Hippocrates' Lunes, Crescents That Equal a Triangle, math that surprises you lesson

Draw a right triangle, put a semicircle on each of its three sides, and shade the two crescents on the short sides that reach beyond the semicircle on the long side. The crescents are bounded by arcs of circles, so their areas would seem to involve the number $\pi$ . Yet together they have exactly the area of the triangle, and $\pi$ disappears completely. This page explains why with the area of a semicircle, the Pythagorean theorem and a careful cancellation, and says what the result does and does not have to do with squaring the circle.

Key terms

Lune of Hippocrates

Definition A lune is a crescent-shaped region bounded by two circular arcs. Start with a triangle $ABC$ whose angle at $C$ is a right angle, with legs $a = BC$ and $b = CA$ and hypotenuse $c = AB$ . On each leg draw a semicircle with that leg as its diameter, on the side of the leg away from the triangle. On the hypotenuse draw a semicircle with diameter $AB$ , on the same side of $AB$ as the triangle. Because the angle at $C$ is right, the midpoint of $AB$ is at the same distance $c/2$ from all three corners, so the arc of this larger semicircle passes through $C$ . This is the converse of Thales’ theorem, the fact that every angle inscribed in a semicircle is a right angle. The larger semicircle is then made of the triangle and two circular segments of the large circle, each cut off by one of the legs, and each segment also lies inside the small semicircle on its leg. The two lunes are the parts of the small semicircles that lie outside the larger one. Their areas $L_1$ and $L_2$ add up to the area of the triangle,

L_1 + L_2 = \frac{ab}{2}.

To see why, let $S_1$ and $S_2$ be the areas of the two segments, $T$ the area of the triangle, $U$ the total area of the two small semicircles and $V$ the area of the large one. The regions fit together so that

\begin{aligned} U &= L_1 + L_2 + S_1 + S_2, \\[0.4em] V &= T + S_1 + S_2, \end{aligned}

and $U = V$ by the Pythagorean theorem. Subtracting the same two segments from both sides leaves $L_1 + L_2 = T$ . The result needs the right angle and exactly this construction. It gives the total of the two lunes, not each lune on its own, and it says nothing about crescents in general. Hippocrates of Chios, a Greek geometer of the fifth century BC, found the area of such a lune in the case of two equal legs, and that single lune is the one usually called the lune of Hippocrates. The version for any right triangle, with two lunes of different sizes, is known as the lunes of Alhazen, after the scholar Ibn al-Haytham, who lived around the year 1000.

In plain words The two small semicircles together have exactly the area of the big one. Both totals contain the same two curved slivers, so removing the slivers from both leaves equal areas behind, the two crescents on one side and the triangle on the other.

Example Put $C = (0, 0)$ , $B = (4, 0)$ and $A = (0, 3)$ , so the legs are $4$ and $3$ and the hypotenuse is $5$ . The large circle has center $(2, 3/2)$ and radius $5/2$ , and it passes through $C$ , since $2^2 + (3/2)^2 = 25/4$ . The triangle has area $\tfrac12 \cdot 3 \cdot 4 = 6$ , so the two lunes have total area exactly $6$ square units. Separately they are not equal. Computing each segment with trigonometry and subtracting it from its semicircle gives about $2.512$ for the lune on the leg of length $3$ and about $3.488$ for the lune on the leg of length $4$ . Neither is half the triangle, and only their total comes out simple.

Pythagorean theorem

Definition In a triangle with a right angle, let $a$ and $b$ be the lengths of the two sides that form the right angle, called the legs, and let $c$ be the length of the side opposite it, the hypotenuse. Then

a^2 + b^2 = c^2 .

Conversely, a triangle whose sides satisfy this equation has a right angle opposite the side of length $c$ . Multiplying the equation by a positive number $k$ keeps it true, so the theorem also holds for any three shapes of the same form built on the three sides and scaled to them, because their areas are $ka^2$ , $kb^2$ and $kc^2$ for one number $k$ . Squares on the sides have $k = 1$ , semicircles on the sides have $k = \pi/8$ and equilateral triangles on the sides have $k = \sqrt3/4$ .

In plain words The squares on the two short sides of a right triangle together have the area of the square on the long side. Any other shape built on the sides in the same proportions keeps this balance.

Example For legs $3$ and $4$ the theorem gives $c^2 = 9 + 16 = 25$ , so $c = 5$ . For legs $5$ and $12$ it gives $c^2 = 25 + 144 = 169$ , so $c = 13$ . For legs $1$ and $1$ it gives $c = \sqrt2$ . In the first triangle the equilateral triangles on the sides have areas $9\sqrt3/4$ , $16\sqrt3/4$ and $25\sqrt3/4$ , and the first two again add up to the third.

Semicircle area

Definition A semicircle of diameter $d$ is the half of a disk of radius $r = d/2$ on one side of a diameter. Its area is half of $\pi r^2$ ,

\frac12 \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{8},

so the area of a semicircle is proportional to the square of its diameter. Here a semicircle means the flat region, and its curved edge is an arc. A circular segment is the region between a chord of a circle and one of the two arcs that the chord cuts off.

In plain words Double the diameter and the semicircle has four times the area. The factor $\pi/8$ is the same for every semicircle, which is why it can be taken out of a comparison of semicircles.

Example The semicircles on diameters $3$ , $4$ and $5$ have areas $9\pi/8$ , $16\pi/8 = 2\pi$ and $25\pi/8$ , and

\frac{9\pi}{8} + \frac{16\pi}{8} = \frac{25\pi}{8}.

In decimals the two smaller areas are about $3.534$ and $6.283$ , and the larger one is about $9.817$ . The equality is the Pythagorean equation $9 + 16 = 25$ multiplied by $\pi/8$ .

Squaring the circle

Definition To square a plane region means to construct, with an unmarked straightedge and a compass in finitely many steps, a square with the same area as the region. Squaring the circle is the ancient problem of doing this for a disk. In 1882 Ferdinand von Lindemann proved that $\pi$ is transcendental, which means it is not a solution of any polynomial equation whose coefficients are integers, not all zero. Every length that can be constructed from a unit length is a solution of such an equation, and if a length $\sqrt\pi$ could be constructed, then so could its square $\pi$ . So the side $\sqrt\pi$ of a square with the area of a unit disk cannot be constructed, and the circle cannot be squared. Every polygon can be squared, and so the lunes of a right triangle can be squared too, since their total area equals a triangle. The lune of Hippocrates is the first region bounded by curves known to have been squared. None of this squares the circle. The $\pi$ terms in the lune comparison cancel only because the semicircles and segments are arranged around a right triangle, and no whole disk is ever matched with a polygon.

In plain words Some regions with curved edges have areas that a straightedge and compass can match exactly with a square, and a whole circle is not one of them. Squares that match a circle’s area to any number of decimal places can still be drawn, so the impossibility concerns exact constructions only.

Example A disk of radius $1$ has area $\pi \approx 3.1416$ , and a square of the same area would need side $\sqrt\pi \approx 1.7725$ , a length that straightedge and compass cannot produce exactly. The two lunes of the 3, 4, 5 triangle have total area $6$ , the area of a square of side $\sqrt6 \approx 2.449$ . That length can be constructed. The diagonal of a square of side $1$ has length $\sqrt2$ , and a right triangle with legs $\sqrt2$ and $2$ has hypotenuse

\sqrt{2 + 4} = \sqrt6 .

Check yourself

Build the figure on a right triangle with legs $5$ and $12$ . What is the total area of the two lunes, and how do the three semicircle areas compare?

Show the answer

The hypotenuse is $\sqrt{25 + 144} = 13$ . The semicircles have areas $25\pi/8$ , $144\pi/8$ and $169\pi/8$ , and the first two add up to the third because $25 + 144 = 169$ . So the cancellation works as before, and the lunes have total area equal to the triangle,

\frac12 \cdot 5 \cdot 12 = 30

square units. Separately they are about $8.14$ on the short leg and $21.86$ on the long leg.

Change a rule. Keep sides $3$ and $4$ but let them meet at an angle of $120^\circ$ , with $C = (0, 0)$ , $B = (3, 0)$ and $A = (-2, 2\sqrt3)$ . Build the three semicircles the same way. Which steps of the argument fail?

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The point $A$ is at distance $\sqrt{4 + 12} = 4$ from $C$ , so the sides are $3$ and $4$ as required. The third side has

\begin{aligned} AB^2 &= (3 + 2)^2 + (2\sqrt3)^2 \\[0.4em] &= 25 + 12 = 37, \end{aligned}

so the small semicircles have total area $25\pi/8$ while the large one has area $37\pi/8$ . The Pythagorean balance $U = V$ is gone.

The way the regions fit together fails as well. The midpoint of $AB$ is $M = (1/2, \sqrt3)$ , and $MC^2 = 1/4 + 3 = 13/4$ , which is less than the squared radius $37/4$ . So $C$ lies inside the large circle instead of on it, the part of the large semicircle outside the triangle is no longer two segments cut off by the legs, and there are no matching pieces to cancel. The result fails too. A numerical calculation shows that the two lunes, the parts of the small semicircles outside the large one, have total area about $2.05$ , while the triangle, with base $3$ and height $2\sqrt3$ , has area $3\sqrt3 \approx 5.20$ .

Extend the pattern to the original lune of Hippocrates. Take a right triangle with both legs of length $2$ and compute the area of each lune directly, without the cancellation argument.

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The hypotenuse is $2\sqrt2$ , so the large circle has center $O$ at the midpoint of the hypotenuse and radius $\sqrt2$ . Each leg is a chord of length $2$ of that circle, and the triangle formed by $O$ and the two ends of the leg has sides $\sqrt2$ , $\sqrt2$ and $2$ . Since $2 + 2 = 4$ , it has a right angle at $O$ . The segment beyond the leg is a quarter of the large disk minus that triangle,

\begin{aligned} S &= \frac{\pi (\sqrt2)^2}{4} - \frac12 \sqrt2 \cdot \sqrt2 \\[0.4em] &= \frac{\pi}{2} - 1 . \end{aligned}

The small semicircle on the leg has area $\pi \cdot 2^2/8 = \pi/2$ , so each lune has area

\frac{\pi}{2} - \left(\frac{\pi}{2} - 1\right) = 1 .

The two lunes total $2$ , the area $\tfrac12 \cdot 2 \cdot 2$ of the triangle, as the general result says. Here each lune has exactly the area of the right triangle formed by its leg and $O$ , and of a square of side $1$ .