What "inscribing" really asks
To inscribe a regular polygon in a circle means to place all its vertices on the circle, evenly spaced, so every side and every angle comes out equal. By the end of guide 2 you can already copy an angle, bisect, and drop a perpendicular — those are the only moves you get. So the real question is not "can I draw a pentagon?" but "can I, with compass and straightedge alone, mark n points that split a circle into n equal arcs?"
Each vertex you place uses up a central angle of exactly 360/n degrees: a hexagon needs 60-degree slices, a pentagon needs 72-degree slices, a 17-gon needs slices of 360/17 (about 21.18) degrees. So inscribing a regular n-gon is the same task as constructing the angle 360/n at the center. That single reframing is the key to the whole guide.
The easy ones: 6, 4, 3, and doubling
The regular hexagon is the gift of the circle: the radius steps around the circumference exactly six times, because a chord equal to the radius subtends a 60-degree arc. No measuring, no guessing — open the compass to the radius and walk six steps. From the hexagon's six points, take every other one and you have an equilateral triangle for free.
The square is just as friendly: draw any diameter, bisect the straight angle with a perpendicular through the center, and the two diameters meet the circle in four evenly spaced points. And here is the engine behind half the table — once you have a regular n-gon, bisecting each central arc doubles the count for free. So 3 gives 6, 12, 24, ...; 4 gives 8, 16, 32, ...; 5 gives 10, 20, 40, ....
n : 3 4 5 6 8 10 12 15 16 17 slice : 120 90 72 60 45 36 30 24 22.5 21.18... status : OK OK OK OK OK OK OK OK OK OK 7, 9, 11, 13, 14, 18, 19 : NOT constructible
Combining and the pentagon
Two more tricks fill in the gaps. First, doubling (bisection) we just saw. Second, combining coprime polygons: if you can build an m-gon and an n-gon and m, n share no common factor, you can build the (m*n)-gon. The reason is plain in angles — having a 360/3 = 120-degree arc and a 360/5 = 72-degree arc, the difference 120 - 72 = 48 degrees equals 360/7.5... no, equals two slices of 360/15 = 24 degrees, so you can step out a regular 15-gon.
But the pentagon itself is the first genuinely surprising case — its 72-degree slice is not a gift of the radius. It hides inside the golden ratio. The half-diagonal of the pentagon relates to its side through the number phi = (1 + sqrt(5))/2, and because that involves only a square root, it is a constructible number. The classical recipe builds a length of sqrt(5)/2 by Pythagoras (a right triangle with legs 1 and 1/2 has hypotenuse sqrt(5)/2), then assembles it into the chord you need.
- Draw a circle with center O and a horizontal radius to a point A on the right; mark the topmost point B.
- Bisect segment OA to get its midpoint M (so |OM| = 1/2 the radius).
- Draw segment MB; its length is sqrt(5)/2 of the radius — a right triangle OMB with legs 1 and 1/2.
- Swing an arc centered at M through B down to the radius OA, landing at a point N; the length |AN| is exactly the side of the inscribed regular pentagon.
- Set the compass to |AN| and walk five steps around the circle starting from B.
Gauss, 17, and Fermat primes
For two thousand years the list of constructible regular polygons seemed stuck at 3, 4, 5, 15, and their doublings. The Greeks tried hard for the heptagon (7) and the 9-gon and never found one — for good reason, as we will see. Then in 1796, a 19-year-old Carl Friedrich Gauss proved that the regular 17-gon is constructible, and he was so proud he asked for one to be carved on his tombstone. He did not draw it; he proved it by reducing the chord to a tower of square roots.
The astonishing fact Gauss found is an exact formula. The cosine of the 17-gon's central slice can be written using nothing but integers, division, and nested square roots — and a square root is precisely the operation a compass performs. Concretely, 16 cos(2 pi / 17) equals the expression below. Because every symbol in it is a constructible number built from sqrt, the angle, and hence the polygon, is constructible.
16 cos(2 pi / 17) =
-1 + sqrt(17)
+ sqrt(34 - 2 sqrt(17))
+ 2 sqrt( 17 + 3 sqrt(17)
- sqrt(34 - 2 sqrt(17))
- 2 sqrt(34 + 2 sqrt(17)) )Gauss did far more than one example. He proved a regular n-gon is constructible exactly when n is a power of 2 times a product of distinct Fermat primes — primes of the form 2^(2^k) + 1. The known ones are 3, 5, 17, 257, and 65537. That is why 3, 5, 15 (= 3 times 5), and 17 work, why 257- and 65537-gons are (in principle) constructible, and why 7, 9 (= 3 times 3, a repeated factor), 11, and 13 do not. The full proof that these are the only ones uses Galois theory and is beyond this rung, but the criterion itself you can now use.
Using the criterion (and its honest edges)
To test any n, strip out the factors of 2, then check whether what remains is a product of distinct Fermat primes. Take n = 20 = 4 times 5: the odd part is 5, a Fermat prime, alone — constructible. Take n = 9: the odd part is 9 = 3 times 3, the same Fermat prime twice — not allowed, so the regular 9-gon is not constructible. Take n = 7: 7 is prime but not a Fermat prime, so no heptagon. This is the same wall that blocks trisecting a general angle, which is exactly the obstacle that kills the 9-gon (a 40-degree angle is one third of 120).
Notice the shape of the whole story: a purely geometric question — which polygons fit a circle — turned out to be answered by numbers, specifically by which lengths live inside a tower of square roots. That bridge from drawing to arithmetic is the heart of the next guide, where you will meet the constructible numbers head-on and see why a square root is all the compass can ever give you.