Inscribe a square, hexagon and equilateral triangle

Try

A circular frame needs six equally spaced attachment points. Guessing six positions may leave a visible closing gap. A compass construction explains why exact steps close. Begin with a circle of positive radius r and known center O. Choose any boundary point A. The triangle formed by O, A and a second boundary point B with AB=r has three equal sides, hence three 60-degree angles.

That one triangle is the engine for a regular hexagon. It uses a length already supplied by the circle, not a measured 60-degree angle. We will also construct an equilateral triangle and a square inside the same circle.

Predict

Predict how many successive radius-length chords return to A. Decide which intersection to choose at each stage so that the construction advances instead of stepping back. Predict what happens if the compass opening is slightly smaller than r, and whether the error can be repaired by declaring the last point equal to A.

Experiment

Draw the given circle. With compass radius r, draw a circle about A and choose one intersection B with the original circle. Continue from B, choosing the intersection other than A. Repeat in the same circular direction until six vertices have been marked. Join neighbors with the straightedge. Label them A₀ through A₅ in order.

For a worked radius-3 circle, the exact hexagon vertices are (3,0), (3/2,3√3/2), (−3/2,3√3/2), (−3,0), (−3/2,−3√3/2), (3/2,−3√3/2). The drawing below has equal horizontal and vertical scales. Use the labels to trace the hexagon and the triangle of alternate vertices.

Understand

Every step makes an equilateral center triangle, so the directed central advance is 60 degrees. Six advances total 360 degrees exactly; the six intermediate vertices are distinct. Rotations through 60 degrees exchange edges and vertices, proving equal sides and equal interior angles. Each side is r and the perimeter is 6r. This proof establishes exact closure; a practical gap reveals drawing error rather than a new geometry theorem.

Select A₀,A₂,A₄ and join them. Their successive central angles are all 120 degrees. A rotation through 120 degrees permutes them and their sides, so the triangle has three equal sides and is equilateral. For r=3 its side length is 3√3: split the 120-degree center triangle along its symmetry line to obtain right triangles with half-chord length 3 sin60°. Its height and perimeter need not be measured to certify equal sides.

For an inscribed square, draw any diameter and construct its perpendicular through O using the preceding lesson. These two perpendicular diameters meet the circle at four points. Join them in circular order. Four congruent right center triangles show the four chords equal. Each vertex angle is the sum of two 45-degree base angles from the center triangles, so it is right. Thus this is a square. Its side squared is 2r², and its diagonal is 2r.

Two arbitrary diameters still give four circle points, but they need not yield equal successive chords. Their joined quadrilateral is a rectangle by the diameter-angle theorem proved later, yet it is a square only when the diameters are perpendicular. This distinction illustrates why a construction must prove every defining property of its claimed figure.

Apply

Construct all three figures from a new circle without reading any angles from a protractor. Explain the role of each radius transfer and perpendicular. If the supplied circle lacks a marked center, construct perpendicular bisectors of two nonparallel chords to recover it first. A zero-radius circle cannot supply distinct boundary vertices, so it is excluded.

A physical opening of 0.99r gives a smaller chord and a smaller central advance than 60 degrees; six steps do not close. Do not hide that discrepancy by moving the last vertex. Report the ideal proof and the accuracy of your drawing separately.

Continue

Next reverse the viewpoint: instead of placing a regular figure inside a given circle, construct circles associated with an arbitrary triangle. Their centers emerge from equal-distance conditions.