Constructing Segments and Angles with a Ruler and Compass

What does it mean to construct a figure in geometry?

A geometric construction is a drawing made with only two tools, an unmarked ruler and a compass, in which every new point has to be a crossing of lines and circles that are already drawn. The ruler (called a straightedge in geometry) draws a straight line through two points; it has no numbers, so it cannot measure. The compass draws a circle, or just an arc of one, around a point, with a radius equal to a distance you already have. Nothing else is allowed: no protractor, no marks on the ruler, no placing a point by eye.

Why work so strictly? Because a construction is exact and provable. A figure drawn with a protractor is only as good as the width of the pencil line, while a construction comes with a reason (usually that two triangles have three equal sides) which shows that the result is exactly right, however carefully or carelessly you draw. This is how Euclid built geometry in the Elements around 300 BC, and why a bisected angle or a perpendicular bisector is still the first thing a geometry course teaches you to draw.

How do you copy, add and subtract line segments?

To copy a segment AB onto a ray that starts at C, open the compass until its point is on A and its pencil on B, then put the point on C and swing an arc across the ray: the crossing D gives CD=AB. The ruler is used only to draw the ray. The compass does the measuring without any numbers, because it keeps its opening, so every point of the arc is exactly the length AB from C.

1. Draw a ray from C (or use the one you have).
2. Open the compass from A to B.
3. Keep that opening, put the point on C and draw an arc that crosses the ray at D.

With this one move you can do arithmetic on lengths:

• Add two segments: copy the first onto a ray from C, ending at D; then put the compass point on D and copy the second one further along the same ray, ending at E. Then CE is the sum.
• Subtract: copy the longer segment onto a ray, then copy the shorter one from its end back towards C. What is left is the difference.
• Multiply by a whole number: copy the same segment end to end, 2, 3, 4 times.

Lengths here are copied, never read: you may not know in centimetres how long AB is, and you do not need to. If you choose one segment as the unit, the same moves give lengths such as 2, 3 or 3/2 times it.

How do you find the midpoint and perpendicular bisector of a segment?

Draw two circles of the same radius around the ends A and B of the segment; the line through their two crossings P and Q is the perpendicular bisector of AB, and it meets AB at its midpoint M. The perpendicular bisector is the line that cuts a segment in half at a right angle; it is also the set of all points that are equally far from A and from B.

1. Open the compass to more than half of AB (the length AB itself is a good choice).
2. Put the point on A and draw an arc above the segment and another below it.
3. Keep the same opening, put the point on B and draw two arcs that cross the first two, at P above and Q below.
4. Draw the line PQ with the ruler. It crosses AB at the midpoint M and is perpendicular to it.

Why it works. Both circles have the same radius, so PA=PB=QA=QB: the quadrilateral APBQ has four equal sides, which makes it a rhombus. The diagonals of a rhombus cut each other in half and meet at a right angle (see the properties of quadrilaterals), and those diagonals are exactly AB and PQ. Another way to see it: the triangles APQ and BPQ have three equal sides, so they are congruent, and the angles at P are equal.

This construction does a lot of work. Finding the midpoint of a side is the first step to the median of a triangle, and the perpendicular bisectors of two sides of a triangle meet at the center of its circumscribed circle (see the centers of a triangle).

How do you construct 60°, bisect an angle and copy an angle?

The three basic angle constructions all use circles of one single radius, and each is proved by two triangles with three equal sides (SSS).

An equilateral triangle gives 60^\circ. This is Euclid's very first proposition. Draw the segment AB, then the circle around A through B and the circle around B through A. They cross at C, and CA=AB=CB: three equal sides. The three angles of an equilateral triangle are equal and add up to 180^\circ, so each is 60^\circ (see the angles of a triangle).

To bisect an angle POQ with vertex O:

1. Put the compass point on O and draw an arc that crosses both arms, at P and Q.
2. Put the point on P and draw an arc inside the angle.
3. Keep the same opening, put the point on Q and draw an arc that crosses the previous one at R.
4. Draw the ray OR. It cuts the angle into two equal angles.

Why: the triangles OPR and OQR have OP=OQ (the first arc), PR=QR (the same opening) and the side OR in common. Three equal sides make them congruent, so the angles POR and QOR are equal. If the second and third arcs also use the radius OP, then O, P, R, Q form a rhombus, and a diagonal of a rhombus bisects its angle.

To copy an angle POQ onto a ray ST:

1. With the point on O, draw an arc that crosses the arms at P and Q. Draw the same arc (the same opening) around S, crossing the ray ST at U.
2. Open the compass to the distance PQ.
3. With the point on U, draw an arc that crosses the first arc at V.
4. Draw the ray SV. The angle USV equals POQ.

Why: the triangles OPQ and SUV have OP=SU, OQ=SV and PQ=UV, all copied with the compass, so they are congruent by SSS and the angles at O and S are equal. Bisecting, copying and adding angles are the only moves you need to build every other angle (see how angles are named and added).

How do you draw a perpendicular line and a parallel line?

A perpendicular is the perpendicular bisector of two points that you mark on the line with a circle, and a parallel is the opposite side of a rhombus, so both come from the constructions above.

A perpendicular at a point C on a line.

1. Put the compass point on C and draw a circle that crosses the line at two points, A on one side and X on the other. They are the same distance from C.
2. Construct the perpendicular bisector of AX (two circles of the same radius around A and X).
3. That bisector goes through the midpoint of AX, which is C, and is perpendicular to the line. To make a right angle of 90^\circ at the end of a ray, extend the ray backwards and take C to be the end.

A perpendicular from a point C that is not on the line.

1. Draw a circle around C that cuts the line at two points, A and X.
2. Construct the perpendicular bisector of AX. Because CA=CX, the point C is on it.
3. The bisector is the perpendicular from C; it meets the line at the foot F, and CF is the shortest way from C to the line.

A parallel to a line AB through a point C that is not on it.

1. Draw the circle around A through C. It crosses the line at X.
2. With the same opening, draw a circle around X and another around C. They cross at A and at a second point Q.
3. Draw the line CQ. The four points A, X, Q, C are the corners of a rhombus (four sides equal to the one opening), and opposite sides of a rhombus are parallel, so CQ\parallel AB.

A second way to make a parallel is to copy an angle: draw any line through C that crosses AB, copy the angle it makes with AB so that it becomes the corresponding angle at C, and use the fact that equal corresponding angles mean parallel lines (see lines cut by a transversal).

Can you watch every circle and line of each construction?

Pick a construction, step through it one circle or line at a time, and measure the finished drawing. The drawing obeys the rule of the three moves: a new point can only appear where two things you have already drawn cross, and the program refuses any other step. Press another drawing to run the same construction on a new segment or angle: the measurements stay exact because the proof does not depend on where the points are.

Which angles can be built with a ruler and compass?

Starting from 60^\circ and 90^\circ and using only halving, adding and subtracting, you can build 30^\circ, 45^\circ, 15^\circ, 75^\circ, 105^\circ, 120^\circ, 135^\circ, 150^\circ and many more, but not 20^\circ or 1^\circ. Each of the rows below is a few moves of the constructions above.

There is one more building block, found by the Greeks: the regular pentagon. Its central angle is 72^\circ, because \cos72^\circ=(√5-1)/(4) and a square root is something a compass can make (a circle of radius (√5)/(2) around the midpoint of a radius, and two more bisections). From 72^\circ you get 36^\circ, 18^\circ and, subtracting 60^\circ, 12^\circ. And then 15^\circ-12^\circ=3^\circ: an angle of one degree is out of reach, but one of three degrees is not.

The whole degrees that can be built are exactly the multiples of 3. Every multiple of 3 is reached by adding and subtracting copies of 3^\circ. For the others, suppose n^\circ could be built with n not a multiple of 3. Then n and 3 have no common factor, so there are whole numbers a and b with an+3b=1, and adding and subtracting copies of n^\circ and 3^\circ would build 1^\circ. But 1^\circ cannot be built, as the section on impossible constructions explains.

Type any whole number of degrees below and you get the cheapest recipe and the finished drawing, or the reason why none exists.

How do you divide a segment into equal parts?

To divide AB into n equal parts, step n equal lengths off a ray from A, join the last mark to B, and draw parallels to that line through the other marks; they cut AB into n equal parts. The reason is the intercept theorem: parallel lines that cut equal pieces from one transversal cut equal pieces from every other. Here is the construction for n=3.

1. Draw a ray from A at any angle to AB.
2. Choose any compass opening and step off three equal lengths along the ray, from A: the marks D_1, D_2, D_3.
3. Draw the line D_3B.
4. Draw the parallels to D_3B through D_2 and D_1 (see the rhombus method above). They cross AB at K_2 and K_1.
5. AK_1=K_1K_2=K_2B.

The same moves divide a segment in a given ratio: to cut AB in the ratio 2:3, divide it into 5 equal parts and take the second mark. Dividing a segment into two equal parts is the perpendicular bisector, which is quicker than this method. And because you can build a/n of any length, the lengths you can make from one unit segment include every fraction.

The drawing below plays the whole construction for 2 to 10 parts; the table at the end measures every mark against k/n of the way from A to B.

What can never be built with a ruler and compass?

A general angle cannot be divided into three equal parts, a cube cannot be doubled and a circle cannot be squared with a ruler and compass; each was proved impossible in the nineteenth century. The three problems were open for more than two thousand years. The answer is not that nobody was clever enough, but that the tools reach only certain numbers.

What numbers can be built? Take a unit length. The three moves can only produce the numbers you get from 1 by adding, subtracting, multiplying, dividing and taking square roots, because crossing a line with a line leads to a linear equation and crossing a line or circle with a circle leads to a quadratic one. Each square root at most doubles the degree of a number over the rationals (the smallest degree of a whole-number polynomial that has it as a root). So a number that can be built has a degree that is a power of 2: 1, 2, 4, 8, … This is Wantzel's theorem (1837).

Why 20^\circ cannot be built. The triple-angle identity says \cos3φ=4\cos^3φ-3\cosφ. Put φ=20^\circ and \cos60^\circ=\frac12, and x=\cos20^\circ satisfies 8x^3-6x-1=0. This cubic has no rational root (the only candidates are ±1, ±\frac12, ±\frac14, ±\frac18, and none works), so it cannot be split, and \cos20^\circ has degree 3. Three is not a power of 2, so 20^\circ cannot be built, and an angle of 60^\circ cannot be trisected. If 1^\circ could be built, so could 20^\circ (twenty copies), which is why no whole degree outside the multiples of 3 can.

The three classical problems.

• Trisecting an angle: possible for special angles (90^\circ gives 30^\circ), impossible in general (60^\circ gives 20^\circ).
• Doubling the cube: a cube of twice the volume needs an edge of 2^(1/3) times the old one, a number of degree 3.
• Squaring the circle: a square of the same area as a circle of radius 1 needs a side √π. Lindemann proved in 1882 that π is transcendental, which means it is not a root of any whole-number polynomial, so √π is out of reach too (see π and the transcendental numbers).

Other tools can do it. Archimedes trisected any angle with a ruler that has two marks one unit apart (a marked ruler, called neusis in Greek). Put the ruler through B so that one mark is on the circle at C and the other is on the line AO extended, at D; the angle at D is exactly one third of the angle at O. Paper folding (origami) and some other curves can also trisect an angle. The impossibility is about the ruler and compass alone.

The drawing below shows Archimedes' trisection for any angle, and tells you whether a ruler and compass alone could have done it.

Which regular polygons can be built with a ruler and compass?

A regular polygon with n sides can be built exactly when n is a power of 2 times different Fermat primes (3, 5, 17, 257, 65537); the first few are 3, 4, 5, 6, 8, 10, 12, 15, 16, 17. This is the Gauss–Wantzel theorem. Carl Friedrich Gauss found at 18 that the regular 17-gon can be built, published it in 1801, and asked for one on his gravestone; Pierre Wantzel proved in 1837 that no other regular polygons can.

Why those numbers? Building the n-gon means building the angle (360^\circ)/(n), that is, the number \cos(2π)/(n), whose degree over the rationals is (φ(n))/(2) for n\ge3. Here φ(n) is Euler's function, the count of whole numbers from 1 to n that share no factor with n. That degree is a power of 2 exactly when φ(n) is, and φ(n) is a power of 2 exactly when n is a power of 2 times different primes of the form 2^2^k+1 (Fermat primes). The only five known are 3, 5, 17, 257 and 65537.

• Pentagon (5): \cos72^\circ=(√5-1)/(4), one square root. Built by the Greeks.
• Heptagon (7) and nonagon (9): φ(7)=6 and φ(9)=6 are not powers of 2, so they cannot be built. The 9-gon would need the trisection of 120^\circ, which is impossible.
• 17-gon: φ(17)=16, and \cos(2π)/(17) needs three nested layers of square roots.
• 257-gon and 65537-gon: built on paper by Richelot (1832) and by Hermes (1894, after about ten years of work).

Type a number of sides to see why a polygon can or cannot be built.

How do you do a construction in code?

In a program a construction is the crossing of two circles (or of a line and a circle) computed from the distance between the centers. For circles of radii r_1 and r_2 whose centers are d apart, the crossings lie at the distance a=(r_1^2-r_2^2+d^2)/(2d) from the first center along the line between the centers, and at the height h=√(r_1^2-a^2) on both sides of it. If d>r_1+r_2 or d<|r_1-r_2| the circles miss each other and there is no point.

Two details matter in floating-point arithmetic (see why floating-point sums are not exact). First, when the circles only touch, r_1^2-a^2 can come out as -10^-17, so take max(h², 0) before the square root. Second, never compare lengths with ==: two distances that are equal on paper differ in the last digit, so compare with a small tolerance, and measure angles with atan2.

The program builds the lens of the perpendicular bisector for A=(0,0) and B=(4,0), checks that the upper crossing makes an equilateral triangle, and tests which whole degrees and which polygons are buildable with Euler's φ function.

import math


def circles_meet(c1, r1, c2, r2):
    """The two points where two circles cross, the one on the left of c1 -> c2 first; None if they miss."""
    dx, dy = c2[0] - c1[0], c2[1] - c1[1]
    d = math.hypot(dx, dy)
    if d == 0 or d > r1 + r2 or d < abs(r1 - r2):
        return None
    a = (r1 * r1 - r2 * r2 + d * d) / (2 * d)
    h = math.sqrt(max(r1 * r1 - a * a, 0.0))      # rounding can leave -1e-17 when the circles only touch
    mx, my = c1[0] + a * dx / d, c1[1] + a * dy / d
    return (mx - h * dy / d, my + h * dx / d), (mx + h * dy / d, my - h * dx / d)


def totient(n):
    return sum(1 for k in range(1, n + 1) if math.gcd(k, n) == 1)


def is_power_of_two(n):
    return n >= 1 and n & (n - 1) == 0


def angle_buildable(deg):
    """A whole number of degrees can be built when cos(deg) has a degree that is a power of two."""
    m = 360 // math.gcd(deg, 360)
    return is_power_of_two(1 if m <= 2 else totient(m) // 2)


def polygon_buildable(n):
    return n >= 3 and is_power_of_two(totient(n))


A, B = (0.0, 0.0), (4.0, 0.0)
P, Q = circles_meet(A, 4, B, 4)                       # the lens of the perpendicular bisector
print([round(v, 4) for v in P], [round(v, 4) for v in Q])    # [2.0, 3.4641] [2.0, -3.4641]
print(round(math.dist(P, A), 4), round(math.dist(P, B), 4), round(math.dist(A, B), 4))    # 4.0 4.0 4.0
print(round(math.degrees(math.atan2(P[1] - A[1], P[0] - A[0])), 4))    # 60.0: the angle PAB
print([d for d in range(1, 31) if angle_buildable(d)])     # [3, 6, 9, 12, 15, 18, 21, 24, 27, 30]
print([n for n in range(3, 41) if polygon_buildable(n)])   # [3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40]
print(circles_meet((0, 0), 1, (5, 0), 1))                  # None: too far apart

const circlesMeet = ([x1, y1], r1, [x2, y2], r2) => {
  const dx = x2 - x1, dy = y2 - y1
  const d = Math.hypot(dx, dy)
  if (d === 0 || d > r1 + r2 || d < Math.abs(r1 - r2)) return null
  const a = (r1 * r1 - r2 * r2 + d * d) / (2 * d)
  const h = Math.sqrt(Math.max(r1 * r1 - a * a, 0))     // rounding can leave -1e-17 when the circles only touch
  const mx = x1 + (a * dx) / d, my = y1 + (a * dy) / d
  return [[mx - (h * dy) / d, my + (h * dx) / d], [mx + (h * dy) / d, my - (h * dx) / d]]
}
const gcd = (a, b) => (b === 0 ? a : gcd(b, a % b))
const totient = (n) => { let t = 0; for (let k = 1; k <= n; k++) if (gcd(k, n) === 1) t++; return t }
const isPowerOfTwo = (n) => n >= 1 && (n & (n - 1)) === 0
const angleBuildable = (deg) => { const m = 360 / gcd(deg, 360); return isPowerOfTwo(m <= 2 ? 1 : totient(m) / 2) }
const polygonBuildable = (n) => n >= 3 && isPowerOfTwo(totient(n))

const [P, Q] = circlesMeet([0, 0], 4, [4, 0], 4)
;[P.map((v) => +v.toFixed(4)), Q.map((v) => +v.toFixed(4))]                 // [[2, 3.4641], [2, -3.4641]]
;+(Math.atan2(P[1], P[0]) * 180 / Math.PI).toFixed(4)                       // 60
Array.from({ length: 30 }, (_, i) => i + 1).filter(angleBuildable)           // [3, 6, 9, 12, 15, 18, 21, 24, 27, 30]
Array.from({ length: 38 }, (_, i) => i + 3).filter(polygonBuildable)         // [3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40]
circlesMeet([0, 0], 1, [5, 0], 1)                                             // null: too far apart

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

double px, py, qx, qy;   // the two crossings, set by circlesMeet

int circlesMeet(double x1, double y1, double r1, double x2, double y2, double r2) {
  double dx = x2 - x1;
  double dy = y2 - y1;
  double d = sqrt(dx * dx + dy * dy);
  if (d == 0 || d > r1 + r2 || d < fabs(r1 - r2)) return 0;
  double a = (r1 * r1 - r2 * r2 + d * d) / (2 * d);
  double h2 = r1 * r1 - a * a;
  double h = h2 > 0 ? sqrt(h2) : 0;       // rounding can leave -1e-17 when the circles only touch
  double mx = x1 + a * dx / d;
  double my = y1 + a * dy / d;
  px = mx - h * dy / d;
  py = my + h * dx / d;
  qx = mx + h * dy / d;
  qy = my - h * dx / d;
  return 1;
}

int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }

int totient(int n) {
  int result = n;
  int m = n;
  for (int p = 2; p * p <= m; p++) {
    if (m % p == 0) {
      while (m % p == 0) m /= p;
      result -= result / p;
    }
  }
  if (m > 1) result -= result / m;
  return result;
}

int isPowerOfTwo(int n) { return n >= 1 && (n & (n - 1)) == 0; }

int angleBuildable(int deg) {
  int m = 360 / gcd(deg, 360);
  return isPowerOfTwo(m <= 2 ? 1 : totient(m) / 2);
}

int main() {
  circlesMeet(0, 0, 4, 4, 0, 4);
  printf("%.4f %.4f %.4f %.4f\n", px, py, qx, qy);
  printf("%.1f\n", atan2(py, px) * 180.0 / 3.14159265358979323846);
  for (int d = 1; d <= 30; d++) if (angleBuildable(d)) printf("%d ", d);
  printf("\n");
  for (int n = 3; n <= 40; n++) if (isPowerOfTwo(totient(n))) printf("%d ", n);
  printf("\n");
  printf("%d\n", circlesMeet(0, 0, 1, 5, 0, 1));
  return 0;
}

/* output:
2.0000 3.4641 2.0000 -3.4641
60.0
3 6 9 12 15 18 21 24 27 30
3 4 5 6 8 10 12 15 16 17 20 24 30 32 34 40
0
*/

Where do ruler-and-compass constructions come from?

Construction is the oldest way of doing geometry, and the question of what it can and cannot do took two thousand years to answer.

• About 300 BC. Euclid's Elements begins with the equilateral triangle and builds on the ruler and compass alone: bisecting an angle and a segment, perpendiculars, parallels, copying an angle, and the regular pentagon (Books I and IV). Book VI divides a segment into equal parts by the intercept theorem.
• About 250 BC. Archimedes trisects an angle with a marked ruler. Greek geometers tried for centuries to trisect an angle, double a cube and square a circle with the ruler and compass alone.
• 1672 and 1797. Georg Mohr and, independently, Lorenzo Mascheroni show that every point that can be built with a ruler and compass can be built with the compass alone.
• 1796. Gauss, aged 18, builds the regular 17-gon, the first new regular polygon since antiquity, and publishes the general rule in 1801.
• 1832. Friedrich Richelot gives the construction of the 257-gon.
• 1837. Pierre Wantzel proves that the doubling of the cube and the trisection of a general angle are impossible, and that Gauss's list of polygons is complete.
• 1882. Ferdinand von Lindemann proves that π is transcendental, which ends the squaring of the circle.
• 1894. Johann Gustav Hermes finishes the construction of the 65537-gon, after about ten years.

What are the common mistakes in constructions?

The most common mistakes in constructions are the eight below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: constructions at a glance

• Rules: an unmarked ruler and a compass; three moves (line through two points, circle around a point, point at a crossing); the construction is exact and has a proof.
• Copy a segment: set the compass to its length and swing an arc from the new point; add and subtract by copying end to end or back.
• Perpendicular bisector and midpoint: two circles of one radius around the ends, join their crossings; justified by a rhombus or by SSS.
• Angles: equilateral triangle →60^\circ; bisect with an arc and two equal arcs; copy by copying the chord; all by SSS.
• Perpendicular and parallel: the perpendicular bisector of two marked points; the opposite side of a rhombus.
• Special angles: 90^\circ, 60^\circ, 30^\circ, 45^\circ, 15^\circ, 75^\circ, 120^\circ, 135^\circ, 150^\circ from halving, adding and subtracting; with 72^\circ, every whole number of degrees that is a multiple of 3.
• Equal parts: equal steps on a ray and parallels (intercept theorem).
• Impossible: trisecting a general angle, doubling the cube, squaring the circle; the buildable numbers have degrees that are powers of 2.
• Polygons: the regular n-gon is buildable exactly when φ(n) is a power of 2, that is, n is a power of 2 times different Fermat primes.
• Code: the crossing of two circles from a=(r_1^2-r_2^2+d^2)/(2d) and h=√(r_1^2-a^2); max(h², 0) and a tolerance.

Glossary: key terms

Geometric construction

A figure drawn with only an unmarked ruler and a compass, where every new point is a crossing of lines and circles already drawn.

Straightedge

A ruler without any marks, used only to draw a straight line through two points.

Perpendicular bisector

The line that cuts a segment in half at a right angle; every point on it is equally far from the two ends.

Angle bisector

The ray from the vertex of an angle that divides it into two equal angles.

Rhombus

A quadrilateral with four equal sides; its diagonals bisect each other at right angles and bisect its angles.

Constructible number

A number that can be made as a length from a unit length with a ruler and compass; it is obtained from 1 by adding, subtracting, multiplying, dividing and taking square roots.

Degree of a number

The smallest degree of a polynomial with whole-number coefficients that has the number as a root; a constructible number has a degree that is a power of 2.

Fermat prime

A prime of the form 2 to the power 2^k, plus 1; only five are known: 3, 5, 17, 257 and 65537.

Euler's totient function

The number of whole numbers from 1 to n that have no common factor with n, written phi of n.

Neusis

A construction that uses a ruler with two marks, which can trisect any angle although a plain ruler and compass cannot.

Frequently asked questions

References

1. Euclid, translated by Thomas L. Heath (1956). Euclid's Elements (Books I, IV and VI). Dover.
2. Richard Courant, Herbert Robbins (1996). What Is Mathematics? An Elementary Approach to Ideas and Methods (chapter III, geometrical constructions). Oxford University Press.
3. Nicholas D. Kazarinoff (2003). The Ruler and the Round: Classic Problems in Geometric Constructions. Dover.
4. Robin Hartshorne (2000). Geometry: Euclid and Beyond. Springer.
5. Ian Stewart (2015). Galois Theory (chapter on geometric constructions). CRC Press.
6. Pierre-Laurent Wantzel (1837). Recherches sur les moyens de reconnaître si un problème de géométrie peut se résoudre avec la règle et le compas. Journal de Mathématiques Pures et Appliquées 2.
7. Carl Friedrich Gauss (1801). Disquisitiones Arithmeticae (section VII, the division of the circle). Gerhard Fleischer.

Related articles

• Angles: Measuring, Types, Pairs and Parallel Lines: Angles explained: degrees, radians and turns, types and pairs of angles, parallel lines and transversals, polygon angles, clock angles and bearings.
• Triangles: Types, Angles, Area, Centers and Congruence: Triangles explained: types by sides and angles, the 180° angle sum, the triangle inequality, area and Heron's formula, centers, congruence and the law of sines.
• Circles: Circumference, Area, Angles and Equations: Circles explained: pi, circumference and area, arcs and sectors, angle theorems, chords and tangents, the equation of a circle, and how lines and circles meet.
• Floating-Point Numbers: How floating-point numbers work: the IEEE 754 bits, spacing and rounding, NaN and infinity, accumulated errors, cancellation, comparisons, and which type to use.