Regular polygons and their circles
Every regular polygon fits exactly inside one circle and exactly around another. Why, and what does the angle at the center between two neighboring corners tell you?
Where you are 15 in Two-dimensional shapes
Why this exists
The angles of a regular polygon are found by working along its edge, one corner at a time. That says how sharply each corner turns, but not how the polygon sits as a whole. A regular polygon also has a middle: one point that is the same distance from every corner and the same distance from every side. Join that point to the corners and the polygon falls apart into identical triangles, all meeting in the middle like slices of a pie.
Those triangles are how almost everything else about a regular polygon is worked out. Its perimeter and its area are n copies of one triangle’s. Its turns onto itself are turns about the middle point. And the idea runs the other way too. To lay out a regular hexagon for the floor of a garden gazebo, or a twelve-sided table top, nobody measures 150^\circ angles along the edge. You draw a circle, share the full turn at its center into equal parts, and join the marks. The angle that does the sharing is the central angle, and this lesson is about it and the two circles that come with it.
Definition
Take a regular polygon with n sides: all its sides are equal and all its angles are equal.
- Its center is the point O that is the same distance from all of its vertices. Every regular polygon has exactly one such point; “Why it is true” shows it.
- The circle with center O through all the vertices is the circumscribed circle. Its radius is the radius of the polygon, and the polygon is inscribed in the circle.
- A central angle is the angle at O between the radii to two neighboring vertices, such as \angle AOB in Figure 1.
- The radii to the vertices cut the polygon into n triangles, one standing on each side, such as triangle OAB. This lesson calls them the triangles at the center; the name is a label for these pages, not a standard term.
- The circle with center O that touches every side is the inscribed circle. “Touches” means it meets each side at exactly one point and never crosses it, so each side is tangent to the circle. Its radius is the distance from O to a side, which is called the The apothem of a regular polygon of the polygon.
Intuition
Stand at O and look at the vertices one after another. Each is the same distance away, and each is the same turn further round than the last, because nothing about a regular polygon changes when you move from one vertex to the next. n equal turns bring you back to where you started, which is one full turn of 360^\circ. So each turn is 360^\circ shared into n equal parts.
That is the polygon’s rotational symmetry seen from the inside: turning the whole polygon about O through one central angle moves every vertex onto the next one, so the polygon lands on itself. The lines of symmetry all pass through O as well.
The two circles are set by the two kinds of point on the edge. The corners stick out further than anything else, so the circumscribed circle passes through them. The midpoint of each side is the nearest the edge comes to O, so the inscribed circle touches there. With more sides the corners flatten out, the two circles close in on each other, and the polygon looks more and more like a circle itself.
Results
The center. Every regular polygon has a center O, one point equally far from all the vertices and equally far from all the sides. So it has a circumscribed circle and an inscribed circle, and the two circles share that center.
The central angle. For a regular polygon with n sides,
The triangles at the center. The radii to the vertices cut the polygon into n congruent isosceles triangles. Each base angle is half an interior angle, and so
which says the central angle is the same size as the exterior angle.
Arcs. Each side cuts off an arc of the circumscribed circle measuring \dfrac{360^\circ}{n}, because an arc has the measure of its central angle. Going round from one vertex to another k places along, the arc measures k \cdot \dfrac{360^\circ}{n}.
The perpendicular from the center. The segment from O perpendicular to a side lands on the midpoint of that side and splits the central angle into two halves of \dfrac{180^\circ}{n}. Its length is the radius of the inscribed circle, and it lies along a line of symmetry.
Inscribing. The other way round: n points on a circle that cut it into n equal arcs are the vertices of a regular polygon. So to inscribe a regular n-gon in a circle, mark off central angles of \dfrac{360^\circ}{n} one after another and join each mark to the next.
Why it is true
There is a center. Let the polygon be ABCDE\ldots, with every interior angle equal to I. Bisect \angle A and \angle B. Each half is \tfrac{I}{2}, which is less than 90^\circ, so the two bisectors lean toward each other and meet at a point O inside the polygon.
In triangle OAB the angles at A and B are both \tfrac{I}{2}. A triangle with two equal angles has equal sides opposite them (the converse of the base angles theorem), so OA = OB.
Now compare triangle OBA with triangle OBC. BA = BC, because the polygon’s sides are equal. \angle OBA = \angle OBC, because OB bisects \angle B. And OB is a side of both. By SAS the two triangles are congruent. So OC = OA, and \angle OCB = \angle OAB = \tfrac{I}{2}, which is half of \angle C: the segment OC bisects \angle C as well.
Triangle OCD is now in exactly the position triangle OBC was in, so the same argument gives OD = OC and shows that OD bisects \angle D, and so on all the way round. Every vertex is the same distance from O, and the n triangles OAB, OBC, OCD, and so on are all congruent.
The central angle. Congruent triangles have equal angles at O. The n triangles fit together round O with no gaps and no overlaps, so their angles at O fill one full turn. That is n equal angles adding to 360^\circ, so each is \dfrac{360^\circ}{n}.
Central angle plus interior angle. In triangle OAB the base angles are each \tfrac{I}{2}, so by the triangle angle sum
As a check against the formula for the interior angle, I = \dfrac{(n-2) \cdot 180^\circ}{n}:
the same answer, reached the other way.
The inscribed circle. Each triangle at the center is isosceles, with OA = OB. Drop the perpendicular from O to \overline{AB}, meeting it at M. Triangles OMA and OMB have a right angle at M, equal hypotenuses OA = OB and the shared leg OM, so they are congruent by HL. That makes AM = MB, so M is the midpoint, and it splits \angle AOB into two equal halves.
All n triangles at the center are congruent, so all these perpendiculars have the same length; call it a. The perpendicular is the shortest way from a point to a line, so every other point of a side is more than a from O. The circle with center O and radius a therefore reaches each side at exactly one point, its midpoint, and stays inside the polygon everywhere else. It is the inscribed circle.
Inscribing. Suppose P_1, P_2, \ldots, P_n lie in order round a circle with center O, and each of the angles \angle P_1OP_2, \angle P_2OP_3, and so on round to \angle P_nOP_1 is \dfrac{360^\circ}{n}. Each triangle OP_kP_{k+1} has two sides that are radii and the angle between them equal to \dfrac{360^\circ}{n}, so by SAS all n triangles are congruent. Their third sides, the sides of the polygon, are therefore all equal, and their base angles are all equal. Each central angle is at most 120^\circ, so O lies inside the polygon, and each angle of the polygon is made of two base angles, one from each triangle beside it. So every angle of the polygon is the same as well. Equal sides and equal angles: the polygon is regular.
Worked examples
The regular dodecagon
A regular dodecagon has 12 sides. Find its central angle, the base angles of each triangle at the center, and each interior angle. Check the interior angle against the angle sum.
SolutionTwelve equal central angles fill the full turn at the center, so each is
Each triangle at the center has two sides that are radii, so it is isosceles and its two base angles are equal. With 30^\circ at the top they share what is left of 180^\circ:
At each vertex, the interior angle is made of two of these base angles, one from the triangle on each side of it, so each interior angle is 75^\circ + 75^\circ = 150^\circ.
Check. The interior angles of a 12-sided polygon add to (12 - 2) \cdot 180^\circ, so each of twelve equal angles is \dfrac{10 \cdot 180^\circ}{12} = 150^\circ. It agrees, and 30^\circ + 150^\circ = 180^\circ as the result says it must.
Radii to vertices further apart
ABCDEFGHI is a regular nonagon with center O, as in Figure 3. Find m\angle AOB, m\angle AOD and m\angle AOF, and the measures of the minor arc \overset{\frown}{AF} and the major arc \overset{\frown}{ABF}.
A nonagon has 9 sides, so each central angle is \dfrac{360^\circ}{9} = 40^\circ. That gives m\angle AOB = 40^\circ straight away.
Going from A to D the radii pass OB and OC, so \angle AOD is made of three central angles side by side. By angle addition, m\angle AOD = 3 \cdot 40^\circ = 120^\circ.
Going from A to F the same way passes five central angles, 5 \cdot 40^\circ = 200^\circ. But an angle between two rays is never more than 180^\circ; 200^\circ is the long way round. The angle \angle AOF is the short way, through I, H and G, which is four central angles:
An arc has the measure of its central angle, so the minor arc \overset{\frown}{AF}, the short way round, measures 160^\circ. The major arc \overset{\frown}{ABF} is the rest of the circle, 360^\circ - 160^\circ = 200^\circ, which is the five steps counted first.
A square in a circle
Square ABCD is inscribed in a circle with center O and radius 5, as in Figure 4. Find the side of the square, its perimeter, and the radius of its inscribed circle.
The central angle of a square is \dfrac{360^\circ}{4} = 90^\circ, so triangle OAB has a right angle at O, with both legs radii of length 5. By the Pythagorean theorem
The side is \displaystyle 5\sqrt{2} \approx 7.07, and the perimeter is \displaystyle 4 \cdot 5\sqrt{2} = 20\sqrt{2} \approx 28.3.
The radius of the inscribed circle is OM, where M is the midpoint of \overline{AB} and \overline{OM} \perp \overline{AB}. The segment OM splits the right angle at O in half, so triangle OMA has 45^\circ at O and 90^\circ at M, which leaves 45^\circ at A. Two equal angles means two equal sides, OM = MA, and MA is half the side:
Check. In the right triangle OMA, OM^2 + MA^2 = 2 \cdot \dfrac{50}{4} = 25 = OA^2.
An equilateral triangle in a circle
Equilateral triangle ABC is inscribed in a circle with center O and radius 6, as in Figure 5. The line AO meets \overline{BC} at M and the circle again at D. Find OM and BC, and the radius of the triangle’s inscribed circle.
Where the line goes. OB = OC, so O is on the perpendicular bisector of \overline{BC}, and AB = AC, so A is on it too. Two points fix a line, so line AO is that perpendicular bisector: M is the midpoint of \overline{BC} and \overline{AD} \perp \overline{BC}.
An equilateral triangle hiding inside. The central angle is \dfrac{360^\circ}{3} = 120^\circ, so m\angle AOB = 120^\circ. A, O and D lie on one line, so \angle AOB and \angle BOD form a linear pair:
Triangle OBD has OB = OD = 6 (both radii) and 60^\circ between them. Its two equal base angles share 120^\circ, so each is 60^\circ as well, and the triangle is equilateral with BD = 6.
The height of that triangle. \overline{BM} is perpendicular to \overline{OD}, so it is the height of the isosceles triangle OBD from B, and the height from the top of an isosceles triangle lands on the midpoint of its base. So M is the midpoint of \overline{OD} and
The side. In the right triangle OMB, BM^2 = 6^2 - 3^2 = 27, so \displaystyle BM = \sqrt{27} = 3\sqrt{3}. M is the midpoint of \overline{BC}, so \displaystyle BC = 2 \cdot 3\sqrt{3} = 6\sqrt{3} \approx 10.4.
The inscribed circle. Its radius is the distance from O to a side, which is OM = 3. Nothing in the argument depended on the radius being 6: for every equilateral triangle, the inscribed circle has exactly half the radius of the circumscribed one.
Coordinates of a regular hexagon
A regular hexagon ABCDEF has its center at the origin, and A = (4, 0), with the vertices lettered counterclockwise, as in Figure 6. Find the coordinates of all six vertices.
The radius is OA = 4, and the central angle is \dfrac{360^\circ}{6} = 60^\circ. Triangle OAB has OA = OB = 4 and 60^\circ between them, so, as in the last example, its base angles are 60^\circ too and it is equilateral.
Drop the perpendicular from B to the x-axis, meeting it at N. In the equilateral triangle OAB this is the height from B, so it lands on the midpoint of \overline{OA}: N = (2, 0). Then in the right triangle ONB,
B is above the axis, so \displaystyle B = (2, 2\sqrt{3}).
The rest come from the lines of symmetry through O. The y-axis is one of them: it reflects B to C and A to D. The x-axis is another, reflecting the top half onto the bottom. So
Check. BC = 2 - (-2) = 4, the same as the side AB, and B is \displaystyle \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{16} = 4 from the origin, on the circle as it should be.
Equal sides on a circle
Five points A, B, C, D, E lie in that order round a circle with center O, and AB = BC = CD = DE = EA. Show that ABCDE is a regular pentagon.
SolutionEach of the triangles OAB, OBC, OCD, ODE and OEA has two sides that are radii and a third side that is one of the five equal sides. By SSS they are all congruent, so their angles at O are equal; call that angle \theta.
The five arcs \overset{\frown}{AB}, \overset{\frown}{BC}, \overset{\frown}{CD}, \overset{\frown}{DE} and \overset{\frown}{EA}, taken between neighbors, go once round the circle and add to 360^\circ. Each is either the minor arc, measuring \theta, or the major arc, measuring 360^\circ - \theta. If even one were the major arc, the five would add to at least 4\theta + (360^\circ - \theta) = 360^\circ + 3\theta, which is more than 360^\circ. So all five are minor arcs and
The five points cut the circle into five equal arcs, and by the inscribing result such points are the vertices of a regular pentagon. Equal sides on a circle are enough; Boundaries shows why neither half of that is enough alone.
Both circles of an equilateral triangle from its side
An equilateral triangle has side 12. Find the radii of its circumscribed and inscribed circles, and compare the areas of the two circles.
SolutionUse the picture of Figure 5, with the radius R unknown this time. Example 4 showed that the inscribed radius OM is half the circumscribed radius, \tfrac{R}{2}, whatever R is. M is the midpoint of the side, so BM = 6. In the right triangle OMB, with hypotenuse OB = R:
A radius is positive, so \displaystyle R = \sqrt{48} = 4\sqrt{3} \approx 6.93, and the inscribed radius is half of it, \displaystyle 2\sqrt{3} \approx 3.46.
The area of a circle is \pi times the square of its radius, so the circles have areas 48\pi and \displaystyle \pi(2\sqrt{3})^2 = 12\pi. The circle round the triangle has 4 times the area of the circle inside it, which is what halving a radius always does to an area.
Check. \displaystyle BM^2 + OM^2 = 6^2 + (2\sqrt{3})^2 = 36 + 12 = 48 = R^2.
Boundaries
Only a regular polygon has all of this at once. Other polygons can have one circle without the other. A rectangle that is not a square has a circumscribed circle (its diagonals are diameters) but no inscribed circle, and its triangles at the center are not all congruent. A rhombus that is not a square has an inscribed circle but no circumscribed circle. Most polygons have neither.
Equal angles on a circle are not enough, and neither are equal sides off one. A rectangle has four equal angles and sits on a circle, but it is not regular. A rhombus has four equal sides and is not regular either. Equal sides and all vertices on one circle together are enough, as Example 6 shows.
The central angle is between neighbors. Radii to vertices k places apart make an angle of k \cdot \dfrac{360^\circ}{n} only while that is at most 180^\circ. Past that, the angle is the short way round, 360^\circ - k \cdot \dfrac{360^\circ}{n}, as with \angle AOF in Example 2.
Which central angles are possible. A polygon has at least 3 sides, so the central angle is at most 120^\circ. It must share 360^\circ into a whole number of equal parts, so 50^\circ is not the central angle of any regular polygon. It need not be a whole number of degrees, though: a regular heptagon has a central angle of \left(\tfrac{360}{7}\right)^\circ.
Join neighbors only. Inscribing joins each marked point to the next one round the circle. Joining every second point of five equally spaced points instead draws a five-pointed star whose edges cross each other, and that is not a polygon in the sense used here.
Where this goes next
The perpendicular from the center to a side has its own name and its own lesson, the The apothem of a regular polygon, and with it the area of a regular polygon is n copies of one triangle at the center: that is the area of a regular polygon. Finding the apothem and the side for a polygon with, say, 9 sides needs the trigonometry of the right triangle OMA, with its angle of \dfrac{180^\circ}{n} at O.
Letting n grow is how the circle itself gets measured. The inscribed polygons hug the circle more and more tightly, so their perimeters close in on the Circumference and their areas on the area of the circle.
With a compass and straightedge, the central angles of 120^\circ, 90^\circ and 60^\circ can be made exactly, and that gives the equilateral triangle, square and regular hexagon in a circle. For triangles that are not regular, the center splits into two different points, the Circumcenter for the circle through the vertices and the Incenter for the circle touching the sides, and the quadrilaterals that fit on a circle are exactly the ones whose opposite angles add to 180^\circ.
Exercises
Central angles
- Find the central angle of a regular pentagon.
- Find the central angle of a regular decagon.
- Find the central angle of a regular heptagon.
- For a regular 18-gon, find the central angle, the base angles of each triangle at the center, and each interior angle.
Working back to the number of sides
- A regular polygon has a central angle of 45^\circ. How many sides does it have?
- A regular polygon has a central angle of 15^\circ. How many sides does it have?
- Each interior angle of a regular polygon is 156^\circ. Find its central angle and its number of sides.
- Is there a regular polygon with a central angle of 50^\circ? Explain.
Angles and arcs between vertices
- ABCDEFGH is a regular octagon with center O. Find m\angle AOC, m\angle AOD and m\angle AOE.
- ABCDEFGHIJ is a regular decagon inscribed in a circle. Find the measures of the minor arc \overset{\frown}{AD} and the major arc \overset{\frown}{AJD}.
- The radii to two vertices of a regular 15-gon make an angle of 96^\circ. How many sides of the polygon lie between those vertices, the short way round?
- A Ferris wheel has 18 cars equally spaced round its rim. Find the angle between the spokes to two neighboring cars, and how far the wheel must turn to bring a car to the place where the car five places ahead of it was.
Lengths in the two circles
- A square is inscribed in a circle of radius 8. Find its side and the radius of its inscribed circle.
- A square has side 10. Find the radii of its circumscribed and inscribed circles.
- A regular hexagon has side 6. Find the radii of its circumscribed and inscribed circles.
- An equilateral triangle is inscribed in a circle of radius 10. Find the radius of its inscribed circle and its side.
- The inscribed circle of an equilateral triangle has radius 4. Find the radius of its circumscribed circle and its side.
- A regular hexagon has its center at the origin and one vertex at (6, 0). Find the coordinates of all six vertices.
- A regular hexagon and an equilateral triangle are inscribed in the same circle of radius r. Find both perimeters, and how many times as long the hexagon’s is.
- Regular polygons with 3, 4 and 6 sides are each inscribed in a circle of radius 1. Find the radius of each one’s inscribed circle.
Reasoning
- Explain why the central angle of a regular polygon is always the same size as its exterior angle.
- Explain why no regular polygon has a central angle larger than 120^\circ.
- A rectangle measuring 6 by 8 is inscribed in a circle. All four of its angles are equal. Use its triangles at the center to explain why it is still not regular.
- Show that a rhombus with no right angle has no circumscribed circle.
- Explain why the inscribed circle of a regular polygon touches each side at the side’s midpoint.
- Look back at the three answers to Exercise 20. What happens to the gap between the two circles as the number of sides grows, and how does the central angle explain it?
Answers
- 72^\circ. Five equal central angles fill 360^\circ: \dfrac{360^\circ}{5} = 72^\circ.
- 36^\circ. \dfrac{360^\circ}{10} = 36^\circ.
- \left(\tfrac{360}{7}\right)^\circ \approx 51.4^\circ. Seven equal parts of a full turn. 360 is not a multiple of 7, so the angle is not a whole number of degrees, which is allowed.
- Central angle 20^\circ, base angles 80^\circ, interior angle 160^\circ. The central angle is \tfrac{360^\circ}{18} = 20^\circ; the two equal base angles share 180^\circ - 20^\circ = 160^\circ, so each is 80^\circ; each interior angle is two base angles. Check: \tfrac{16 \cdot 180^\circ}{18} = 160^\circ.
- 8 sides. \dfrac{360}{n} = 45 gives n = 8.
- 24 sides. n = 360 \div 15 = 24.
- Central angle 24^\circ, 15 sides. The central angle is 180^\circ - 156^\circ = 24^\circ, and 360 \div 24 = 15. Check: \tfrac{13 \cdot 180^\circ}{15} = 156^\circ.
- No. 360 \div 50 = 7.2, and a polygon cannot have 7.2 sides. The nearest are 7 sides, with about 51.4^\circ, and 8 sides, with 45^\circ.
- m\angle AOC = 90^\circ, m\angle AOD = 135^\circ, m\angle AOE = 180^\circ. Each central angle is \tfrac{360^\circ}{8} = 45^\circ, and C, D, E are 2, 3 and 4 places from A. \angle AOE is a straight angle, so \overline{AE} is a diameter.
- m\overset{\frown}{AD} = 108^\circ and m\overset{\frown}{AJD} = 252^\circ. Each side cuts off \tfrac{360^\circ}{10} = 36^\circ, D is 3 places from A, and the major arc is 360^\circ - 108^\circ.
- 4 sides. Each central angle is \tfrac{360^\circ}{15} = 24^\circ, and 96 \div 24 = 4. The long way round has the other 11.
- 20^\circ between neighboring spokes, and a turn of 100^\circ. The 18 spokes share the full turn, 360 \div 18 = 20, and five places is 5 \cdot 20^\circ = 100^\circ.
- Side \displaystyle 8\sqrt{2} \approx 11.3, inscribed radius \displaystyle 4\sqrt{2} \approx 5.66. The central angle is 90^\circ, so AB^2 = 8^2 + 8^2 = 128. As in Example 3, the inscribed radius is half the side.
- Circumscribed radius \displaystyle 5\sqrt{2} \approx 7.07, inscribed radius 5. The perpendicular from O lands on the midpoint of a side, and, as in Example 3, it is as long as half a side, 5. Then OA^2 = 5^2 + 5^2 = 50 in the right triangle OMA.
- Circumscribed radius 6, inscribed radius \displaystyle 3\sqrt{3} \approx 5.20. The central angle is 60^\circ, so each triangle at the center is equilateral with side 6, and OA = 6. The inscribed radius is its height: OM^2 = 6^2 - 3^2 = 27.
- Inscribed radius 5, side \displaystyle 10\sqrt{3} \approx 17.3. As in Example 4, the inscribed radius is half of 10. Then BM^2 = 10^2 - 5^2 = 75, so \displaystyle BM = 5\sqrt{3}, and the side is twice that.
- Circumscribed radius 8, side \displaystyle 8\sqrt{3} \approx 13.9. The inscribed radius is half the circumscribed one, so that is 8. Then BM^2 = 8^2 - 4^2 = 48, so \displaystyle BM = 4\sqrt{3}, and the side is twice that.
- (6, 0), \displaystyle (3, 3\sqrt{3}), \displaystyle (-3, 3\sqrt{3}), (-6, 0), \displaystyle (-3, -3\sqrt{3}), \displaystyle (3, -3\sqrt{3}). As in Example 5 with 6 for 4: the foot of the perpendicular from B is (3, 0), and the height is \displaystyle \sqrt{6^2 - 3^2} = 3\sqrt{3}. The axes reflect B to the others.
- Hexagon 6r, triangle \displaystyle 3\sqrt{3}\,r, and the hexagon’s is \displaystyle \tfrac{2\sqrt{3}}{3} \approx 1.15 times as long. The hexagon’s triangles at the center are equilateral, so its side is r. For the triangle, as in Example 4, half a side is \displaystyle \sqrt{r^2 - \left(\tfrac{r}{2}\right)^2} = \tfrac{\sqrt{3}}{2}r, so a side is \displaystyle \sqrt{3}\,r. Then \displaystyle \tfrac{6r}{3\sqrt{3}\,r} = \tfrac{2}{\sqrt{3}} = \tfrac{2\sqrt{3}}{3}.
- \tfrac{1}{2}, \displaystyle \tfrac{\sqrt{2}}{2} \approx 0.71 and \displaystyle \tfrac{\sqrt{3}}{2} \approx 0.87. Triangle: half the radius, as in Example 4. Square: OM = MA and OM^2 + MA^2 = 1, so OM^2 = \tfrac{1}{2}. Hexagon: the height of an equilateral triangle with side 1, OM^2 = 1 - \tfrac{1}{4} = \tfrac{3}{4}.
- In a triangle at the center the two base angles are each half an interior angle I, so the central angle is 180^\circ - I. The exterior angle at a vertex forms a linear pair with the interior angle, so it is also 180^\circ - I. Both equal 180^\circ - I, so they are equal.
- A polygon has at least 3 sides, and the central angle \tfrac{360^\circ}{n} gets smaller as n gets larger. The largest it can be is for n = 3, which gives 120^\circ.
- Its diagonals are both \displaystyle \sqrt{6^2 + 8^2} = 10 and cross at the center O, 5 from every vertex. The triangles at the center have sides 5, 5, 6 on the short sides and 5, 5, 8 on the long ones, so they are not all congruent, and its four central angles are not all equal. A regular polygon’s triangles at the center are all congruent, so the rectangle is not regular: equal angles were not enough, because its sides are not equal.
- Opposite angles of a rhombus are equal, since a rhombus is a parallelogram. If all four vertices were on one circle, opposite angles would add to 180^\circ, as for every cyclic quadrilateral. Two equal angles adding to 180^\circ are each 90^\circ, so the rhombus would have a right angle, which it does not.
- Each triangle at the center is isosceles, so the perpendicular from O to its side cuts that side into two equal parts (the two halves of the triangle are congruent by HL). All these perpendiculars have the same length a, and every other point of a side is further than a from O, because the perpendicular is the shortest distance. So the circle of radius a meets each side only at the foot of the perpendicular, which is the midpoint.
- The ratio of the inscribed radius to the circumscribed one goes 0.5, about 0.71, about 0.87, so the two circles get closer together as n grows. The central angle \tfrac{360^\circ}{n} shrinks, so each triangle at the center gets narrower, its side gets shorter compared with the radius, and the height OM gets closer to the slanted side OA. The ratio comes as close to 1 as you like but never reaches it, since OM is always a leg and OA the hypotenuse of a right triangle.