The diameter of the circle circumscribed about a regular hexagon is 35 cm. Inside each side of the hexagon we have chosen one point through which we draw a perpendicular line of length 20 cm inside the hexagon. Is it possible that none of the six drawn perpendiculars will intersect with any other perpendiculars?
I have found that the length of each side is 17.5 cm. If we were to draw a perpendicular from each of the two adjacent angles of the hexagon, the intersection would give us a right-angled triangle with angles of 30 and 60 degrees (where one leg is 17.5 and the hypotenuse=17.5/sin60=20.2, so that is slightly longer than the length of the perpendicular). But I need to choose exactly the interior point on the side, it's harder to prove that so far.
An explicit construction of these perpendicular lines is needed for a rigorous proof.
EDIT: JUST REALIZED $2k\neq{m+h}$, but $k\sqrt{3}=m+h$, so $$2.5\sqrt{3}=0.5r-h$$resulting in $$2.5cm\sqrt{3}=0.5(17.5cm)-h$$so $$h=\left(8.75-2.5\sqrt{3}\right)cm\approx 4.42cm$$ THE VALUE FOR $h$ ON THE STICKY NOTE IS WRONG.
The distance each line can be translated along its respective perpendicular side of the hexagon from its respective vertex (assuming they are moved the same distance) before they intersect is $h$, shown on the sticky notes.