3-4. Construct triangle $ABC$ given points $M$ and $N$ - the feet of the altitudes $AM$ and $BN$ - and the line on which side $AB$ lies.
Sol 4. Points $M$ and $N$ lie on a circle with diameter $AB$. The center $O$ of this circle is the intersection of line $AB$ and the perpendicular bisector of segment $NM$, so we can construct it. Points $A$ and $B$ are the points of intersection of line $AB$ and the circle centered at $O$ passing through point $M$. Therefore, points $A$ and $B$ can be constructed. Point $C$ is then constructed as the intersection of lines $AN$ and $BM$.

3 - part 5. Solve the equation:

$$
|x+1|-|x|+3|x-1|-2|x-2|=x+2
$$

Sol 5. Answer: $x=-2$ or $x \geqslant 2$.

If $x \geqslant 2$, we get an identity.

If $1 \leqslant x<2$, we get the equation $4x=8$, which has no solutions in this interval.

If $0 \leqslant x<1$, we get the equation $-2x=2$, which has no solutions in this interval.

If $-1 \leqslant x<0$, we get $0=2$, which is impossible.

If $x<-1$, we get the root $x=-2$.

3 - part 6. How many roots does the equation

$$
\sin x=\frac{x}{100} ?
$$

Sol 6. Answer: 63.

First, note that the number of positive roots is equal to the number of negative roots, and there is also the root 0. Therefore, it is sufficient to verify that the number of positive roots is 31. If $\sin x=\frac{x}{100}$, then $|x|=100|\sin x| \leqslant 100$. Consider the graphs of the functions $y=x / 100$ and $y=\sin x$. The segment of the $Ox$ axis from 0 to 100 contains 15 segments of length $2\pi$ and one segment of length less than $2\pi$. By examining the specified graphs, it is easy to see that on the first segment of length $2\pi$ there is one root of the given equation, and on each of the other 14 segments of length $2\pi$ there are two roots. Calculations show that the length of the last segment is greater than $\pi$, so there are also two roots on it. In total, we get 31 positive roots.

## Second Round

## 7 - 8 grades