Geometry · G.CO.D.13

Equilateral triangle, square and hexagon in a circle

With only a compass and a straightedge that has no markings, how can you draw a triangle with three exactly equal sides, or a square or a regular hexagon with every corner on a given circle?

You need

Regular polygons and their circlesGeometry Perpendicular and parallel lines through a pointGeometry Isosceles and equilateral trianglesGeometry

Where you are 8 in Compass and straightedge constructions

Geometry11

Compass and straightedge constructions8

▴ 3 above Perpendicular and parallel lines through a point Why a construction works Constructions with circles

Equilateral triangle, square and hexagon in a circle

Dividing a segment into equal parts

When you reach for it

You need a 60^\circ angle and have no protractor. Or you need a triangle whose three sides are exactly equal, or a regular polygon drawn exactly inside a circle: a hexagon for a tiling pattern, a square, a triangle. The tools are the two used in copying a segment and bisecting: a straightedge, which draws the line through two points and has no marks, and a compass, which holds one width.

All three constructions come from one fact about the compass: every point it draws is the same distance from its center. A triangle with three equal sides is a pair of circles, and a regular polygon in a circle is a set of equal angles at the center. For a regular polygon with n sides in a circle, the angle at the center between two neighboring corners is

\frac{360^\circ}{n}

which is 120^\circ, 90^\circ and 60^\circ for n = 3, 4 and 6. Building one of these polygons means building that angle at the center, and the compass makes 60^\circ and the perpendicular makes 90^\circ.

The procedure

An equilateral triangle on a given side

Start with a segment \overline{AB}. The triangle will have \overline{AB} as one side.

Open the compass to AB: the point on A, the pencil on B. Draw the circle about A. Every point of it is the distance AB from A. It passes through B.

Without changing the width, draw the circle about B. Every point of it is the distance AB from B, and it passes through A.

Call one of the two crossings C, and draw \overline{CA} and \overline{CB}. The point C is on the circle about A, so CA = AB. It is also on the circle about B, so CB = BA. So

CA = AB = BC

and \triangle ABC has three equal sides. Only the short pieces of the circles near C are needed on paper; the full circles are drawn here to show where the points come from.

Figure 1Each circle has radius AB. The point C is on both, so CA = AB and CB = AB, three equal sides. The second crossing D gives a second equilateral triangle below.

The circles do cross. Their common width is AB, which is more than half of AB, and that is exactly the condition for two equal arcs from A and B to meet, as in bisecting a segment. They cross twice, at C on one side of \overline{AB} and D on the other, and either crossing gives an equilateral triangle.

The angles. An equilateral triangle is isosceles three ways over. By the base angles theorem, CA = CB gives \angle A \cong \angle B, and BA = BC gives \angle A \cong \angle C. So the three angles are equal, and since they add to 180^\circ (the triangle angle sum), each is

\frac{180^\circ}{3} = 60^\circ

So this is also the construction of a 60^\circ angle: the angle \angle CAB is one.

This is the first construction in Euclid’s Elements, Book I, Proposition 1. Euclid’s proof takes it as evident that the two circles meet; later treatments of geometry add an assumption to cover that step.

A regular hexagon in a circle

Start with a circle, its center O and its radius r.

Open the compass to the radius: the point on O, the pencil on the circle. From now on the width stays r.

Pick any point A on the circle. With the point on A, mark where an arc crosses the circle, and call it B. Move the point to B and mark C the same way, at the crossing that is not A, then D, E and F, always going round in the same direction.

Join A, B, C, D, E, F and back to A.

Why it works. OA = OB = r, because A and B are on the circle, and AB = r, because the compass was set to r. So \triangle OAB is equilateral and m\angle AOB = 60^\circ. Every step is the same, so each step turns 60^\circ about O. Six steps turn

6 \times 60^\circ = 360^\circ

which is all the way round, so the arc from F lands exactly on A and the six chords close up.

Figure 2With the compass held at the radius, each step marks a chord equal to the radius. Triangle OAB is equilateral, so each step turns 60^\circ about O, and six steps close up.

The hexagon is regular. Each of its six sides is a compass width, r. Each of its six angles is two angles of equilateral triangles side by side; at B, for instance, the ray \overrightarrow{BO} runs between \overrightarrow{BA} and \overrightarrow{BC}, so the angle measures add:

\begin{aligned} m\angle ABC &= m\angle ABO + m\angle OBC \\ &= 60^\circ + 60^\circ = 120^\circ \end{aligned}

So the side of a regular hexagon equals the radius of its circle, a fact Euclid records after Book IV, Proposition 15. The figure is six equilateral triangles around one point.

A steadier way to mark the same points. Each step starts from a point the previous step made, so a small drawing error is copied forward and grows. Draw the line through A and O instead; it meets the circle again at D, since m\angle AOD = 3 \times 60^\circ = 180^\circ makes D the far end of the diameter. Then arcs of width r about A cut the circle at B and F, and arcs of width r about D cut it at C and E. Every point now comes straight from A or D.

An equilateral triangle in a circle

Mark the six points of the hexagon and join every other one, A, C and E.

Why it works. Each of the angles \angle AOC, \angle COE and \angle EOA spans two steps, so each is 2 \times 60^\circ = 120^\circ. The triangles AOC, COE and EOA each have two sides r around a 120^\circ angle, so they are congruent by SAS, and their third sides match:

AC = CE = EA

Only three of the six points are needed, and they can be found with one diameter and one arc: draw the diameter \overline{AD}, and an arc of width r about D cuts the circle at C and E. Example 5 proves that this triangle is equilateral.

Figure 3One diameter and one arc. Triangle OCD is equilateral, which makes each of the three angles at O between A, C and E equal to 120^\circ (Example 5).

A square in a circle

Start with a circle and its center O.

Draw a diameter. Lay the straightedge through O and any point A of the circle; the line meets the circle again at C. Then OA = OC, so O is the midpoint of \overline{AC}.

Construct the perpendicular to \overleftrightarrow{AC} at O. This is the perpendicular at a point, and the circle has already done its first step: A and C are the two points the same distance either side of O. Arcs of one width, more than r, about A and about C cross at a point on the perpendicular bisector of \overline{AC}, which passes through the midpoint O. The line through O and that crossing meets the circle at B and D.

Join A, B, C, D and back to A.

Figure 4Two perpendicular diameters. The four triangles at O are congruent by SAS, so the four sides are equal, and each corner is 45^\circ + 45^\circ = 90^\circ.

Why it works. OA, OB, OC and OD are all radii, length r, and the four angles at O are right angles. So the triangles AOB, BOC, COD and DOA each have two sides r around a 90^\circ angle, and are congruent by SAS. Their third sides are the four sides of ABCD, so

AB = BC = CD = DA

Each of those triangles is isosceles with a right angle at O, so its two base angles are equal and add to 90^\circ, which makes each 45^\circ. At B, then,

\begin{aligned} m\angle ABC &= m\angle ABO + m\angle OBC \\ &= 45^\circ + 45^\circ = 90^\circ \end{aligned}

and the same holds at every corner. Four equal sides and four right angles make ABCD a square. By the Pythagorean theorem in \triangle AOB, its side is \displaystyle \sqrt{r^2 + r^2} = r\sqrt{2}. This is Euclid’s Book IV, Proposition 6.

The three side lengths

In a circle of radius r:

Polygon	Angle at the center	Side
equilateral triangle	120^\circ	\displaystyle r\sqrt{3}
square	90^\circ	\displaystyle r\sqrt{2}
regular hexagon	60^\circ	r

The hexagon and square sides are shown above; the triangle’s is worked out in Example 3. Fewer sides means longer sides, since each side spans a bigger angle at the center.

Worked examples

Example 1

An equilateral triangle and its twin

An equilateral triangle is constructed on a segment \overline{AB} of length 6 cm, and both crossings C and D are marked. Find the perimeter of \triangle ABC, its height from C, the distance CD, and the four angles of ACBD.

Solution

Every side of \triangle ABC is the compass width, 6 cm, so the perimeter is 3 \times 6 = 18 cm.

C is the same distance from A and from B, so it lies on the perpendicular bisector of \overline{AB}. So does D. The line \overleftrightarrow{CD} is therefore that bisector, and it crosses \overline{AB} at a right angle at the midpoint M, with AM = 3 cm. The height from C is CM, a leg of the right triangle AMC whose hypotenuse is CA = 6:

\begin{aligned} CM &= \sqrt{6^2 - 3^2} = \sqrt{27} \\ &= 3\sqrt{3} \approx 5.20 \text{ cm} \end{aligned}

D is the same distance below M, because the same right triangle sits below the segment, so

CD = 2 \times 3\sqrt{3} = 6\sqrt{3} \approx 10.39 \text{ cm}

All four sides of ACBD are 6 cm, so it is a rhombus made of two equilateral triangles. Its angles at C and D are single angles of those triangles, 60^\circ each. Its angles at A and B are each two such angles side by side, 60^\circ + 60^\circ = 120^\circ.

Check. 60^\circ + 120^\circ + 60^\circ + 120^\circ = 360^\circ, the angle sum of a quadrilateral.

Example 2

The hexagon in a circle of radius 5

A regular hexagon is constructed in a circle of radius 5 cm. Find its side, its perimeter, each of its angles and its area.

Solution

The side is the compass width, which is the radius: 5 cm. The perimeter is 6 \times 5 = 30 cm. Each angle is 60^\circ + 60^\circ = 120^\circ, as shown in the procedure.

For the area, the radii cut the hexagon into six equilateral triangles of side 5. One of them has height \displaystyle \sqrt{5^2 - 2.5^2} = \sqrt{18.75} = \tfrac{5\sqrt{3}}{2}, by the same right triangle as in Example 1, so its area is

\tfrac{1}{2} \times 5 \times \tfrac{5\sqrt{3}}{2} = \frac{25\sqrt{3}}{4}

Six of them make

6 \times \frac{25\sqrt{3}}{4} = \frac{75\sqrt{3}}{2} \approx 64.95 \text{ cm}^2

The height of each triangle, \displaystyle \tfrac{5\sqrt{3}}{2}, is the hexagon’s apothem, and this is a first case of the area of a regular polygon.

Example 3

Three polygons in one circle

A circle of radius 8 cm holds an inscribed equilateral triangle, square and regular hexagon. Find the side and perimeter of each, and compare with the circumference.

Solution

The hexagon’s side is the radius, 8 cm, so its perimeter is 48 cm.

The square’s side is the hypotenuse of a right triangle with legs 8 and 8:

\sqrt{8^2 + 8^2} = 8\sqrt{2} \approx 11.31 \text{ cm}

and its perimeter is \displaystyle 4 \times 8\sqrt{2} = 32\sqrt{2} \approx 45.25 cm.

For the triangle, take three neighboring hexagon points A, B, C; the triangle’s side is \overline{AC}. The quadrilateral OABC has four sides of 8 (two radii and two hexagon sides), so it is a rhombus, and the diagonals of a rhombus are perpendicular and cut each other in half. So \overline{OB}, of length 8, crosses \overline{AC} at a right angle at a point M with OM = 4. In the right triangle OMA,

\begin{aligned} AM &= \sqrt{8^2 - 4^2} = 4\sqrt{3} \\ AC &= 2 \times 4\sqrt{3} = 8\sqrt{3} \approx 13.86 \text{ cm} \end{aligned}

and the perimeter is \displaystyle 3 \times 8\sqrt{3} = 24\sqrt{3} \approx 41.57 cm.

The circumference is 2\pi \times 8 = 16\pi \approx 50.27 cm. The perimeters, about 41.57, 45.25 and 48 cm, climb toward it as the number of sides grows, because each side is a short cut across an arc and more sides means shorter cuts.

Example 4

The hexagon on a coordinate grid

A circle has center O = (0, 0) and radius 2, and the hexagon construction starts at A = (2, 0). Show that the first mark can be \displaystyle B = (1, \sqrt{3}), list all six points, and find the side of the equilateral triangle ACE and of the square that has a corner at A.

Solution

The mark B must be 2 from O, to be on the circle, and 2 from A, to be a compass width from A. By the distance formula,

\begin{aligned} OB &= \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2 \\ AB &= \sqrt{(2 - 1)^2 + (0 - \sqrt{3})^2} = \sqrt{4} = 2 \end{aligned}

The other point that fits is \displaystyle (1, -\sqrt{3}); going round the other way would mark it first.

The next mark C is 2 from B and from O, and \displaystyle C = (-1, \sqrt{3}) fits: BC = 2 across, and \displaystyle OC = \sqrt{1 + 3} = 2. The points three steps apart are ends of a diameter, since three steps turn 180^\circ, and the far end of a diameter through (x, y) is (-x, -y). So the six points are

\begin{aligned} &A = (2, 0), \; B = (1, \sqrt{3}), \; C = (-1, \sqrt{3}), \\ &D = (-2, 0), \; E = (-1, -\sqrt{3}), \; F = (1, -\sqrt{3}) \end{aligned}

For triangle ACE, C and E are straight above each other, so \displaystyle CE = \sqrt{3} - (-\sqrt{3}) = 2\sqrt{3}, and

AC = \sqrt{(2 + 1)^2 + (0 - \sqrt{3})^2} = \sqrt{12} = 2\sqrt{3}

matching \displaystyle r\sqrt{3} with r = 2.

The square with a corner at A uses the diameter \overline{AD} along the x-axis and the perpendicular diameter along the y-axis, so its corners are (2, 0), (0, 2), (-2, 0) and (0, -2). Its side is \displaystyle \sqrt{2^2 + 2^2} = 2\sqrt{2}, matching \displaystyle r\sqrt{2}.

Example 5

The triangle from one diameter and one arc

A circle has center O and diameter \overline{AD}. An arc about D, with the compass set to the radius, crosses the circle at C and E. Prove that \triangle ACE is equilateral.

Solution

The plan is to show that the three angles at O between A, C and E are equal, and then use SAS.

Statements	Reasons
OC = OD = DC = r	C and D are on the circle; C is on the arc of width r about D
m\angle COD = 60^\circ	\triangle OCD is equilateral
m\angle AOC = 180^\circ - 60^\circ = 120^\circ	\angle AOC and \angle COD form a linear pair along the diameter
m\angle EOD = 60^\circ, m\angle AOE = 120^\circ	the same steps with E in place of C
m\angle COE = 60^\circ + 60^\circ = 120^\circ	\overrightarrow{OD} runs between \overrightarrow{OC} and \overrightarrow{OE}
\triangle AOC \cong \triangle COE \cong \triangle EOA	SAS: two radii around a 120^\circ angle
AC = CE = EA	corresponding parts of congruent triangles

Check. 120^\circ + 120^\circ + 120^\circ = 360^\circ, all the way round O.

These are the hexagon points A, C, E found without stepping round: C and E are the two hexagon points next to D. The method needs only the straightedge once and the compass once after the circle.

Example 6

From the square to the octagon

A square ABCD is constructed in a circle of radius 10 cm with center O. Show how to add four more points to make a regular octagon, and find the octagon’s angle at the center, each of its angles, and its side.

Solution

Bisect each right angle at O, or, which draws the same line, construct the perpendicular bisector of each side; since OA = OB, the perpendicular bisector of \overline{AB} passes through O. Each bisector meets the circle at a new point between two corners, four new points in all. Every angle at the center between neighbors is now

\frac{90^\circ}{2} = 45^\circ = \frac{360^\circ}{8}

and eight equal angles at the center give eight congruent triangles by SAS, so the octagon is regular.

Let P be the new point between A and B. Triangle OAP has OA = OP = 10 and m\angle AOP = 45^\circ, so its base angles are equal and each is \displaystyle \frac{180^\circ - 45^\circ}{2} = 67.5^\circ. The octagon’s angle at P is two of these: 67.5^\circ + 67.5^\circ = 135^\circ.

For the side, use coordinates: O = (0, 0), A = (10, 0) and B = (0, 10). The bisector of the right angle \angle AOB is the line y = x, and the point on it 10 from O is \displaystyle P = (5\sqrt{2}, 5\sqrt{2}), since \displaystyle (5\sqrt{2})^2 + (5\sqrt{2})^2 = 50 + 50 = 100. Then

\begin{aligned} AP^2 &= (10 - 5\sqrt{2})^2 + (5\sqrt{2})^2 \\ &= 100 - 100\sqrt{2} + 50 + 50 \\ &= 200 - 100\sqrt{2} \end{aligned}

so \displaystyle AP = \sqrt{200 - 100\sqrt{2}} \approx 7.65 cm. The perimeter is about 8 \times 7.65 \approx 61.23 cm, closer to the circumference 20\pi \approx 62.83 cm than the square’s \displaystyle 40\sqrt{2} \approx 56.57 cm.

The same doubling works from the hexagon: bisecting its six angles at the center gives a regular 12-gon with angle 30^\circ at the center, and doubling again gives 24 sides.

Boundaries

One width, and it must be the right one. For the triangle on \overline{AB}, both circles use the width AB. With some other width s used from both ends, the crossing gives a triangle with sides s, s and AB: isosceles, but equilateral only when s = AB. For the hexagon and the triangle in a circle, the width must be the circle’s own radius.

The center must be known. Every circle construction here starts from O. A circle drawn with no center marked needs its center found first, by crossing the perpendicular bisectors of two chords, as in constructions with circles.

Exact in principle. The reasoning is exact; a drawing is as exact as the pencil. Stepping round the circle copies each small error into the next step, which is why the diameter version is the steadier one.

Only some regular polygons come this directly. From 3, 4 and 6 sides, bisecting the angles at the center gives 8, 12, 16, 24 and so on. A regular pentagon can also be constructed, by a longer method. A regular heptagon, with 7 sides, cannot be constructed with compass and straightedge at all; that was proved in the nineteenth century.

In a plane. Two circles in a plane meet in at most two points. In space, two spheres with the same radius AB about A and B meet in a whole circle, and every point of it is the third corner of an equilateral triangle on \overline{AB}.

Where this goes next

The 60^\circ angle, bisected with the angle bisector, gives 30^\circ and 15^\circ, and the equilateral triangle cut by its height is the source of the exact values in the special right triangles: the height \displaystyle \tfrac{\sqrt{3}}{2} of a triangle of side 1 is where \sin 60^\circ comes from. The hexagon’s six points on a circle of radius 1 are the 60^\circ marks of the unit circle.

The six equilateral triangles inside a hexagon lead to the apothem and the area of any regular polygon, and to its interior angle. Equilateral triangles, squares and hexagons are also the three regular polygons that tile the plane on their own.

Changing the two equal widths into three given lengths turns this construction into constructing a triangle from three sides. And every argument here, radii that the compass makes equal and the congruent triangles they form, is the pattern for justifying any construction.

Exercises

An equilateral triangle on a segment

An equilateral triangle ABC is constructed on \overline{AB} with AB = 8 cm. Find its perimeter and its height from C.
On \overline{AB} with AB = 10 cm, both crossings C and D are marked. Find CD and the perimeter of ACBD.
A = (0, 0) and B = (4, 0). Find both crossings of the circles of radius AB about A and about B.
An equilateral triangle ABC is constructed on \overline{AB}, with AB = 5 cm. A second equilateral triangle BDC is then constructed on \overline{BC}, with D on the other side of \overline{BC} from A. Find m\angle ABD and the perimeter of ABDC.

The hexagon

A regular hexagon is constructed in a circle of radius 7 cm. Find its side and perimeter.
A regular hexagon constructed in a circle has perimeter 54 cm. What was the radius?
Regular hexagon ABCDEF is constructed in a circle with center O. Find m\angle AOB, m\angle OAB and m\angle ABC.
Find the area of a regular hexagon constructed in a circle of radius 4 cm.
Regular hexagon ABCDEF is constructed in a circle of radius 4 cm. Find AD and AC.

The triangle and the square in a circle

A circle has radius 10 cm. Find the side of the equilateral triangle and of the square constructed in it.
A square constructed in a circle has side 6 cm. Find the radius of the circle.
An equilateral triangle constructed in a circle has side 9 cm. Find the radius of the circle.
A circle has center (0, 0) and radius 5, and a square is constructed in it with one corner at A = (3, 4). Find the other three corners and the side.
A square is constructed in a circle of radius 5 cm. Find the area of the square, and what fraction of the circle’s area it covers.

More sides by bisecting

A regular 12-gon is made by bisecting the angles at the center of a regular hexagon. Find its angle at the center and each of its angles.
A circle has center (0, 0) and radius 2, with hexagon points A = (2, 0) and \displaystyle B = (1, \sqrt{3}). Show that the point of the 12-gon between them is \displaystyle P = (\sqrt{3}, 1), and find AP.
Starting from the hexagon, the angles at the center are bisected three times in turn. How many sides does each new polygon have, and what is the angle at the center of the last one?

Reasoning

In the hexagon construction, explain why the sixth arc lands exactly on the starting point, rather than a little before or after it.
Someone draws the circle about A with width AB and the circle about B with a different width. Explain why the crossing is not the corner of an equilateral triangle on \overline{AB}.
In the triangle construction on \overline{AB} with crossings C and D, prove that \overleftrightarrow{CD} is the perpendicular bisector of \overline{AB}.
Prove that joining every other point of a regular hexagon in a circle gives an equilateral triangle.
Explain why the square construction gives a square and not just a rhombus.

Answers

Perimeter 24 cm, height \displaystyle 4\sqrt{3} \approx 6.93 cm. Each side is 8; the height meets \overline{AB} at its midpoint, so it is \displaystyle \sqrt{8^2 - 4^2} = \sqrt{48} = 4\sqrt{3}.
\displaystyle CD = 10\sqrt{3} \approx 17.32 cm, perimeter 40 cm. Each crossing is \displaystyle \sqrt{10^2 - 5^2} = 5\sqrt{3} from the midpoint, on opposite sides, and all four sides of ACBD are 10.
\displaystyle (2, 2\sqrt{3}) and \displaystyle (2, -2\sqrt{3}). The crossings are on the perpendicular bisector x = 2, at height \displaystyle \sqrt{4^2 - 2^2} = \sqrt{12} = 2\sqrt{3} above and below. Check: \displaystyle \sqrt{2^2 + 12} = 4.
m\angle ABD = 120^\circ, perimeter 20 cm. \overrightarrow{BC} runs between \overrightarrow{BA} and \overrightarrow{BD}, so 60^\circ + 60^\circ = 120^\circ. All four sides of ABDC are 5, so the perimeter is 4 \times 5 = 20.
Side 7 cm, perimeter 42 cm. The side equals the radius.
Radius 9 cm. The side is \displaystyle \frac{54}{6} = 9, and the side equals the radius.
m\angle AOB = 60^\circ, m\angle OAB = 60^\circ, m\angle ABC = 120^\circ. \triangle OAB is equilateral, and the angle at B is two of its angles.
\displaystyle 24\sqrt{3} \approx 41.57 cm^2. Six equilateral triangles of side 4, each with height \displaystyle 2\sqrt{3} and area \displaystyle \tfrac{1}{2} \times 4 \times 2\sqrt{3} = 4\sqrt{3}.
AD = 8 cm and \displaystyle AC = 4\sqrt{3} \approx 6.93 cm. A and D are three steps apart, 180^\circ, so \overline{AD} is a diameter. OABC is a rhombus of side 4, so \overline{OB} crosses \overline{AC} at a right angle 2 from O, giving \displaystyle \tfrac{1}{2}AC = \sqrt{4^2 - 2^2} = 2\sqrt{3}.
Triangle \displaystyle 10\sqrt{3} \approx 17.32 cm, square \displaystyle 10\sqrt{2} \approx 14.14 cm. These are \displaystyle r\sqrt{3} and \displaystyle r\sqrt{2} with r = 10.
\displaystyle 3\sqrt{2} \approx 4.24 cm. The side is \displaystyle r\sqrt{2}, so \displaystyle r = \frac{6}{\sqrt{2}} = 3\sqrt{2}.
\displaystyle 3\sqrt{3} \approx 5.20 cm. The side is \displaystyle r\sqrt{3}, so \displaystyle r = \frac{9}{\sqrt{3}} = 3\sqrt{3}.
(-4, 3), (-3, -4) and (4, -3), side \displaystyle 5\sqrt{2}. The opposite corner is the far end of the diameter, (-3, -4). The other diameter is perpendicular: (-4, 3) is 5 from the center, and 3(-4) + 4(3) = 0 shows the two radii are perpendicular; its far end is (4, -3). The side is \displaystyle \sqrt{(3 + 4)^2 + (4 - 3)^2} = \sqrt{50} = 5\sqrt{2}.
Area 50 cm^2, a fraction \displaystyle \frac{2}{\pi} \approx 0.64 of the circle. The side is \displaystyle 5\sqrt{2}, so the area is \displaystyle (5\sqrt{2})^2 = 50, and the circle’s area is 25\pi.
Angle at the center 30^\circ, each angle 150^\circ. \displaystyle \frac{60^\circ}{2} = 30^\circ; each triangle at the center has base angles \displaystyle \frac{180^\circ - 30^\circ}{2} = 75^\circ, and each corner is two of them.
\displaystyle AP = \sqrt{8 - 4\sqrt{3}} \approx 1.04. P is on the circle, since 3 + 1 = 4. It is the same distance from A and B: \displaystyle AP^2 = (2 - \sqrt{3})^2 + 1^2 = 8 - 4\sqrt{3} and \displaystyle BP^2 = (\sqrt{3} - 1)^2 + (1 - \sqrt{3})^2 = 8 - 4\sqrt{3}. So P is on the perpendicular bisector of \overline{AB}, which is the bisector of \angle AOB, and on the circle between A and B.
12, 24 and 48 sides; the last has angle 7.5^\circ at the center. Each bisection doubles the sides, and \displaystyle \frac{360^\circ}{48} = 7.5^\circ.
Each step makes an equilateral triangle with O, because the chord and both radii are r, so each step turns exactly 60^\circ about O. Six steps turn 360^\circ, all the way round, so the sixth chord ends at the starting point. Any other width would turn some other angle, and 360^\circ would not in general be a whole number of steps.
The crossing C is AB from A but some other distance from B, so CA \neq CB. A triangle with two unequal sides is not equilateral.
C is on both circles, so CA = AB = CB; likewise DA = DB. Each of C and D is equidistant from A and B, so each lies on the perpendicular bisector of \overline{AB}. They are different points, so the line through them is that bisector.
With the hexagon ABCDEF and center O, each of \angle AOC, \angle COE and \angle EOA spans two 60^\circ steps, so each is 120^\circ. The triangles AOC, COE and EOA have two radii around equal angles, so they are congruent by SAS, and AC = CE = EA.
A rhombus needs only four equal sides; a square also needs right angles. The four triangles at O are isosceles with a right angle at O, so each base angle is 45^\circ, and each corner of ABCD is two of them, 90^\circ. Four equal sides and four right angles make a square. (Also, its diagonals are both diameters, so they are equal, which a rhombus that is not a square never has.)