We have a simple hortizonal straight line $AB$ and we want divide it to 4 equal parts.
So at least how many arcs we need to do that?
Actually we must draw 3 vertical lines that divide the line into 4 equal parts but we need at least 12 arcs to draw 3 vertical equitable lines.
Any ideas?
Options are 5 or 2 or 3 or 4 arcs!
Best Answer
Not all of the circles have to be unique.
Construct $DC$, the perpendicular bisector of $AB$ so that $E$ is the midpoint of $AB$. This uses $2$ circles.
Construct $FG$, the perpendicular bisector of $AE$ so that $J$ is the midpoint of $AE$. This uses $2$ more circles.
Construct $IH$, the perpendicular bisector of $EB$ so that $K$ is the midpoint of $EB$. However, since the left circle has already been constructed, we only need $1$ more circle.
In total, this uses $2+2+1=5$ circles.