A road 1 km long is fully illuminated by street lamps, with each lamp lighting a section of the road 1 m long. What is the maximum number of lamps that can be on the road, given that after turning off any lamp, the road will no longer be fully illuminated?
If the segments illuminated by the n-th and (n+2)-th street lamps intersect, then the (n+1)-th street lamp can be turned off.

## Solution

We will number the street lamps with natural numbers in the order along the road. If the segments illuminated by the n-th and (n+2)-th street lamps intersect, then the (n+1)-th street lamp can be turned off. Therefore, segments with different odd numbers do not intersect. On a segment of length 1000 m, no more than 999 non-intersecting segments of length 1 m can be placed. Thus, there are no more than 1998 street lamps. We will place 1998 street lamps so that the centers of the illuminated segments form an arithmetic progression, the first term of which is 0.5 m, and the 1998-th term is 999.5 m. Between the n-th and (n+2)-th segments, there remains a gap of 1/1997 m. It is illuminated only by the (n+1)-th street lamp. Therefore, no street lamp can be turned off.

## Answer

1998.00

Given points \( A(2 ; 4), B(6 ;-4) \) and \( C(-8 ;-1) \). Prove that triangle \( ABC \) is a right triangle.

## Hint

Calculate the sides of the given triangle and apply the converse of the Pythagorean theorem (or calculate the dot product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)).

## Solution

## First method.

Using the distance formula between two points, we find that

\[
\begin{gathered}
AB=\sqrt{(6-2)^{2}+(-4-4)^{2}}=\sqrt{16+64}=\sqrt{80} \\
AC=\sqrt{(-8-2)^{2}+(-1-4)^{2}}=\sqrt{100+25}=\sqrt{125} \\
BC=\sqrt{(-8-6)^{2}+(-1+4)^{2}}=\sqrt{196+9}=\sqrt{205}
\end{gathered}
\]

Since \( AB^{2}+AC^{2}=80+125=205=BC^{2} \), triangle \( ABC \) is a right triangle.

Second method.

Since

\[
\begin{aligned}
\overrightarrow{AB}=\overrightarrow{(6-2 ;-4-4)} & =\overrightarrow{(4 ;-8)}, \overrightarrow{AC}=\overrightarrow{(-8-2 ;-1-4)}=\overrightarrow{(-10 ;-5)} \\
\overrightarrow{BC} & =\overrightarrow{(-8-6 ;-1+4)}=\overrightarrow{(-14 ; 3)}
\end{aligned}
\]

then

\[
\overrightarrow{AB} \cdot \overrightarrow{AC}=4 \cdot(-10)+(-8) \cdot(-5)=-40+40=0
\]

thus \( \overrightarrow{AB} \perp \overrightarrow{AC} \). Therefore, triangle \( ABC \) is a right triangle.