The Spiral of Theodorus

A chain of right triangles, each with one leg of length 1, that constructs the square root of every whole number — and a 2,400-year-old puzzle about why it stops at √17.

By the PythagoreanTheorem.net teamUpdated 9 min read

The spiral of Theodorus (also called the square root spiral, Pythagorean spiral or wheel of Theodorus) is a chain of right triangles. Each has an outer leg of length 1, and its hypotenuse becomes a leg of the next triangle. The hypotenuses measure √2, √3, √4, …, √17, so the spiral constructs square roots using only right angles and unit lengths.

It is named after Theodorus of Cyrene, a Greek mathematician of the 5th century BCE, although — as explained below — the spiral itself is a modern reconstruction. This page shows how to build it, the numbers behind it, why the classic picture ends at √17, how to construct any √n with ruler and compass, and a classroom art project.

How the spiral of Theodorus is built

  1. Start with a unit right triangle. Draw a right triangle with both legs equal to 1. By the Pythagorean theorem its hypotenuse is √12 + 12 = √2.
  2. Build on the hypotenuse. At the outer end of that hypotenuse, draw a segment of length 1 at a right angle to it. Join its end to the center. The new hypotenuse is √(√2)2 + 12 = √3.
  3. Repeat. Each new triangle uses the previous hypotenuse as one leg and a new unit segment as the other, always turning the same way.

The whole construction rests on one line of algebra. If a triangle has legs √n and 1, then

(√n)2 + 12 = n + 1  ⇒  hypotenuse = √n + 1

So each step adds exactly 1 to the square of the spoke length. The spokes from the center are √1, √2, √3 and so on, and every perfect square appears as a whole number: √4 = 2, √9 = 3, √16 = 4.

√2 √3 √4 √5 √6 √7 √8 √9 √10 √11 √12 √13 √14 √15 √16 √17 1 1 Every triangle has an outer leg 1 and an inner leg √n, so its hypotenuse is √(n + 1) (√n)² + 1² = n + 1
The spiral of Theodorus with 16 triangles. Each spoke is the hypotenuse of one triangle and a leg of the next; the outer edge (indigo) is made of unit segments.

The spokes and angles

Each triangle is thinner than the one before. The triangle with legs √n and 1 has an angle at the center of

θn = arctan1 over √n

because the side opposite that angle is 1 and the side next to it is √n. Two of the triangles are old friends: the first is a 45-45-90 triangle (θ1 = 45°), and the third, with sides 1, √3 and 2, is a 30-60-90 triangle (θ3 = 30°).

TriangleInner legHypotenuseAngle at centerRunning total
11√2 ≈ 1.41445.00°45.00°
2√2√3 ≈ 1.73235.26°80.26°
3√3√4 = 230.00°110.26°
42√5 ≈ 2.23626.57°136.83°
5√5√6 ≈ 2.44924.09°160.92°
6√6√7 ≈ 2.64622.21°183.13°
7√7√8 ≈ 2.82820.70°203.84°
8√8√9 = 319.47°223.31°
93√10 ≈ 3.16218.43°241.74°
10√10√11 ≈ 3.31717.55°259.29°
11√11√12 ≈ 3.46416.78°276.07°
12√12√13 ≈ 3.60616.10°292.17°
13√13√14 ≈ 3.74215.50°307.67°
14√14√15 ≈ 3.87314.96°322.64°
15√15√16 = 414.48°337.11°
164√17 ≈ 4.12314.04°351.15°
17 (overlaps)√17√18 ≈ 4.24313.63°364.78°

Decimals are rounded to three places and angles to the nearest hundredth of a degree. If you want to simplify any of these radicals — √8 = 2√2, √12 = 2√3 — see square roots and radicals or the simplify square root calculator.

Example 1: the area of the spiral

Triangle n has legs √n and 1, so its area is 1 over 2√n. The 16 triangles of the classic spiral cover

1 over 2(√1 + √2 + ⋯ + √16) ≈ 1 over 2(44.469) ≈ 22.23

About 22.23 square units — roughly the area of a circle of radius 2.66.

Example 2: when does a spoke pass 3?

The spoke after triangle n has length √n + 1. It exceeds 3 when n + 1 > 9, that is, from triangle 9 onward.

Triangle 8 has hypotenuse exactly √9 = 3; triangle 9 is the first whose hypotenuse, √10 ≈ 3.162, is longer than 3.

Why the spiral stops at √17

Add up the angles at the center. After 16 triangles — the last one having hypotenuse √17 — the total is about 351.15°, just short of a full turn. The 17th triangle (legs √17 and 1, hypotenuse √18) needs another 13.63°, but only 8.85° remain before the starting line. It would push the total to about 364.78° and overlap the very first triangle.

√17 1st 17th starting line After 16 triangles: 351.15° Room left: 8.85° 17th triangle needs 13.63° → it overlaps the first
The 17th triangle (dashed) crosses the starting line, so the spiral can no longer be drawn without overlapping. The last hypotenuse that fits is √17.

This is a neat geometric fact. Whether it explains anything about Theodorus is a separate — and much harder — question.

Theodorus, Theaetetus and Plato: what we actually know

Theodorus of Cyrene (about 465–398 BCE, dates approximate) was a mathematician whose pupils included the young Athenian Theaetetus. None of his writings survive. Almost everything we know about his work comes from one passage in Plato's dialogue Theaetetus, written about 369 BCE. In it, the young mathematician Theaetetus recalls a lesson: Theodorus showed, one case at a time, that the sides of squares with areas 3, 5 and so on up to 17 square feet are incommensurable with a side of 1 foot — that is, √3, √5, … cannot be written as ratios of whole numbers — and there, for some reason, he stopped. (Plato does not credit him with √2, presumably because it was already known; see why √2 is irrational.) According to later tradition Plato also studied with Theodorus.

Theaetetus goes on in the dialogue to state the general rule: the side of a square whose area is a whole number but not a perfect square is always incommensurable with the unit.

Legend versus evidenceNo ancient source mentions a spiral. The idea that Theodorus drew one, and stopped at 17 because the triangles began to overlap, was proposed in 1941 by J. H. Anderhub. It is an attractive guess, but many historians doubt it: an overlap tells you nothing about irrationality, and other reconstructions explain the stopping point through the proof method instead. Scholars even disagree about how to translate the Greek — whether 17 was the last case Theodorus completed or the first one where he "got into difficulties". The honest summary is that nobody knows why he stopped.

So the name is an homage rather than an attribution. The spiral is still a superb way to see the square roots that Theodorus studied, and it ties his name to the long history of the Pythagorean theorem.

Constructing √n with ruler and compass

You can draw the spiral exactly, without measuring angles, using only a straightedge and compass. The one skill you need is erecting a perpendicular at the end of a segment.

  1. Extend the spoke. Draw the line through the center O and the outer end P of the current spoke, continuing a little beyond P.
  2. Mark two points. With the compass point on P, draw an arc that crosses the line on both sides of P, at X and Y.
  3. Find the perpendicular. Open the compass wider and draw equal arcs from X and from Y. The line through P and the crossing point of these arcs is perpendicular to the spoke.
  4. Mark the unit. Set the compass to your unit length and mark it along the perpendicular from P to get Q. Then OQ is the next spoke.

There is also a direct way to construct √n for any n in one go, using the geometric mean theorem. Lay off segments of lengths n and 1 end to end, draw a semicircle on the whole segment, and raise a perpendicular at the join.

ACBD n = 5 1 √5
With AC = n and CB = 1, the perpendicular CD has length √n · 1 = √n. Here n = 5.

Why it works: the angle at D is a right angle (it stands on a diameter), and the altitude CD splits triangle ADB into two similar triangles, so CD over AC = CB over CD, which gives CD2 = AC · CB = n.

What happens if you keep going

Nothing stops you from adding triangles past the 17th; they simply lap over the first turn and start a second winding outside it. Several patterns emerge:

  • The angles shrink slowly. arctan(1/√n) is close to 1/√n radians for large n, so the total angle after n triangles grows roughly like 2√n radians (minus a constant). A full turn takes more and more triangles: 16 for the first, then about 37 more for the second.
  • The windings settle down. Because the radius is √n and the angle is about 2√n, the radius grows by about 2π over 2 = π per turn. The distance between successive windings quickly approaches π ≈ 3.14, so far from the center the spiral looks like an Archimedean spiral (evenly spaced turns), not a logarithmic one like a nautilus shell.
  • A smooth version exists. The mathematician Philip J. Davis found a continuous curve passing through all the corner points, in the same way that the gamma function interpolates factorials.
  • Other right-triangle fractals. Repeating squares on the sides of a right triangle instead of unit legs produces a different classic picture, the Pythagoras tree.

Classroom activity: square root spiral art

Drawing the spiral is a favorite middle- and high-school project because it combines the Pythagorean theorem, careful construction, measurement error and art. It takes one or two class periods. For more ideas see Pythagorean theorem activities.

Materials: plain paper (letter or A4), a ruler, a sharp pencil, an index card or set square for right angles (or a compass), and colored pencils or markers.

  1. Choose the unit. One inch (2.54 cm) works well: the 16-triangle spiral then measures about 6.7 by 5.3 inches and fits on a sheet. Put the center about 3 inches from the left edge and 2½ inches from the top, draw the first leg pointing right, and turn counterclockwise.
  2. Draw the first triangle. Two 1-inch legs at a right angle, then the hypotenuse.
  3. Add triangles. Use the corner of the index card to make each outer leg exactly perpendicular to the previous spoke, measure 1 inch, and connect to the center. Label each spoke √2, √3, …
  4. Measure and compare. Measure the √17 spoke. It should be about 4.12 inches (10.47 cm). Compute your percent error: small errors in each right angle add up, which is a lesson in itself.
  5. Make it art. Color the triangles in a repeating pattern, keep going into a second winding, or decorate each region. Many classes display the results as a gallery.

Teacher tipAsk students to predict, before drawing, which spoke will be exactly 3 and which will first point "backwards" past the starting line. Older students can check the running angle total with the arctangent and compare it with their drawing; the table above is the answer key.

Frequently asked questions

What is the spiral of Theodorus used for?

It is mainly a teaching and art tool: it shows that √n can be constructed exactly for every whole number n, using only right angles and unit lengths. It also gives a memorable picture of how the square roots grow — fast at first, then more and more slowly.

Why does the spiral of Theodorus stop at √17?

The angles at the center of the first 16 triangles add up to about 351°. The next triangle needs about 13.6°, which would carry the spiral past its starting line and make it overlap the first triangle. Whether this is why Theodorus himself stopped at 17 is unknown.

Who was Theodorus of Cyrene?

A Greek mathematician of the 5th century BCE, born in Cyrene (in modern Libya), whose pupils included the Athenian mathematician Theaetetus. Plato's dialogue Theaetetus reports that he proved the irrationality of the square roots of 3, 5 and further non-square numbers up to 17. None of his own writings survive.

Is the spiral of Theodorus a golden spiral?

No. A golden spiral is a logarithmic spiral whose turns get farther apart as it grows. The spiral of Theodorus is closer to an Archimedean spiral: far from the center, its turns are separated by an almost constant distance of about π.

How do you construct √5 with a compass and ruler?

One way: draw a right triangle with legs 2 and 1; its hypotenuse is √5. Another: mark segments of 5 and 1 units end to end, draw a semicircle on the whole 6-unit segment, and raise a perpendicular at the join — it meets the semicircle at a height of √5.

Sources and further reading

  1. Plato, Theaetetus, 147c–148b (the lesson of Theodorus).
  2. MacTutor History of Mathematics: Theodorus of Cyrene.
  3. Wikipedia: Spiral of Theodorus.
  4. Philip J. Davis, Spirals: From Theodorus to Chaos, A K Peters, 1993.
  5. Wilbur R. Knorr, The Evolution of the Euclidean Elements, Reidel, 1975 (on Theodorus and the early theory of incommensurables).

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