5. Similarity¶

A map is a small copy of a country, a photograph is a small copy of a face, and a model aircraft in a wind tunnel is a small copy of a real one. When is a copy faithful? Which numbers survive the shrinking and which change, and by how much? This chapter makes "same shape, different size" precise. You will prove the one theorem everything else rests on, the intercept theorem, and from it the three tests for similar triangles. Then you will see why areas grow as the square and volumes as the cube of the scale, why that keeps giants out of the real world, how a right triangle hides a geometric mean, and what the golden ratio really is once the myths are cleared away.

In this chapter

You will learn:

what a dilation (a scaling from a centre) does to lengths and angles
the intercept theorem and its converse, with a proof that works for every length, rational or not
the AA, SAS and SSS tests for similar triangles, each proved from the intercept theorem
why areas scale by \(k^2\) and volumes by \(k^3\), and what the square-cube law says about animals, on Earth and on Mars
the geometric mean, the altitude theorem in a right triangle, and how to construct \(\sqrt{n}\)
the golden ratio: its definition, its true home in the regular pentagon, and which famous claims about it are false

You need: Chapter 2 (the congruence tests SSS, SAS, ASA), Chapter 3 (angles at parallel lines, the angle sum) and Chapter 4 (area of a triangle as half base times height). For a gentler version, see the Topics lessons Dilations and scale drawings and Similar triangles and scale factors.

Levels: 5.1 is beginner; 5.2 to 5.4 are intermediate; the proofs in 5.2 and the pentagon in 5.5 are intermediate to advanced.

5.1 Scaling¶

A photograph printed at 10 by 15 centimetres is enlarged to 20 by 30. Everything in it doubles: the width of a face, the length of a nose, the distance between two eyes. Nothing turns, nothing bends, and a right angle in the small print is still a right angle in the big one. That is the idea of scaling, and geometry makes it exact with one simple map.

Definition (dilation). Fix a point \(O\), the centre, and a positive number \(k\), the scale factor. The dilation with centre \(O\) and factor \(k\) sends each point \(P\) to the point \(P'\) on the ray from \(O\) through \(P\) with

\[OP' = k \cdot OP .\]

The centre itself stays where it is. If \(k > 1\) the figure is enlarged, if \(k < 1\) it is shrunk, and \(k = 1\) changes nothing.

In the figure, \(O\) is the centre and \(k = 2\): each of \(A'\), \(B'\), \(C'\) is on the ray from \(O\) through \(A\), \(B\), \(C\), twice as far out. Measure the sides and you find \(A'B' = 2\,AB\), \(B'C' = 2\,BC\), \(C'A' = 2\,CA\), and the angles of the two triangles are equal. We will prove both facts in 5.2. For now, take them as the definition of what we want.

Definition (similar). Two figures are similar if one can be turned into the other by a dilation followed by a congruence (a rigid motion: sliding, turning, flipping). In words: same shape, possibly a different size, possibly moved. We write \(\triangle ABC \sim \triangle A'B'C'\) and always list the vertices in matching order, so that \(A\) goes to \(A'\), \(B\) to \(B'\) and \(C\) to \(C'\).

For polygons this comes down to two conditions:

corresponding angles are equal, and
corresponding sides are in the same ratio \(k\): \(A'B'/AB = B'C'/BC = C'A'/CA = k\).

Both conditions are needed for polygons in general. A square and a long rectangle have the same angles (all right angles) but are not similar; a square and a rhombus with 60 degree angles have all sides in ratio 1 but are not similar either. One of the surprises of this chapter is that for triangles, either condition alone is enough.

Worked example (a map). A hiking map has scale 1 : 25 000, meaning that 1 centimetre on the map stands for 25 000 centimetres on the ground. Two huts are 8.4 cm apart on the map. How far apart are they?

Multiply by the scale factor: \(8.4 \times 25\,000 = 210\,000\) cm.
Convert: \(210\,000\) cm \(= 2100\) m \(= 2.1\) km.

Check: going back, \(2.1\) km \(= 210\,000\) cm, and \(210\,000 / 25\,000 = 8.4\) cm.

The map is a dilation (factor \(1/25\,000\)) of the land, flattened. Angles on the map, for example the angle between two straight roads, are the same as on the ground. That is why you can take a compass bearing from a map. (A map of a small region can be almost exactly similar to the land; a map of the whole Earth cannot, because the Earth is curved. That story belongs to Chapter 16.)

Worked example (shadows). At the same moment of a sunny day, a 1.5 m stick standing upright casts a shadow 2 m long, and a tree casts a shadow 14 m long. How tall is the tree?

The sun is so far away that its rays arrive parallel. Stick, shadow and ray make a right triangle; tree, shadow and ray make another, with the same angle where the ray meets the ground. So the triangles have the same angles, and (we are about to prove) they are similar. The ratio height : shadow is the same for both:

\[\frac{h}{14} = \frac{1.5}{2}, \qquad h = 14 \times 0.75 = 10.5 \text{ m}.\]

Check: \(10.5 / 14 = 0.75 = 1.5/2\).

Greek writers of later centuries told the story that Thales of Miletus (sixth century BCE) measured the height of an Egyptian pyramid this way, waiting for the moment when a stick's shadow equalled its height. The story is told by authors writing centuries after Thales, so we cannot be sure it happened, but the method is perfectly sound.

5.2 Similar triangles¶

The whole theory of similarity rests on one theorem about a triangle cut by a line parallel to one side.

The intercept theorem¶

Theorem (intercept theorem). In triangle \(ABC\), let \(D\) be a point on side \(AB\) and \(E\) a point on side \(AC\) with \(DE\) parallel to \(BC\). Then

\[\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}.\]

You may meet this theorem as the "basic proportionality theorem", the "side splitter theorem" or, in Europe, as Thales' theorem (a name that the English-speaking world gives to a different theorem, the one in Chapter 6).

Before proving it, notice what makes it hard. If \(AD = 2\) and \(AB = 5\), you could try to cut \(AB\) into five equal pieces, draw parallels to \(BC\) through each cut point, and show the parallels also cut \(AC\) into five equal pieces. That works whenever the ratio \(AD/AB\) is a fraction. But what if \(AD/AB\) is \(1/\sqrt{2}\)? There is no common unit that measures both, and the "count the pieces" proof falls apart. The Greeks discovered such incommensurable lengths (the side and diagonal of a square are the classic pair) and it forced them to rebuild the theory of ratio; Eudoxus' answer is Book V of Euclid's Elements. We will avoid the problem entirely with Euclid's own trick from Book VI: compare areas.

Two facts about area (both from Chapter 4). Write \([XYZ]\) for the area of triangle \(XYZ\).

Two triangles with the same base and equal heights have equal area, since each is half base times height.
Two triangles with the same height have areas in the ratio of their bases. In particular, if \(D\) lies on side \(AB\) of triangle \(ABE\), then \([ADE]/[ABE] = AD/AB\): both triangles have apex \(E\) and their bases lie on the same line, so they share the height from \(E\).

Proof of the intercept theorem. Draw the segments \(BE\) and \(CD\).

Step 1. Triangles \(DBE\) and \(DCE\) have the same base \(DE\). Their third vertices \(B\) and \(C\) lie on the line \(BC\), which is parallel to \(DE\), so \(B\) and \(C\) are the same distance from the line \(DE\). The heights are equal, so \([DBE] = [DCE]\).

Step 2. Add the triangle \(ADE\) to each. Since \(D\) lies on \(AB\), triangles \(ADE\) and \(DBE\) together make triangle \(ABE\). Since \(E\) lies on \(AC\), triangles \(ADE\) and \(DCE\) together make triangle \(ADC\). So

\[[ABE] = [ADE] + [DBE] = [ADE] + [DCE] = [ADC].\]

Step 3. By fact 2, using apex \(E\) and bases on line \(AB\): \([ADE]/[ABE] = AD/AB\). Using apex \(D\) and bases on line \(AC\): \([ADE]/[ADC] = AE/AC\). The two left-hand sides are equal by Step 2, so

\[\frac{AD}{AB} = \frac{AE}{AC}.\]

Step 4 (the third ratio). Through \(E\) draw the line parallel to \(AB\); it meets \(BC\) at a point \(F\). Now \(DEFB\) has both pairs of opposite sides parallel, so it is a parallelogram, and \(DE = BF\). Apply Step 3 to triangle \(CAB\) with the line \(EF\) parallel to \(AB\) (now \(C\) plays the part of the apex): \(CE/CA = CF/CB\). Then

\[\frac{DE}{BC} = \frac{BF}{BC} = \frac{BC - CF}{BC} = 1 - \frac{CF}{CB} = 1 - \frac{CE}{CA} = \frac{CA - CE}{CA} = \frac{AE}{AC}.\]

That proves all three ratios equal. \(\blacksquare\)

Nothing in the proof assumed the ratios were fractions. Areas are real numbers, and the argument is the same for every length. The same proof works, with small changes, when \(D\) and \(E\) lie on the sides extended beyond \(B\) and \(C\), or beyond \(A\); in every case the parallel line cuts the two sides (or their extensions) in the same ratio.

Converse. If \(D\) is on \(AB\) and \(E\) is on \(AC\) with \(AD/AB = AE/AC\), then \(DE\) is parallel to \(BC\).

Proof. Through \(D\) draw the line parallel to \(BC\) (by the parallel postulate, there is exactly one); it meets \(AC\) at some point \(E'\). By the theorem, \(AE'/AC = AD/AB = AE/AC\), so \(AE' = AE\). Two points on the segment \(AC\) at the same distance from \(A\) are the same point, so \(E' = E\) and \(DE\) is the parallel. \(\blacksquare\)

Worked example. In triangle \(ABC\), \(DE \parallel BC\) with \(D\) on \(AB\), \(E\) on \(AC\), \(AD = 3\), \(DB = 5\), \(AE = 4.5\) and \(BC = 12\). Find \(EC\) and \(DE\).

\(AB = 3 + 5 = 8\), so \(AD/AB = 3/8\).
\(AE/AC = 3/8\), so \(AC = 4.5 \times 8/3 = 12\) and \(EC = 12 - 4.5 = 7.5\).
\(DE/BC = 3/8\), so \(DE = 12 \times 3/8 = 4.5\).

Check: \(AE/EC = 4.5/7.5 = 0.6\) and \(AD/DB = 3/5 = 0.6\). (Splitting the sides in the same ratio is an equivalent way to state the theorem.)

The midpoint theorem is the special case \(AD/AB = 1/2\): the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long.

Dilations, at last¶

Now we can prove what we saw in 5.1. Let the dilation with centre \(O\) and factor \(k\) send \(P\) to \(P'\) and \(Q\) to \(Q'\). If \(O\), \(P\), \(Q\) are not on one line, then in triangle \(OP'Q'\) we have \(OP/OP' = OQ/OQ' = 1/k\), so by the converse of the intercept theorem \(PQ \parallel P'Q'\), and by the theorem itself \(PQ/P'Q' = 1/k\), that is,

\[P'Q' = k \cdot PQ .\]

(If \(O\), \(P\), \(Q\) are on one line, the same equation follows by subtracting distances along the line.) So a dilation multiplies every distance by \(k\) and sends every line to a parallel line. Since it sends lines to parallel lines, it keeps angles: the angle between two lines equals the angle between their parallels (corresponding angles, Chapter 3).

The three similarity tests¶

Theorem (AA similarity). If two angles of one triangle equal two angles of another, the triangles are similar.

Two equal angles force the third to be equal too, since the angles of each triangle add to 180 degrees. What is not obvious is that the sides must then be in proportion.

Proof. Let \(\angle A = \angle A'\) and \(\angle B = \angle B'\). By renaming the triangles if needed, assume \(A'B' \le AB\). Mark \(D\) on side \(AB\) with \(AD = A'B'\), and draw through \(D\) the parallel to \(BC\), meeting \(AC\) at \(E\).

\(\angle ADE = \angle ABC\) (corresponding angles at the parallels \(DE\) and \(BC\)), and \(\angle ABC = \angle A'B'C'\).
So triangles \(ADE\) and \(A'B'C'\) have \(\angle A = \angle A'\), \(AD = A'B'\) and \(\angle ADE = \angle A'B'C'\). By ASA (Chapter 2) they are congruent: \(AE = A'C'\) and \(DE = B'C'\).
By the intercept theorem, \(AD/AB = AE/AC = DE/BC\). Replacing each length by its equal,

\[\frac{A'B'}{AB} = \frac{A'C'}{AC} = \frac{B'C'}{BC}.\]

The sides are in proportion and the angles are equal, so the triangles are similar. \(\blacksquare\)

Theorem (SAS similarity). If \(\angle A = \angle A'\) and \(A'B'/AB = A'C'/AC\), the triangles are similar.

Proof. Again assume \(A'B' \le AB\). Mark \(D\) on \(AB\) with \(AD = A'B'\) and \(E\) on \(AC\) with \(AE = A'C'\). Then \(AD/AB = AE/AC\), so by the converse of the intercept theorem \(DE \parallel BC\). Triangle \(ADE\) is congruent to \(A'B'C'\) by SAS (two sides and the angle between them). And triangle \(ADE\) has the same angles as \(ABC\) (the angle at \(A\) is shared, and the parallels give \(\angle ADE = \angle ABC\)), so by AA it is similar to \(ABC\). A triangle congruent to one similar to \(ABC\) is similar to \(ABC\). \(\blacksquare\)

Theorem (SSS similarity). If \(A'B'/AB = B'C'/BC = C'A'/CA\), the triangles are similar.

Proof. Call the common ratio \(k\) and assume \(k \le 1\). Mark \(D\) on \(AB\) with \(AD = A'B' = k\,AB\) and draw the parallel to \(BC\) through \(D\), meeting \(AC\) at \(E\). By the intercept theorem, \(AE = k\,AC = A'C'\) and \(DE = k\,BC = B'C'\). So triangle \(ADE\) has the same three sides as \(A'B'C'\), and by SSS they are congruent. Since \(ADE\) has the same angles as \(ABC\), so does \(A'B'C'\). \(\blacksquare\)

Each test mirrors a congruence test with "equal sides" replaced by "sides in the same ratio". There is no "ASA similarity" separate from AA, because the side in ASA plays no part once only the shape matters. And just as SSA is not a congruence test, "SSA similarity" fails as well.

Worked example (similar triangles inside a figure). Two segments \(AC\) and \(BD\) cross at \(P\), with \(AB \parallel DC\), \(AB = 6\), \(DC = 10\) and \(AP = 4.5\). Find \(PC\).

\(\angle PAB = \angle PCD\) (alternate angles at the parallels), and \(\angle APB = \angle CPD\) (vertical angles).
By AA, \(\triangle PAB \sim \triangle PCD\), with ratio \(DC/AB = 10/6 = 5/3\).
So \(PC = \tfrac53 \times AP = \tfrac53 \times 4.5 = 7.5\).

Check: \(AP/PC = 4.5/7.5 = 0.6 = 6/10 = AB/DC\).

Worked example (a ramp). A wheelchair ramp rises 0.5 m over a horizontal run of 6 m. A support post stands 2.4 m (horizontally) from the bottom of the ramp. How tall must it be?

The post, the ground and the ramp make a small right triangle that shares the angle at the bottom with the whole ramp's triangle. By AA they are similar, with ratio \(2.4/6 = 0.4\), so the post is \(0.4 \times 0.5 = 0.2\) m tall.

5.3 Areas and volumes¶

A pizza 30 cm across costs twice as much as one 20 cm across. Which is the better deal? The diameter grew by a factor of \(1.5\), but the amount of pizza grew by \(1.5^2 = 2.25\), so the big one gives 2.25 times the pizza for 2 times the price. Areas do not scale like lengths.

Theorem (area scaling). If two figures are similar with scale factor \(k\), their areas are in the ratio \(k^2\).

Proof for triangles. Take \(\triangle ABC \sim \triangle A'B'C'\) with ratio \(k\). Let \(h\) be the height from \(C\) to line \(AB\), with foot \(H\). The dilation part of the similarity sends \(H\) to a point \(H'\) on line \(A'B'\), and it keeps the right angle at \(H\), so \(C'H'\) is the height of the image, of length \(kh\). The base is \(k \cdot AB\). So

\[[A'B'C'] = \tfrac12 \cdot (k\,AB) \cdot (k\,h) = k^2 \cdot \tfrac12\, AB \cdot h = k^2\,[ABC].\]

For polygons: cut the polygon into triangles by diagonals; the similarity sends this cutting to a cutting of the image into triangles, each \(k^2\) times as large; add.

For curved figures: cover the figure by a fine grid of small squares. Each square of side \(s\) becomes a square of side \(ks\), with \(k^2\) times the area. The area of a curved figure is the limit of the areas of the grid squares inside it as the grid gets finer (Chapter 4), so its area is multiplied by \(k^2\) as well. \(\blacksquare\)

Volumes. The same argument in three dimensions, with small cubes in place of squares: a cube of side \(s\) becomes a cube of side \(ks\), made of \(k^3\) copies of the old one when \(k\) is a whole number, and of volume \(k^3 s^3\) in any case.

scale factor \(k\)	lengths	areas	volumes
2	\(\times 2\)	\(\times 4\)	\(\times 8\)
3	\(\times 3\)	\(\times 9\)	\(\times 27\)
10	\(\times 10\)	\(\times 100\)	\(\times 1000\)
\(\tfrac12\)	\(\times \tfrac12\)	\(\times \tfrac14\)	\(\times \tfrac18\)

Worked example (a statue). A sculptor makes a 1 : 10 model of a bronze statue. The model has a mass of 3 kg and needed 0.2 litres of varnish to coat. If the full statue is solid bronze like the model, what are its mass and the varnish it needs?

Mass goes with volume: \(3 \times 10^3 = 3000\) kg.
Varnish goes with surface area: \(0.2 \times 10^2 = 20\) litres.

The statue is a thousand times heavier but needs only a hundred times the varnish.

The square-cube law¶

Galileo set out this argument in his Two New Sciences (1638), and J. B. S. Haldane made it famous in his 1926 essay "On Being the Right Size". Scale an animal up by a factor \(k\), keeping its shape and its materials.

Its weight is proportional to its volume: it grows by \(k^3\).
The strength of a bone, how much load it carries before it breaks, is proportional to the area of its cross section: it grows by \(k^2\).
So the stress on the bone, load per unit of cross section, grows by \(k^3/k^2 = k\).

A giant ten times as tall as a person, built exactly like one, would put ten times the stress on every bone. Human leg bones are strong, but not ten times stronger than they need to be; the giant would break its thighs at its first step. That is why large animals are not scaled-up small ones: an elephant's legs are far thicker for its length than a gazelle's, and the largest animals of all, whales, let the water carry their weight.

The same law works the other way for heat and for falling. Heat leaves a body through its skin (area, \(k^2\)) but is made by its whole volume (\(k^3\)), so a mouse loses heat fast for its size and must eat constantly. A falling mouse meets air resistance in proportion to its area but is pulled down in proportion to its volume, which is why Haldane could write that a mouse dropped down a mine shaft walks away.

A Mars aside. On Mars the surface gravity is about 0.38 of Earth's (3.7 metres per second per second against 9.8). Weight is mass times gravity, so on Mars every bone carries 0.38 of its Earth load. The stress in a scaled animal is proportional to \(k \times g\); to keep it at the Earth value we need

\[k \times 0.38 = 1, \qquad k \approx 2.6 .\]

So the square-cube law alone would allow a body about 2.6 times as tall to stand on Mars as safely as a person stands on Earth. That is a statement about bones under a static load and nothing more: muscles, hearts and lungs scale differently, and a real body is not a scaled copy of another. But it shows how a single exponent decides questions that look like biology.

Worked example. The surface area of a sphere is proportional to \(r^2\) and its volume to \(r^3\). A raindrop of radius 1 mm and one of radius 2 mm: the bigger one has \(2^2 = 4\) times the surface and \(2^3 = 8\) times the water, so half as much surface per litre. Small things are mostly surface; big things are mostly inside.

5.4 The geometric mean and the altitude theorem¶

The arithmetic mean of two positive numbers \(a\) and \(b\) is \((a+b)/2\), the number halfway between them. The geometric mean is \(\sqrt{ab}\): the side of the square with the same area as the rectangle \(a \times b\). For 4 and 9 the arithmetic mean is 6.5 and the geometric mean is \(\sqrt{36} = 6\). The geometric mean is the right average for things that multiply: if an investment grows by a factor 4 one year and 9 the next, the steady factor that gives the same result after two years is 6, not 6.5. Another way to say it: \(g = \sqrt{ab}\) is the number with

\[\frac{a}{g} = \frac{g}{b},\]

"\(a\) is to \(g\) as \(g\) is to \(b\)". The Greeks called \(g\) the mean proportional between \(a\) and \(b\), and they found it inside every right triangle.

Let \(ABC\) have a right angle at \(C\). Drop the perpendicular from \(C\) to the hypotenuse \(AB\), with foot \(H\). Write \(a = BC\), \(b = AC\), \(c = AB\), \(h = CH\), \(p = AH\), \(q = HB\), so \(p + q = c\).

Lemma. The altitude cuts the right triangle into two triangles similar to it and to each other: \(\triangle ACB \sim \triangle AHC \sim \triangle CHB\).

Proof. Triangles \(AHC\) and \(ACB\) share the angle at \(A\) and both have a right angle (at \(H\) and at \(C\)), so they are similar by AA. In the same way \(CHB\) and \(ACB\) share the angle at \(B\) and have a right angle. Two triangles similar to a third are similar to each other. Matching the vertices: in \(AHC\) the right angle is at \(H\), the angle equal to \(\angle A\) is at \(A\), so the remaining angle at \(C\) equals \(\angle B\). \(\blacksquare\)

Theorem (altitude theorem). \(h^2 = pq\): the altitude is the geometric mean of the two pieces of the hypotenuse.

Proof. From \(\triangle AHC \sim \triangle CHB\), matching \(A \to C\), \(H \to H\), \(C \to B\) (angles \(\angle A\), right angle, \(\angle B\) in each), the ratio of short leg to long leg agrees:

\[\frac{AH}{CH} = \frac{CH}{HB}, \qquad \frac{p}{h} = \frac{h}{q}, \qquad h^2 = pq. \qquad \blacksquare\]

Theorem (leg theorem). \(b^2 = cp\) and \(a^2 = cq\): each leg is the geometric mean of the hypotenuse and the piece of the hypotenuse next to it.

Proof. From \(\triangle AHC \sim \triangle ACB\): \(AH/AC = AC/AB\), so \(p/b = b/c\) and \(b^2 = cp\). The other is the same with \(B\) in place of \(A\). \(\blacksquare\)

Add the two equations of the leg theorem:

\[a^2 + b^2 = cq + cp = c(p + q) = c^2 .\]

That is Pythagoras' theorem again, with a proof that uses only similarity. It is one of the proofs in Chapter 4; here you can see where it comes from.

Worked example. The altitude to the hypotenuse of a right triangle cuts the hypotenuse into pieces \(p = 4\) and \(q = 9\). Find the altitude and the legs.

\(h = \sqrt{4 \times 9} = 6\).
\(c = 13\), \(b = \sqrt{13 \times 4} = \sqrt{52} = 2\sqrt{13} \approx 7.21\), \(a = \sqrt{13 \times 9} = 3\sqrt{13} \approx 10.82\).

Check: \(a^2 + b^2 = 117 + 52 = 169 = 13^2\). And the area two ways: \(\tfrac12 ab = \tfrac12 \cdot 3\sqrt{13} \cdot 2\sqrt{13} = 39\) and \(\tfrac12 c h = \tfrac12 \cdot 13 \cdot 6 = 39\).

Constructing a square root¶

The altitude theorem gives a construction, with compass and straightedge, of \(\sqrt{n}\) for any length \(n\). Lay off \(AH = 1\) and then \(HB = n\) along a line. Draw the circle with diameter \(AB\) (bisect \(AB\) to find its centre, as in Chapter 2). Raise the perpendicular to \(AB\) at \(H\); it meets the circle at \(C\). The angle \(ACB\) is a right angle, because it stands on a diameter (Thales' theorem, proved in Chapter 6). So \(CH^2 = 1 \cdot n\) and \(CH = \sqrt{n}\).

The construction is the first sign of something Chapter 14 proves in full: a compass can take square roots, and nothing more exotic.

The AM-GM inequality, by a picture¶

In the same figure the circle has radius \((p+q)/2\), the arithmetic mean of \(p\) and \(q\), and the half-chord \(CH\) is their geometric mean \(\sqrt{pq}\). A half-chord can never be longer than a radius (the radius \(OC\) is the hypotenuse of the right triangle \(OHC\)). So

\[\sqrt{pq} \le \frac{p+q}{2},\]

with equality exactly when \(H\) is the centre \(O\), that is when \(p = q\). This is the AM-GM inequality for two numbers. Algebra proves it in one line, \((\sqrt p - \sqrt q)^2 \ge 0\); the picture shows why it is true.

Worked example (the best rectangle). Of all rectangles with perimeter 20, which has the greatest area? If the sides are \(p\) and \(q\), then \(p + q = 10\), and the area \(pq\) satisfies \(\sqrt{pq} \le 5\), so \(pq \le 25\), with equality only when \(p = q = 5\). The square wins.

5.5 The golden ratio, honestly¶

Book VI of Euclid's Elements defines a way of cutting a segment that it calls cutting in extreme and mean ratio: cut \(AB\) at a point \(P\) so that the whole is to the larger part as the larger part is to the smaller,

\[\frac{AB}{AP} = \frac{AP}{PB}.\]

Euclid needed it for one purpose: to construct the regular pentagon and, in Book XIII, the dodecahedron and icosahedron. The name "golden section" is much later; it appears in German in the 1830s.

Finding the ratio. Let \(AB = 1\) and \(AP = x\), so \(PB = 1 - x\). Then

\[\frac{1}{x} = \frac{x}{1-x}, \qquad x^2 = 1 - x, \qquad x^2 + x - 1 = 0, \qquad x = \frac{\sqrt5 - 1}{2} \approx 0.618 .\]

(The other root of the quadratic is negative.) The ratio \(AB/AP\) is

\[\varphi = \frac{1}{x} = \frac{2}{\sqrt5 - 1} = \frac{1 + \sqrt5}{2} \approx 1.6180 .\]

This number \(\varphi\) (the Greek letter phi) is the golden ratio. Because \(x = 1/\varphi\) satisfies \(x^2 + x - 1 = 0\), dividing by \(x^2\) shows \(\varphi\) satisfies

\[\varphi^2 = \varphi + 1, \qquad \text{and so} \qquad \frac{1}{\varphi} = \varphi - 1 \approx 0.618 .\]

Square it and you add 1; take its reciprocal and you subtract 1. Since \(\sqrt5\) is irrational, so is \(\varphi\): the two parts of a golden cut are incommensurable.

Its real home: the regular pentagon¶

Theorem. In a regular pentagon, the ratio of a diagonal to a side is \(\varphi\). Moreover two diagonals cut each other in the golden ratio.

Proof. Let \(ABCDE\) be a regular pentagon with side \(s\) and diagonal \(d\). Its angles are each \(108°\) (the angle sum of a pentagon is \(540°\), Chapter 3).

Angles. Triangle \(ABC\) is isosceles (\(AB = BC = s\)) with apex angle \(108°\), so its base angles are \((180° - 108°)/2 = 36°\): \(\angle BAC = \angle BCA = 36°\). The same holds for triangle \(BCD\): \(\angle CBD = 36°\).

Let the diagonals \(AC\) and \(BD\) meet at \(P\).

Triangle \(BPC\). Its angles at \(B\) and \(C\) are \(\angle PBC = \angle DBC = 36°\) and \(\angle PCB = \angle ACB = 36°\). So it is isosceles with \(PB = PC\), and its third angle is \(108°\).

Triangle \(ABP\). \(\angle BAP = \angle BAC = 36°\), and \(\angle ABP = \angle ABC - \angle PBC = 108° - 36° = 72°\), so \(\angle APB = 180° - 36° - 72° = 72°\). Two equal angles: \(AP = AB = s\).

Similarity. Triangles \(ABC\) (angles \(36°, 108°, 36°\)) and \(BPC\) (angles \(36°, 108°, 36°\)) are similar by AA. In \(ABC\) the base is \(AC = d\) and the legs are \(s\); in \(BPC\) the base is \(BC = s\) and the legs are \(PC = AC - AP = d - s\). So

\[\frac{d}{s} = \frac{s}{d - s}.\]

Write \(r = d/s\). Dividing top and bottom of the right side by \(s\) gives \(r = 1/(r - 1)\), so \(r^2 - r - 1 = 0\) and \(r = \varphi\) (the positive root). And \(AC/AP = d/s = \varphi\), \(AP/PC = s/(d - s) = \varphi\): the point \(P\) cuts the diagonal \(AC\) in extreme and mean ratio. \(\blacksquare\)

A bonus: the cosine of 36 degrees. Drop the perpendicular from \(B\) to \(AC\) in triangle \(ABC\); it bisects \(AC\) (the triangle is isosceles), so \(d/2 = s \cos 36°\) (you will meet cosine properly in Chapter 7). Hence

\[\cos 36° = \frac{d}{2s} = \frac{\varphi}{2} = \frac{1 + \sqrt5}{4} \approx 0.8090 .\]

The golden rectangle and Fibonacci¶

A golden rectangle has sides in ratio \(\varphi : 1\). Cut off a square of side 1 and what remains is a rectangle \(1\) by \(\varphi - 1 = 1/\varphi\), with sides in ratio \(1 : 1/\varphi = \varphi : 1\), a smaller golden rectangle. The golden rectangle is the only rectangle with this property: if removing a square from a \(r \times 1\) rectangle (\(r > 1\)) leaves a similar one, then \(r/1 = 1/(r-1)\), which is the golden equation again.

The Fibonacci numbers \(1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots\) (each the sum of the two before) have ratios \(2/1, 3/2, 5/3, 8/5, 13/8, \ldots = 2, 1.5, 1.667, 1.6, 1.625, \ldots\) that close in on \(\varphi\). Using \(\varphi^2 = \varphi + 1\) repeatedly gives \(\varphi^n = F_n \varphi + F_{n-1}\), where \(F_n\) is the \(n\)-th Fibonacci number; the Algebra book proves Binet's formula, which explains why the ratios converge.

Try it in Python.

phi = (1 + 5 ** 0.5) / 2
a, b = 1, 1
for n in range(10):
    a, b = b, a + b
    print(b, a, b / a, b / a - phi)

The error shrinks by a factor of about \(\varphi^2 \approx 2.6\) each step, and changes sign every time.

What is true and what is not¶

The golden ratio is real mathematics: it is in the pentagon and the pentagram, in the coordinates of the icosahedron and dodecahedron (Chapter 11), in Penrose's aperiodic tilings, and, through the "golden angle" \(360°/\varphi^2 \approx 137.5°\), in models of how some plants place successive leaves and seeds. Many popular claims, however, do not survive measurement.

The Parthenon. There is no ancient record that its builders used \(\varphi\), and the "golden rectangles" drawn over photographs depend on which edges one chooses to measure; with different choices the ratios differ.
The Mona Lisa and other paintings. Rectangles drawn on paintings can be made to fit almost any ratio near 1.6. There is no evidence Leonardo planned one.
The human body. Ratios such as height to navel height vary from person to person and cluster around 1.6 only loosely; 1.6 is not 1.618.
The nautilus shell. It is a logarithmic spiral, but measured shells grow at rates different from the "golden spiral".
"The most beautiful rectangle." Experiments asking people to pick their favourite rectangle, from Fechner in the 1870s onward, have given mixed results that depend strongly on how the question is asked.

George Markowsky's 1992 article listed in the Further reading checks many of these claims. The lesson is a general one: a ratio of about 1.6 between two lengths you chose after the fact is evidence of nothing.

A short history¶

Similarity is one of the oldest ideas in geometry. Babylonian scribes and Egyptian builders used proportional triangles to compute heights and slopes long before anyone proved anything. The proofs came with the Greeks. When the Pythagoreans found that the diagonal and side of a square have no common measure, the simple theory of ratios as fractions broke; Eudoxus of Cnidus (fourth century BCE) built a theory of ratio that works for any magnitudes, and Euclid set it out in Book V of the Elements, then used it in Book VI for similar figures, including the intercept theorem (VI.2), the similarity tests (VI.4 to VI.7), the mean proportional (VI.13) and the extreme and mean ratio (VI.30). Our proofs replace Eudoxus' theory with real numbers and areas, which is how modern courses do it; the full treatment of length and area as real numbers belongs to analysis.

Check yourself¶

1. A map has scale 1 : 50 000. Two towns are 7.2 cm apart on the map. How far apart are they, and how large is a lake that covers 3 square centimetres of the map?

Distance: \(7.2 \times 50\,000 = 360\,000\) cm \(= 3.6\) km. Area scales by the square of the factor: \(3 \times 50\,000^2 = 7.5 \times 10^9\) square cm. One square kilometre is \(10^5 \times 10^5 = 10^{10}\) square cm, so the lake covers \(0.75\) square km.

2. A 2 m pole casts a 3 m shadow at the same moment that a building casts a 27 m shadow. How tall is the building?

The sun's rays are parallel, so the two right triangles have equal angles and are similar (AA). \(h/27 = 2/3\), so \(h = 18\) m. Check: \(18/27 = 2/3\).

3. In triangle \(ABC\), \(DE \parallel BC\) with \(D\) on \(AB\) and \(E\) on \(AC\). If \(AD = 4\), \(DB = 6\), \(AE = 5\) and \(BC = 15\), find \(EC\) and \(DE\).

\(AD/AB = 4/10 = 0.4\). By the intercept theorem \(AE/AC = 0.4\), so \(AC = 12.5\) and \(EC = 7.5\); and \(DE = 0.4 \times 15 = 6\). Check: \(AD/DB = 4/6\) and \(AE/EC = 5/7.5\), both \(2/3\).

4. Are triangles with sides 6, 8, 10 and 9, 12, 15 similar? What is the ratio of their areas? Check with the actual areas.

\(9/6 = 12/8 = 15/10 = 1.5\), so they are similar by SSS. The area ratio is \(1.5^2 = 2.25\). Both are right triangles (\(6^2 + 8^2 = 10^2\)), with areas \(\tfrac12 \cdot 6 \cdot 8 = 24\) and \(\tfrac12 \cdot 9 \cdot 12 = 54\), and \(54/24 = 2.25\).

5. Prove the midpoint theorem: the segment joining the midpoints \(M\) of \(AB\) and \(N\) of \(AC\) is parallel to \(BC\) and half as long.

\(AM/AB = AN/AC = 1/2\) and the triangles \(AMN\) and \(ABC\) share the angle \(A\), so they are similar by SAS with ratio \(1/2\). Hence \(MN = BC/2\). Corresponding angles are equal, \(\angle AMN = \angle ABC\), so \(MN \parallel BC\) (equal corresponding angles give parallel lines, Chapter 3). Or use the converse of the intercept theorem directly.

6. A 1 : 20 model of a sculpture has a mass of 1.5 kg and a surface area of 900 square cm. What are the mass and surface area of the full-size solid sculpture made of the same material?

Mass (volume): \(1.5 \times 20^3 = 12\,000\) kg. Surface: \(900 \times 20^2 = 360\,000\) square cm \(= 36\) square metres.

7. In a right triangle the altitude to the hypotenuse cuts it into pieces 2 and 8. Find the altitude, the two legs, and check with Pythagoras.

\(h = \sqrt{2 \cdot 8} = 4\). The hypotenuse is 10. Legs: \(\sqrt{10 \cdot 2} = 2\sqrt5 \approx 4.47\) and \(\sqrt{10 \cdot 8} = 4\sqrt5 \approx 8.94\). Check: \(20 + 80 = 100 = 10^2\), and the area is \(\tfrac12 \cdot 2\sqrt5 \cdot 4\sqrt5 = 20 = \tfrac12 \cdot 10 \cdot 4\).

8. Trapezoid \(ABCD\) has \(AB \parallel CD\), \(AB = 8\), \(CD = 12\); its diagonals meet at \(P\). If \(AC = 15\), find \(AP\). Prove your method.

\(\angle PAB = \angle PCD\) and \(\angle PBA = \angle PDC\) (alternate angles at the parallels), so \(\triangle PAB \sim \triangle PCD\) by AA, with \(AP/PC = AB/CD = 8/12 = 2/3\). So \(AP = \tfrac25 \cdot 15 = 6\) and \(PC = 9\). In general the diagonals of a trapezoid cut each other in the ratio of the parallel sides.

9. A regular pentagon has side 10 cm. How long is its diagonal? The diagonal \(AC\) is crossed by the two diagonals \(BD\) and \(BE\). How long are the three pieces they cut it into?

The diagonal is \(10\varphi \approx 16.18\) cm. By 5.5, the diagonal \(AC\) is cut at \(P\) (by \(BD\)) with \(AP = s = 10\) and \(PC = d - s = 6.18\). The reflection of the pentagon in the line through \(B\) and the centre swaps \(A\) with \(C\) and \(E\) with \(D\), so \(BE\) cuts \(AC\) at a point \(Q\) with \(QC = AP = 10\) and \(AQ = 6.18\). So the pieces are \(AQ = 6.18\), \(QP = 10 - 6.18 = 3.82\), \(PC = 6.18\) (that is, \(10/\varphi\), \(10/\varphi^2\), \(10/\varphi\), adding up to \(10\varphi\)).

10. Show that \(\varphi^n = F_n\varphi + F_{n-1}\) for \(n = 2, 3, 4, 5\) (with \(F_1 = F_2 = 1\), \(F_0 = 0\)), and use it to compute \(\varphi^5\).

\(\varphi^2 = \varphi + 1 = F_2\varphi + F_1\). Multiply by \(\varphi\): \(\varphi^3 = \varphi^2 + \varphi = 2\varphi + 1\), \(\varphi^4 = 2\varphi^2 + \varphi = 3\varphi + 2\), \(\varphi^5 = 3\varphi^2 + 2\varphi = 5\varphi + 3\). Each step sends \(F_n\varphi + F_{n-1}\) to \(F_n(\varphi + 1) + F_{n-1}\varphi = (F_n + F_{n-1})\varphi + F_n\), which is the induction step. \(\varphi^5 = 5(1.6180) + 3 \approx 11.090\). Check: \(1.6180^5 \approx 11.090\).

11. (Angle bisector theorem.) The bisector of angle \(A\) in triangle \(ABC\) meets \(BC\) at \(D\). Prove \(BD/DC = AB/AC\), and find \(BD\) when \(AB = 6\), \(AC = 9\), \(BC = 10\).

Through \(C\) draw the parallel to \(AD\); it meets line \(BA\) (extended beyond \(A\)) at \(E\). Then \(\angle AEC = \angle BAD\) (corresponding angles) and \(\angle ACE = \angle DAC\) (alternate angles). The bisector makes \(\angle BAD = \angle DAC\), so triangle \(AEC\) has two equal angles and \(AE = AC\). In triangle \(BEC\), \(AD \parallel EC\), so by the intercept theorem \(BD/DC = BA/AE = AB/AC\). With the numbers: \(BD/DC = 6/9 = 2/3\) and \(BD + DC = 10\), so \(BD = 4\) and \(DC = 6\).

12. On Mars (\(g\) about 0.38 of Earth's), by what factor could a body be scaled up and keep the same static stress in its bones as a person on Earth? And on the Moon (\(g\) about 0.165 of Earth's)?

Stress is proportional to \(k \cdot g\). Setting \(k \cdot 0.38 = 1\) gives \(k \approx 2.6\); on the Moon \(k = 1/0.165 \approx 6.1\). (Only bones under a static load: a real design problem has many other constraints.)

What you learned¶

A dilation with factor \(k\) multiplies every distance by \(k\), sends lines to parallel lines and keeps angles; similar figures differ by a dilation and a rigid motion.
The intercept theorem (a parallel cuts two sides in the same ratio) has an area proof that works for all lengths, rational or not; its converse also holds.
For triangles, AA, SAS and SSS each prove similarity, and each follows from the intercept theorem plus a congruence test.
Under scaling by \(k\), areas change by \(k^2\) and volumes by \(k^3\); the square-cube law limits the size of animals and structures, and Mars's lower gravity loosens the limit by about 2.6.
The altitude to the hypotenuse is the geometric mean of the pieces it cuts (\(h^2 = pq\)); this gives Pythagoras, a construction of \(\sqrt n\) and a picture proof of AM-GM.
The golden ratio \(\varphi = (1+\sqrt5)/2\), with \(\varphi^2 = \varphi + 1\), is the diagonal-to-side ratio of the regular pentagon; many popular claims about it in art and nature are not supported by evidence.

Next: Chapter 6 studies the circle: inscribed angles, tangents, the power of a point, and how Archimedes trapped \(\pi\).