Compass and Straightedge Constructions

Two tools and nothing else: a straightedge that draws a line through two points, and a compass that draws a circle about one point through another. Play the classic constructions step by step, test which regular polygons and which trisections are possible, or construct something yourself. The table under the drawing works out the lengths and the algebra.

Worked example — Compass and Straightedge Constructions

Compass operations	2
Straightedge operations	1
Total operations	3
Radius of both circles	1.0000
Distance MA	0.5000
Distance MB	0.5000
Angle between the lines	90.0000°
Largest deviation from exact	0

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Compass and Straightedge Constructions

A construction uses two operations: draw the line through two points already built, and draw the circle about one built point through another. Where lines and circles cross, new points are built. A length is constructible when some finite sequence of these steps produces it. Starting from a unit length, the constructible lengths are exactly those reached by the four arithmetic operations and square roots, so a number needing a cube root, or anything else, cannot be built.

What each symbol means

n: the number of sides of a regular polygon
φ(n): Euler's totient, the count of whole numbers from 1 to n that have no common factor with n other than 1
F_k: a Fermat prime, 2^(2^k) + 1. Only 3, 5, 17, 257 and 65537 are known
cos θ: the length that builds an angle θ. Constructing the angle and constructing this number are the same problem
deg: the degree of cos(360°/n) over the rationals, φ(n) ÷ 2. The n-gon can be built exactly when it is a power of 2

Gauss built the 17-gon at 18, and no one will ever build the 7-gon 🖖

Press The 17-gon and watch it built: 41 circles and 30 lines in this construction, and the seventeen vertices come out equally spaced to fifteen decimal places. Carl Friedrich Gauss found in 1796, at eighteen, that it can be done, and is said to have asked for one on his gravestone. He showed that a regular n-gon can be built whenever n is a power of 2 times a product of different Fermat primes (the known ones are 3, 5, 17, 257 and 65537), and Pierre Wantzel proved in 1837 that no other n works. Press Why not the 7-gon. Seven is prime but not a Fermat prime, and the folded table lists every n from 3 to 100: only 24 of them can be built. The 9-gon fails too, because 3 appears twice.

You can trisect 90° in five moves and 60° never 🖖

Press Trisect 90°: 30° is an angle of the equilateral triangle, and five operations split the right angle into three. Now press Trisect 60°. A third of it is 20°, and cos 20° is a root of the cubic 8c³ − 6c − 1 = 0. A compass and straightedge can only ever produce lengths reached by the four arithmetic operations and square roots, and a cubic with rational coefficients and no rational root has no root that square roots can reach. The table checks all eight rational candidates and none works. Wantzel published that proof in the same 1837 paper, closing a problem the Greeks had left open for two thousand years. On this page's angles, whole multiples of 3°, an angle can be trisected exactly when it is a multiple of 9°.

A compass that forgets its width loses nothing 🖖

Euclid's third postulate lets you draw a circle only about one point through another, a compass that, as it is now put, collapses the moment it is lifted, so it cannot carry a distance from one place to another. It seems a crippling limit, and his second proposition shows it is not: any length can be copied anyway, with a few extra circles and lines. Open Copy an angle and switch between the rigid and collapsing compass. In this recording the rigid one takes 3 circles and 1 line, and the collapsing one 11 circles and 5 lines, to do exactly the same job. Everything one compass can construct, the other can too. The forgetful one only makes each construction longer.

Problem solved in full

Where cos 72° comes from in the pentagon 5 steps
A regular pentagon is inscribed in a circle of radius 1 by the construction on this page: take the midpoint M of the radius OH₂, draw the circle about M through A, the top of the circle, and let it cross the diameter at N. Find the length ON, cos 72°, and the side of the pentagon. Press Regular pentagon to check.

Open this problem in the tool
1. M is the midpoint of a radius, so:
  
  $O M = \frac{1}{2}$
2. A is at the top of the circle, a right angle away from H₂, so Pythagoras gives MA:
  
  $M A = 1^{2} + (\frac{1}{2})^{2} = \frac{5}{2} = 1.1180$
3. N lies on the circle about M through A, so MN = MA, and N is on the far side of O:
  
  $O N = M N - O M = \frac{5}{2} - \frac{1}{2} = \frac{5 - 1}{2} = 0.6180$
4. ON is the side of the regular decagon, and half of it is the cosine of the pentagon's central angle:
  
  $cos 7 2^{\circ} = \frac{O N}{2} = \frac{5 - 1}{4} = 0.3090$
5. The pentagon's side AN is the hypotenuse of the right triangle OAN:
  
  $A N = 1^{2} + O N^{2} = 1 + 0.3820 = 1.1756 = 2 sin 3 6^{\circ}$
Answer

ON = 0.6180, cos 72° = 0.3090 = (√5 − 1) ÷ 4, and the pentagon's side is 1.1756, which is 2 sin 36°. Every step used only a circle, a line and a square root, which is the whole reason the pentagon can be built: its cosine needs √5 and nothing worse.

$cos 7 2^{\circ} = \frac{5 - 1}{4} = 0.3090, A N = 1.1756$

By the 42Math team · Reviewed against published sources. Corrections welcome.