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Solutions

2L Solution

  1. Construct the perpendicular bisector P1 of two arbitrary points on the circle
  2. Construct the perpendicular bisector P2 of two other arbitrary points on the circle, intersecting P1 at A
  3. Point at A
1.6-2L.PNG

5E Solution

  1. Construct a circle C1 with an arbitrary center A on the circle and an arbitrary radius, intersecting the given circle at B and C
  2. Construct the circle C2 with center B and radius AB, intersecting circle C1 at D and E
  3. Construct the circle C3 with center C and radius AC, intersecting circle C1 at F and G
  4. Construct line DE
  5. Construct line FG, intersection DE at H
  6. Point at H
1.6-5E.PNG

Explanation

Given any two points on a circle, the perpendicular bisector of the segment made up of those two points passes through the center of the circle. Therefore, by constructing two perpendicular bisectors, you'll find the center. The 5E solution uses the same reasoning, except you need to construct the perpendicular bisector without using the Perpendicular Bisector tool. As a reminder, in order to build a perpendicular bisector, you just need to construct two circles of equal radius with their centers on the circle and construct the line that passes through the intersections of those two circles. The trick to save 1 Elementary operation is to use one of the existing circles to construct your second perpendicular bisector.

List of Levels
Alpha Angle of 60°Perpendicular BisectorMidpointCircle in SquareRhombus in RectangleCircle CenterInscribed Square
Beta W.I.P.
Gamma W.I.P.
Delta W.I.P.
Epsilon W.I.P.
Zeta W.I.P.
Eta W.I.P.
Theta W.I.P.
Iota W.I.P.
Kappa W.I.P.
Lamda W.I.P.
Mu W.I.P.
Nu W.I.P.
Xi W.I.P.
Omicron W.I.P.
Tutorials
Equalateral Triangle • Intersect ToolPerpendicular BisectorMove ToolAngle BisectorPerpendicularParallel LineCompass
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