(a-\frac{a}{3}\right)^{2}=70^{2}
\end{array}\right.
$$

We solve the system and find $B M=\left(\frac{a}{3}\right)^{2}+\left(\frac{c}{4}\right)^{2}=\frac{100^{2}}{7}$. Therefore, $B M=\frac{100}{\sqrt{7}}$. We write the nearest integer in the answer.

Answer: 38.