116 Normat 57:3, 116–128 (2009)

Alternative route: from van Schooten to Ptolemy

Andrew Percy and D. G. Rogers

SASE, Monash University

Northways Road, CHURCHILL

AUSTRALIA, Vic 3842

andrew.percy@sci.monash.edu.au

1 Van Schooten’s Theorem (1646)

Frans van Schooten (1615–1660) was inﬂuential in the mathematical circle of his

time in several ways, but not least in the selection or formulation of choice re-

sults. In this regard, van Schooten acted as a signiﬁcant conduit for the ideas of

contemporaries, notably René Descartes (1596–1650), but also Pierre de Fermat

(1601–1665), to reach a wider audience — Isaac Newton (1642–1727), for one, read

van Schooten’s work with close attention in the mid-1660s, drawing on it in his

Lucasian lectures at Cambridge in the decade from 1673 (compare [22, esp. (a),

n. 7, p. x]; another Dutch source for Newton was the algebra textbook of 1661 by

Gerard Kinckhuysen (1625–1666)).

One of van Schooten’s theorems, in his De Organica Conicarum Sectionum In

Plano Descriptione, Tractatus [19], of 1646, brings to light a curious property of

equilateral triangles (see also [14, (b)] and [11]). Let 4ABC be an equilateral

triangle. If P is a point on the arc AB of the circumcircle of 4ABC opposite C,

as in Figure 1(i), then van Schooten shows that

CP = AP + BP. (1)

The fact that AP BC is a cyclic quadrilateral is something with which to set to

work. We might recall a staple in school geometry textbooks in years gone by (see,

for example, [12, 7, 8]; and compare [6]), the theorem of Ptolemy (c. 85–c. 165)

that once served as the foundation of trigonometric computation: the rectangle

contained by the diagonals of a cyclic quadrilateral is equal to the rectangles con-

tained by the pairs of opposite sides. So, for the cyclic quadrilateral AP BC, we

have

AB.CP = BC.AP + AC.BP,

or, in the notation in Figure 1(i),

az = ax + ay.

Normat 3/2009 Andrew Percy and D. G. Rogers 117

Hence,

z = x + y;

that is, (1) follows quickly as an easy exercise on Ptolemy’s Theorem. Thus, de-

pending on our prior knowledge, van Schooten’s Theorem might not detain us long.

(i)

(ii)

Figure 1: Van Schooten’s Theorem

However, recollecting the historical context, it is also curious to note how (1) is

suggested fairly readily in considering the geometry of a more celebrated problem,

the challenge issued by Fermat in the early 1640s to those who, like Descartes,

doubted his methods to ﬁnd a point in a triangle that minimises the sum of the

distances to the vertices. The late Folke Eriksson (1927–1999) contributed an in-

formative and erudite account [9, (a)] on this subject to Normat in 1991 (see also

[14, (a)] and [1, (c) §6.2]). Drawing on this article or otherwise, it is known that

if P is the required point in the triangle 4ABC

, as in Figure 2(i), then in the

ﬁrst place the sides of the triangle subtend equal angles at P . But, secondly, as

a construction for P , if an equilateral triangle 4ABC is placed externally on the

edge AB of 4ABC

, then P is the point in this triangle where the circumcircle of

4ABC intersects CC

, as shown in Figure 2(ii). Moreover, the length of CC

the minimum sought by Fermat. Consequently, in the knowledge of this answer to

Fermat’s challenge, (1) follows immediately.

Van Schooten had been much closer personally to Descartes than to Fermat (com-

pare [15, pp. 58, 25, 67, 56]). Indeed, Descartes in conversation with van Schooten

had tended to put Fermat down, saying Fermat had discovered “many pretty, spe-

cial things” in contrast to Descartes’ own preference for general results with numer-

ous applications. On the other hand, Fermat had been mortiﬁed that his earlier

eﬀorts went unmentioned when van Schooten issued a reconstruction of Apollo-

nius’ Plane Loci — van Schooten’s student, Christian Huygens (1629–1695), even

118 Andrew Percy and D. G. Rogers Normat 3/2009

C’

(i) characterisation

C’

(ii) construction

Figure 2: Fermat’s Point

expressed the view that much of Fermat’s restoration of this work was thin and

perfunctory. But van Schooten is also known to have collected copies of Fermat’s pa-

pers while in Paris. So, perhaps Fermat’s challenge had not escaped van Schooten’s

notice when he advanced (1).

Certainly, (1) has all the appearance of a “pretty, special thing” of the sort that

typiﬁed Fermat’s work according to Descartes. Yet, as we shall see, it also has an

aspect perhaps more satisfying to Descartes’ tastes. For, a simple proof of this par-

ticular case of Ptolemy’s Theorem pushes through, not only to establish Ptolemy’s

Theorem in full generality, but also to answering Fermat’s challenge. Thus, van

Schooten’s modest case in (1) is actually equivalent to these seemingly more mo-

mentous results.

Some indication of this alternative line of development is pieced together in a couple

of geometry textbooks [10] by Henry George Forder (1889–1981), at the start of the

1930s (the relevant passages are reproduced in Figure 7 by permission of the Forder

Estate). Forder’s approach has an aﬃnity with the celebrated solution to Fermat’s

challenge to be found, for example, in [5, §1.8], but which had then appeared only

a little earlier in a note [13, (a)] by Joseph Ehrenfried Hofmann (1900–1973). Van

Schooten is mentioned neither by Forder nor in the current common reference [16,

(a) §§1.5–6] for the link between Ptolemy’s Theorem and Fermat’s challenge (a

fuller treatment of this link, including the equivalence of Fermat’s principle of least

time for light rays with Snell’s law of diﬀraction, appears in a subsequent book

[16, (b) §24] by the same author). However, Hofmann went on to write with great

authority on the mathematics of both Fermat [13, (c)] and van Schooten [13, (b)],

besides also returning more speciﬁcally to Fermat’s challenge in [13, (d)]

Hofmann’s demonstration might well go back further. For instance, Lothar Wolf-

gang von Schrutka, Edler von Rechtenstamm (1881–1945) presented something of

a prototype [20, (a)] in 1914. Of late, there has been renewed interest in a general-

isation of van Schooten’s theorem noted in 1936 by Dimitrie Pompeiu (1873–1954)

and to which we come in Section 5 (see, [3]; compare also [14, (c)] and [18]). But

Normat 3/2009 Andrew Percy and D. G. Rogers 119

our starting point, in Section 2, is a recent variant of Forder’s observation that

appears in [11].

As it happens, the centenary volume [17] in which this note appears contains an

essay [6] on Ptolemy’s Theorem, although no connection is drawn between these

contributions. More curiously in this respect, the volume includes a brief account

[4] of a result of Ghiyath al-Din Jamsh¯ıd Mas’ud al-K¯ash¯ı (1390–1450) working at

the apogee of late Arabic astronomical computation. Yet, despite al-K¯ash¯ı’s result

being ﬁrmly in Ptolemy’s tradition of table making, it is neither presented nor

recognised as a special case of Ptolemy’s theorem, one indeed already anticipated

by Euclid (?325–?265) in a proposition towards the end of Data [21], as we no-

tice on re-encountering it in passing in Section 3. This ﬂuctuating attention down

the centuries, here encapsulated within a single volume, is suggestive alike of the

workings of mathematical enquiry and of the historiography of mathematics.

Ptolemy presented his theorem on cyclic quadrilaterals in Almagest I.10. The in-

stances where a side or a diagonal is a diameter of the circumcircle were instrumen-

tal in trigonometric computations for astronomical purposes. The classical proof

in terms of similar triangles, touched on in Section 6, is rehearsed in such older

textbooks as [12, 7], and anew with greater attention to visual presentation in [1,

(a)]. But the Law of Cosines may be called in aid, as in [8], to develop algebraic

expressions for the squares of the diagonals in terms of the sides, expressions that

are themselves of considerable antiquity and from which Ptolemy’s Theorem is

an immediate consequence. This argument is turned around with the help of fur-

ther visual aids in [1, (b)] to deduce the result on the diagonals given Ptolemy’s

Theorem.

2 The equilateral triangle

Now, the equilateral triangle is suﬃciently special as to invite further thought in

place of an appeal to another result, however well-known; perhaps ingenuity will

come to our aid, if knowledge is deﬁcient. The introduction of P on the circumcircle

of 4ABC, cuts AP BC into two pieces 4ACP and 4BCP that ﬁt together along

CP . But, since the sides of 4ABC are equal, following [11], it looks as though

we can rotate the triangle 4ACP about C so that AC matches up with BC. As

suggested in Figure 1(ii), this seems to produce a new equilateral triangle 4P P

If so, then (1) will simply express the equality of the sides CP and P P

of this

triangle.

We can achieve the eﬀect of rotation by cutting oﬀ the triangle 4ACP from AP BC

and juxtaposing a triangle 4BCP

congruent with it, thus conserving the angle

at C:

∠P CP

= ∠ACB = π/3.

Now, Elements III.22 tells us that the opposite angles of a cyclic quadrilateral add

to two right angles. In particular, then

∠CBP

+ ∠CBP = ∠P AC + ∠CBP = π.

120 Andrew Percy and D. G. Rogers Normat 3/2009

So, this property in Figure 1(i) translates into the collinearity of P , B and P

in Figure 1(ii), ensuring that our surgery on the cyclic quadrilateral AP BC does

result in a triangle, viz 4P P

C. We also have equal angles abounding, in view of

Elements III.21, that angles subtended by a chord in the same arc of a circle are

equal, and the fact that 4ABC is equiangular:

∠P

P C = ∠BP C = ∠BAC = π/3

and

∠P P

C = ∠BP

C = ∠AP C = ∠ABC = π/3.

Hence, our new triangle 4P P

C is indeed equiangular and so equilateral.

Thus, our sense that dissecting AP BC into two pieces and interchanging them

relative to one another to form another equilateral triangle proves well-founded,

and we can push through to answer van Schooten’s question. Put another way, the

dissection demonstration veriﬁes Ptolemy’s Theorem in the special case of a cyclic

quadrilateral AP BC where 4ABC is equilateral.

3 The isosceles triangle

It is clear that, with results such as Ptolemy’s Theorem, or even Elements III.21

or III.22, we needs must know them in order to be in a position to use them;

knowledge is a common hurdle that everyone has to surmount. But what can be

a source of insight for one person can be an impediment to progress for another.

After all, acquaintance with, say, Ptolemy’s Theorem may not be lacking, but we

still have to have some intuition to employ it in a given instance — sometimes

knowing too much leaves us uncertain where to start.

What makes the dissection argument work so well with the equilateral triangle?

It is important to understand the mechanics involved, if our intuition is not to

become yet another stumbling block for those anxious about thinking through an

answer to van Schooten’s problem for themselves. Rather, approached in a spirit

of critical research, the dissection demonstration might perhaps serve as a building

block encouraging further investigation.

The crucial step in the dissection argument in the previous section is the initial one.

Once we are assured that the sides AC and BC of 4ABC are equal, everything else

falls neatly into place, on account of standard results in circle geometry. But this

assessment means that the demonstration will also work when 4ABC is isosceles

with equal sides AC and BC and so with equal base angles ∠BAC and ∠ABC

(see Figure 2). For, the angle at C remains conserved; P , B and P

continue to be

collinear; and we still have plenty of equal angles:

∠P

P C = ∠BP C = ∠BAC; ∠P P

C = ∠BP

C = ∠AP C = ∠ABC. (2)

Since ∠ABC = ∠BAC, we also have ∠P

P C = ∠P P

C. The upshot is that the

transposition of the two pieces in our dissection produces another isosceles triangle

Normat 3/2009 Andrew Percy and D. G. Rogers 121

(i)

(ii)

Figure 3: Isosceles triangle

4P P

C with equal sides P C and P

C. Moreover, in view of (2), the base angles in

the isosceles triangle 4ABC are equal to the base angles in the isosceles triangle

4P P

C, ensuring that these triangles are similar.

It follows that

AB : AC = P P

: P C; (3)

that is, in the notation of Figure 2,

x + y

. (4)

Hence,

cz = ax + ay,

thereby verifying Ptolemy’s Theorem in the case of a cyclic quadrilateral AP BC

in which 4ABC is isosceles with AC = BC.

We might note, as an instance of the isosceles case, the right kite, where in addition

CP is a diameter of the circumcircle of 4ABC so that ∠CAP and ∠CBP are both

right angles. In this instance, we can also verify Ptolemy’s Theorem by computing

the area of the right kite in two ways, on the one hand because it consists of two

congruent right triangles and on the other because the diagonals are at right angles.

The isosceles case is, in fact, the historical curiosity to which we alluded in the

penultimate paragraph of Section 1. For, in one formulation, it predates Ptolemy’s

Theorem as such, having been given by Euclid in Data. But, in another, it reappears

much later as the “fundamental theorem” of al-K¯ash¯ı, as it is called in [2].

122 Andrew Percy and D. G. Rogers Normat 3/2009

4 The general triangle

This success encourages the further question: what happens if conditions are relaxed

completely and 4ABC is an arbitrary triangle? There is no problem in carrying out

the surgery on AP BC, deleting 4ACP and adjoining a triangle 4A

congruent

with it so that A

C lies along BC (see Figure 3). But now there is no guarantee

that A

will be coincident with B, as was, in eﬀect, the case for equilateral and then

isosceles triangles. However, as before, the angle at C is conserved; and at least we

know that A

is parallel to P B, because opposite angles of a cyclic quadrilateral

add to two right angles. Thus, if we rescale 4A

by a factor of a/b to produce

a similar triangle 4BCP

, then P , B and P

will be collinear (see Figure 3(ii)).

With this extra step, we have once more created a new triangle, 4P P

(i)

P˝

xa/b

(ii)

Figure 4: Ptolemy’s Theorem

Now, the signiﬁcant feature of 4P P

C is that it inherits the angles of 4ABC,

extending our ﬁndings in the special cases of equilateral triangles and then isosceles

triangles. Not only is the angle at C in the former the same as that in the latter

by construction, but also, corresponding to (2), we have

∠P

P C = ∠BP C = ∠BAC

and

∠P P

C = ∠BP

C = ∠A

C = ∠AP C = ∠ABC.

Thus, 4P P

C and 4ABC are similar, since the three angles of one are pairwise

equal to the three angles of the other.

In this more general case, instead of (3), we have

AB : AC = P P

: P C.

Normat 3/2009 Andrew Percy and D. G. Rogers 123

Adopting the notation of Figure 3, this gives

= (

+ y)/z,

which, rewritten as

cz = ax + by, (5)

can then be recognised as a restatement of Ptolemy’s Theorem. So, at this point, our

discussion tips over, from an exploration of special cases, into an alternative proof of

Ptolemy’s Theorem, since now the cyclic quadrilateral AP BC is perfectly general.

But had we known Ptolemy’s Theorem, and been content in that knowledge, we

might never have entered upon this diﬀerent avenue of approach.

5 The general convex quadrilateral

Indeed, we now see that the only thing that goes wrong with the construction in

Figure 3 when the quadrilateral AP BC is not necessarily cyclic is that P , B and

need not be aligned. However, triangles ∆ABC and ∆P P

C remain similar

since

∠ACB = ∠P CP

; AC : BC = P C : P

C, (6)

by construction. So our previous argument goes through except that, instead of

having P P

explicitly, we must resort to the triangle inequality:

P P

≤ P B + BP

with equality if and only if P , B and P

are collinear, that is, if and only if AP BC

is a cyclic quadrilateral, as before. Hence, as an extension of Ptolemy’s Theorem,

(5) becomes an inequality:

cz ≤ ax + by, (7)

with equality if and only if P lies on the arc AB of the circumcircle of ∆ABC

opposite C. It is exactly this extension that provides one approach among others

to answer Fermat’s challenge, as mentioned in Section 1. Since, as we saw there,

the solution to the challenge implies (1), we now see that we can also ﬁnd our way

back from that solution to Ptolemy’s Theorem and this extension as an inequality.

Returning to Figure 1, it is now apparent that, if P is not on the circumcircle of

∆ABC, then P , B and P

are the vertices of a triangle with sides equal to x, y

and z. This triangle has sometimes been named after Pompeiu, in honour of his

work in the 1930s. So, putting the triangle P BP

to work in the last step in our

argument generalises Pompeiu’s triangle into the bargain.

124 Andrew Percy and D. G. Rogers Normat 3/2009

6 The right triangle

Of course, Ptolemy’s Theorem contains within it Pythagoras’ Theorem that, for

a right triangle, the square on the hypotenuse is equal in area to the sum of the

squares on the legs, as can be seen by restricting the cyclic quadrilateral to be a

rectangle. On the other hand, the classical proof of Ptolemy’s Theorem procedes

rather as an outworking of Euclid’s demonstration of Elements VI.31 extending

Pythagoras’s Theorem from squares to similar ﬁgures, in particular similar trian-

gles, similarly situated on the edges of the right triangle, starting with the case

in Figure 5(i) where the cyclic quadrilateral AP BC is a rectangle. In this case,

let L be the foot of the perpendicular from C onto AB. Then, following Euclid,

the triangles 4ACL and 4CBL are both similar to 4ABC and so respectively to

4P CB and 4CP A. From these pairings, we infer that

AL.CP = AC.BP, LB.CP = CB.AP,

which taken together yield

AB.CP = (AL + LB).CP = AC.BP + CB.AP (8)

(i) rectange

(ii) cyclic quadrilateral

(iii) convex quadrilateral

Figure 5: From Pythagoras to Ptolemy

At this juncture, (8) is no more than a restatement of Pythagoras’ Theorem, as

opposite sides of the rectangle are equal, AB = CP , AC = BP , and so are the

Normat 3/2009 Andrew Percy and D. G. Rogers 125

diagonals, BC = AP . The trick in turning (8) into Ptolmey’s Theorem is to allow

L to vary on AB so as to retain one pairing of the similar triangles, say, 4ACL

with 4P CB (suggested by the shading in Figure 5(ii)). For, this ensures that

∠BCL = ∠BCP + ∠P CL = ∠P CL + ∠LCA = ∠P CA, (9)

while ∠CBL = ∠CP A by Elements III.21. Thus, 4CBL and 4CP A, having

pairwise equal angles, remain similar. Hence, we recapture (8) for the general

cyclic quadrilateral, and with it a motivated presentation of the classical proof

of Ptolemy’s theorem. In the spirit of our earlier argument using a single pivot

with scaling, each pair of similar triangles can be seen as a separate pivot with

scaling, although this might seem artiﬁcial without our motivation (in [1, (a)], the

argument precedes a separate treatment of Fermat’s point).

(i)

P˝

ac/b

(ii)

Figure 6: From van Schooten to Pythagoras?

To cap this argument for a general convex quadrilateral after the manner of Section

5, we can take L so that 4ACL and 4P CB are similar (see Figure 5(iii)). We shall

still have, in addition to (9),

AC.P C = LC.BC.

This combination is enough to ensure that 4CBL and 4CP B are similar (compare

(6)). However, L is no longer restricted to lie on AB, so the triangle inequality turns

(8) into (7) (compare [1, §6.1]).

It is of further interest to see what becomes of our proof in Section 4 when AP BC

is the rectangle in Figure 6(i). The result, shown in Figure 6(ii), is that 4P P

is a right triangle similar to 4ABC, with BC the altitude at C dividing 4P P

into two further similar right triangles. This arrangement of similar right triangles

returns us, in eﬀect, to Elements VI.31 (compare Figure 5(i)). At least in theory,

we could start with the ﬁgure for VI.31 and imagine ourselves reversing the steps

to devise the alternative proof of Ptolemy’s Theorem by dissection and rescaling.

However, the right triangle seems a less plausible starting point for this than van

Schooten’s equilateral triangle, perhaps because it requires a greater sense of what

to do next at each stage. Some doors are trap doors.

126 Andrew Percy and D. G. Rogers Normat 3/2009

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Cambridge, 1972). MR0437261.

128 Andrew Percy and D. G. Rogers Normat 3/2009

Figure 7 : Facsimiles from Henry Forder’s books School Geometry (1930) and Higher

Course Geometry (1931), reproduced by permission of the Forder Estate