0

Is there a method to slice a circle's surface area in three equal parts by slicing the circle using two straight lines whose common point of origin is located on the circumference of the circle?

Niko
  • 1
  • 1
  • I am guessing that there must be some symmetry principle employed in the solution. – Niko Jul 12 '19 at 14:04
  • yes, there is, although the solution may not be analytical. – Vasili Jul 12 '19 at 14:15
  • By "Is there a method", are you asking if such lines exist at all, or are you asking if you can construct the two straight lines in some restricted fashion, e.g. using ruler and compass? – Magma Jul 12 '19 at 14:36
  • Draw the picture and find the relation between areas enclosed by strings – Moti Jul 12 '19 at 15:30
  • I'm just guessing that it is possible - in principle - to trisect the circle area in equal parts either by analytical means or by compass and ruler by starting the trisection at a point on the circumference. The reality of the problem actually dawned on me when I needed to cut a circular pill of 3 mg in two pieces, one of 1 mg and the other of 2 mg. It is then that I realized that I didn't know of any mathematical or geometrical way that could help me tackle this problem in a simple or straightforward way! – Niko Jul 15 '19 at 01:54
  • Is the center of the circle given? – Moti Jul 19 '19 at 14:57

1 Answers1

0

Here you find instructions - some work left for you: enter image description here

The goal is to find angle CBD. Given BC, f, the triangle ABC is well defined for a given radius. We will create an equation for f in which the section BCD of the circle will be equal to the segments under the arcs. The sector ABD ,ay be calculated as function of f - angle CAD is related to angle BAC. Angle BAC = Angle BAD. Angels BAC + BAD + CAD = 2$\pi$. Does this detailed enough?

Moti
  • 1,928