Bogosnov I.I.

On the plane, the curves $y=\cos x$ and $x=100 \cos (100 y)$ were drawn, and all points of their intersection with positive coordinates were marked. Let $a$ be the sum of the abscissas, and $b$ be the sum of the ordinates of these points. Find $a / b$.
After replacing $x=100 u$, the equations will become: $y=\cos (100 u), u=\cos (100 y)$. The ordinates of the corresponding intersection points of the new curves will be the same, while the abscissas will be reduced by a factor of 100. Let $c$ be the sum of the abscissas of the new intersection points (with positive coordinates). The new curves are symmetric with respect to the line $y=u$, so their intersection points are symmetrically located with respect to this line. Therefore, $c=b$, and $a=100 b$.

## Answer

100.