Can You Square a Circle?

The ancient Greeks played this game with two tools, and so do we:

• a straightedge: it draws a straight line through two points you already have, as long as you like, but it has no marks, so it can’t measure;
• a compass (the rope and peg): it draws a circle with its center at one point you have, through another point you have.

New points are made only where these lines and circles cross. That’s all. No measuring, no guessing “about here”, no sliding the plank until something looks right.

These two tools are enough for a surprising amount: copying a length or an angle, bisecting a line or an angle, a perpendicular, a line through a point parallel to another. These are the steps of Grade 10’s constructions, where a quadrilateral becomes a triangle of the same area with one parallel line.

Start with an easier shape: a rectangle, 4 meters by 9. Its area is 36 m², so a square with sides of 6 m has the same area. But what about 3 m by 2 m? The square needs a side of √6, which is 2.449489… and never ends. The unmarked plank can’t measure that. Can the compass make it anyway?

It can, in three moves. Lay the two sides end to end along a line: BE is one side, a, and EF the other, b. Draw a semicircle on BF. At E, draw a line perpendicular to BF, up to the semicircle at H. Then EH is the side of the square.

Why it works. The semicircle’s center G is halfway along BF, so its radius is $GH = \frac{a + b}{2}$, and G is $GE = \frac{a - b}{2}$ from E (or $\frac{b - a}{2}$, whichever is positive). The triangle GEH has a right angle at E, so by Pythagoras’ theorem

$$EH^2 = GH^2 - GE^2 = \left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2 = \frac{4ab}{4} = ab.$$

The square on EH has area ab, the rectangle’s area, exactly, whether or not its side is a whole number. The length EH is called the mean proportional (or geometric mean) of a and b: a:EH = EH:b.

This is Proposition 14 of Book II of Euclid’s Elements, written in Greece around 300 BCE. India’s Shulba Sutras, the rope manuals for building fire altars, have their own exact way of turning a rectangle into a square, centuries older on most datings.

Now chain it with Grade 10. Any straight-sided shape becomes a triangle of the same area (a parallel line removes one corner at a time); a triangle becomes a rectangle on half its base; and the rectangle becomes a square. So every straight-sided shape can be squared exactly, with only a straightedge and compass. The Greeks did exactly this; finding an area meant making a square, and the word for it was quadrature.

Back to the courtyard. The circle has radius 1 m, so its area is $\pi \times 1^2 = \pi$ square meters, and the square must have side $\sqrt{\pi} = 1.7724538\ldots$ m. That is the number the widget at the top was hunting.

Making $\sqrt{\pi}$ and making $\pi$ are really the same job. If you could draw a line $\pi$ meters long, the semicircle trick on a rectangle $\pi$ by 1 would give $\sqrt{\pi}$. And if you could draw $\sqrt{\pi}$, a square on it would have area $\pi$.

So squaring the circle comes down to one question about lengths: starting from a length of 1, can a straightedge and compass make a length of exactly $\pi$?

Start with a line 1 unit long. Here is what the tools can do to lengths:

• add and subtract: lay two lengths end to end, or one back along the other;
• multiply and divide: with a pair of similar triangles (draw lengths 1 and a along one arm of an angle and b along the other; a parallel line cuts off ab, or b/a);
• take a square root: the semicircle trick on a rectangle a by 1 gives $\sqrt{a}$.

Starting from 1, these make every whole number, every fraction, $\sqrt{2}$, $\sqrt{2} + \sqrt{3}$, $\sqrt{5 + \sqrt{7}}/3$, roots of roots, as deep as you like. Lengths made this way are called constructible numbers.

Every new point in a construction is where two of your lines and circles cross. Put the drawing on graph paper, with coordinates, and each kind of crossing is an equation to solve.

• Line meets line. Each line is an equation like y = mx + c. Solving two of them together needs only adding, subtracting, multiplying and dividing the numbers you already have.
• Line meets circle. A circle with center (p, q) and radius r is $(x - p)^2 + (y - q)^2 = r^2$. Put the line’s y = mx + c into it and you get a quadratic equation in x. The quadratic formula solves it with one square root.
• Circle meets circle. Write out both circles’ equations and subtract one from the other: the $x^2$ and $y^2$ cancel, leaving a straight line’s equation. So two circles meet where one of them meets that line: the case before.

So however long a construction runs, every length in it is built from 1 by +, −, ×, ÷ and square roots, nothing else: exactly the constructible numbers. The widget’s five moves are not a simplification; they are the whole power of the tools.

Now comes the key step. A number built from square roots can be unwound into an equation with whole numbers. Take $x = \sqrt{2 + \sqrt{3}}$. Square it: $x^2 = 2 + \sqrt{3}$. Move the 2 and square again: $(x^2 - 2)^2 = 3$, that is

$$x^4 - 4x^2 + 1 = 0.$$

Each squaring removes a root. When several roots sit side by side, as in the next puzzle, it takes more steps, and proving that it always works takes more care than we give here, but it does: any constructible number, however deep its roots, is a solution of some equation like this, a polynomial equation with whole-number coefficients. A number that solves such an equation is called an algebraic number. Every constructible number is algebraic.

So the courtyard question becomes: is π a solution of some polynomial equation with whole-number coefficients? If it isn’t, it isn’t constructible, and the circle can’t be squared.

Being irrational is not enough. Johann Lambert proved in the 1760s that π is not a fraction (as √2 isn’t), but √2 is irrational and the plank and rope make it easily, as the diagonal of a square. What was needed was a much stronger fact.

In 1882 Ferdinand von Lindemann proved it: π solves no polynomial equation with whole-number coefficients at all. A number like that is called transcendental (it “goes beyond” algebra). His proof built on Charles Hermite’s of 1873, that the number e is transcendental, and on the link between e and π, $e^{i\pi} = -1$. It uses calculus and complex numbers and fills a few pages even in today’s tidied-up versions, so we won’t give it here; but everything else in the chain you have now seen for yourself:

1. Every length the tools can make is built from 1 by +, −, ×, ÷ and $\sqrt{\ }$.
2. Every such number solves a polynomial equation with whole-number coefficients.
3. π doesn’t (Lindemann).
4. So no construction makes a length π, or √π, and no straightedge-and-compass construction squares the circle.

Before and after the exact problem was settled, mathematicians made a sport of short constructions that come astonishingly close. The most famous are by the Indian mathematician Srinivasa Ramanujan, in his 1914 paper “Modular equations and approximations to π”. His second starts from a circle with diameter AB and makes a length OS with a ruler and compass alone.

That mean proportional is the semicircle trick again. Our script followed his steps with a radius of 1 and found the length it gives makes

$$\pi \approx \sqrt[4]{9^2 + \frac{19^2}{22}} = 3.14159265258\ldots$$

against π = 3.14159265358…: eight decimals right. Ramanujan wrote that on a circle 8,000 miles across, about the size of the Earth, the error would be “less than a twelfth of an inch”. He printed his value as 3.14159265262…, and with those decimals the error is 0.082 inch, just under a twelfth. The true value of his formula ends …258, which makes it 0.085 inch, a hair over. Either way, about 2 millimeters on a circle the size of a planet.

The Greeks left three famous construction problems. The same reasoning settled the other two, even before π, in 1837, when the French mathematician Pierre Wantzel proved both impossible.

Doubling the cube. Make a cube with twice the volume of a given one. Its side must be ∛2 = 1.2599… times as long, a solution of $x^3 = 2$. That is a polynomial equation, so ∛2 is algebraic; the trouble is the 3. Each square root you add can at most double the size of the simplest equation a number solves: $\sqrt{2}$ needs $x^2 = 2$, $\sqrt{2 + \sqrt{3}}$ needed $x^4$. So constructible numbers have simplest equations of degree 1, 2, 4, 8, 16…, always a power of 2. The simplest equation for ∛2 is $x^3 - 2 = 0$ itself: if it broke into smaller ones, one of them would have degree 1 and give a fraction that solves it, and none does, and 3 is not a power of 2. Wantzel’s achievement was to make “the size of the simplest equation” precise and prove that it really only doubles.

Trisecting an angle. Cut any angle into three equal parts. Some angles can be (90∘° gives 30∘°, which you can construct), but not 60∘°. The angle 20∘° comes with the length $\cos 20^\circ$, and the triple-angle formula $\cos 3A = 4\cos^3 A - 3\cos A$ turns $\cos 60^\circ = \frac{1}{2}$ into

$$8x^3 - 6x - 1 = 0, \quad x = \cos 20^\circ = 0.9396\ldots$$

Again degree 3, which can’t be broken down (a piece of degree 1 would give a fraction that solves it, and none of $\pm 1, \pm\frac{1}{2}, \pm\frac{1}{4}, \pm\frac{1}{8}$, the only fractions that could solve it, does), so 20∘° can’t be constructed and 60∘° can’t be trisected.

Twists: where “impossible” breaks

Change the tools. Archimedes trisected any angle with a ruler carrying two marks, sliding it until the marks sat on a line and a circle. Paper folding can trisect angles and double cubes too: a fold can solve cubic equations, which a compass can’t. But no extra tool of that kind squares the circle, because π isn’t the solution of any polynomial equation with whole-number coefficients, cubic or otherwise.

A curved shape that can be squared. In the 400s BCE, Hippocrates of Chios found a crescent made of two circular arcs, a lune, whose area equals a triangle’s exactly.

Take a quarter circle with center O and radius OA = OB = 1, and draw a semicircle outwards on the chord AB. By Pythagoras, $AB = \sqrt{2}$, so the semicircle has radius $\frac{\sqrt{2}}{2}$ and area $\frac{1}{2}\pi \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{\pi}{4}$: exactly the quarter circle’s area. Both contain the same sliver between the chord AB and the quarter circle’s arc. Take that sliver away from each, and what is left is equal: the lune from the semicircle, the triangle OAB from the quarter circle. The π cancels, and a triangle can be squared.

Hippocrates must have hoped the circle would fall next. It never did, but the lune shows the trouble is not with curves; it is with π.

A polygon nobody expected. A regular hexagon is easy and a regular pentagon was known to Euclid. In 1796 Carl Friedrich Gauss showed that a regular 17-sided polygon can be constructed, because

$$\cos\frac{2\pi}{17} = \frac{-1 + \sqrt{17} + \sqrt{34 - 2\sqrt{17}} + 2\sqrt{17 + 3\sqrt{17} - \sqrt{34 - 2\sqrt{17}} - 2\sqrt{34 + 2\sqrt{17}}}}{16},$$

nothing but square roots. Yet a regular 7-sided polygon can’t be constructed. Gauss found which ones can, and Wantzel proved in 1837 that no others can: those whose number of sides is a power of 2 times different primes of a special kind, the Fermat primes (3, 5, 17, 257 and 65,537 are the ones known).

• Can You Prove √2 Is Not a Fraction? Irrational, the first step on the way to transcendental, and proof by contradiction.
• How Did Madhava Find π Without a Calculator? An endless sum for π found in Kerala two centuries before Europe.
• Are Some Infinities Bigger Than Others? Georg Cantor showed in 1874 that the algebraic numbers can be listed and the real numbers can’t, so almost every number is transcendental.
• Why Are There Only Five Perfect Dice? Another proof that only so many of something can exist.
• Is the Golden Ratio Really Everywhere? A constructible number with a famous construction, and the claims made about it.
• How Many Ways Can You Prove Pythagoras’ Theorem? (Grades 6–8) The theorem behind the semicircle trick and the lune.
• Grade 10 Math: Construction Turning shapes into triangles and parallelograms of the same area, step by step.
• Computing, Going Further: How Can Throwing Rice Find π? π from random throws instead of a ruler and compass. (Coming soon.) And How Does a Calculator Find a Square Root? The semicircle trick’s job, done by arithmetic. (Coming soon.)