78 Normat 57:2, 78–89 (2009)

A minimum requiring angle trisection

Trond Steihaug and D. G. Rogers

Institutt for Informatikk, Universitetet i Bergen

PB7803, N5020, Bergen, Norge

trond.steihaug@ii.uib.no

For Audun Holme,

Editor, Normat, 2005–2006,

On his seventieth birthday, 1 December, 2008

1 Paul Nahin’s envelope-folding problem

The solution to a problem can be too perfect, winning our admiration, but not our

engagement. In contrast, a solution that is still rough and ready round the edges

retains the power to draw us in, making us want to try our own hand. This was our

response to reading Paul Nahin’s account of what he calls the “envelope-folding

problem” and to which he devotes a six-page section in his recent book When

Least is Best [10, §3.3]. Perhaps attempting to share our enjoyment with readers of

Normat is ill-advised, merely depriving them of similar pleasurable diversion. But

the stimulus we should most like to transmit, in the hope that readers can resolve

matters further, is that of discovering more than we can explain.

(i)

(ii)

(iii)

Figure 1: Folding a right triangle

Normat 2/2009 Trond Steihaug and D. G. Rogers 79

Nahin’s problem seems simple enough. We have a right triangle 4ABC with right

angle at C, with legs a and b and hypotenuse c, where a

= c

and without loss

of generality we may suppose that a ≤ b, as indicated in Figure 1(i). The vertex C

is folded onto the hypotenuse AB at C

to create a crease UV . The envelope-folding

problem asks: what is the least possible area of the folded right triangle 4U V C?

For instance, we can always fold parallel to the hypotenuse, as in Figure 1(ii).

In this case, 4UV C has an area one quarter that of 4ABC, that is, ab/8. Again,

we can also fold along a diagonal on the inscribed square, as in Figure 1(iii). Since

the inscribed square of 4ABC is well-known to have side s = ab/(a + b), this gives

a folded triangle 4UV C with area s

/2. The inequality between the arithmetic

and geometric means implies that this second area is always at least as large as

the ﬁrst, with equality only in the case of the isosceles right triangle. Symmetry

considerations suggest that, in the isosceles case, the common fold does give the

minimum area of the folded right triangle 4UV C.

(i) "

min

(ii) "

max

Figure 2: Angle bisectors as extreme folds

But otherwise there seems little in the way of intuition to guide us, so we must have

recourse to a more analytical investigation. To this end, Nahin lets θ = ∠CU V be

the angle that the crease UV makes with AC (see Figure 1(i)). If the crease is to

cut oﬀ a triangle 4UV C, then it varies from the internal angle bisector of ∠BAC

to the internal angle bisector of ∠ABC (compare Figure 2). Thus, writing as usual

α = ∠BAC,

≤ θ ≤

. (1)

Nahin then builds expectation through a circuitous derivation of the area ∆ of

4UV C in terms of the parameter θ:

∆ = ∆(θ) =

(ab)

8 cos θ sin θ[a sin θ + b cos θ]

. (2)

Now, it is possible to derive (2) much more directly, as we do in the next section. But

that is not the point. Rather, the minimisation of (2) is easier than Nahin seems to

suggest, amounting to the maximisation of the denominator. Thus, recalling only

double and then treble angle formulae, we go on to locate the minimum of (2)

80 Trond Steihaug and D. G. Rogers Normat 2/2009

subject to (1) with startling simplicity by angle trisection:

θ =

(π/2 + α). (3)

Now it is well-known that, in general, angle trisection involves the solution of a

cubic equation, and is therefore not amenable to construction by straightedge and

compasses. However, any angle can be trisected by paper-folding — an introduction

[11] to paper-folding, including this result, was presented by Benedetto Scimemi

in Normat in 1998. David Cox, who contributes an article [2, (c)] to the present

issue, provides related results in his book [2, §10.3] (there is a connection between

[11] and [2, (c)] via [2, (b)]). So, Nahin’s envelope-folding problem has the happy

outcome that the minimising solution can itself be obtained by folding. Readers

who follow Normat closely may also be interested to note that the results in [11]

also lead to foldings for the roots of the quartic equations arising in the ladder

problems reviewed by Kent Holing in his series [8, esp. Pt. III] (compare [10, §3.4]

and [5, Ex. 18]).

Unfortunately, Nahin misses these developments. For, his purpose in introducing

the problem was to illustrate “how a computer can play a highly useful role in

minimization analyses”. So, having achieved (2) — and challenged his readers to

try the “nasty business” of setting d∆/dθ = 0 — he opts immediately to use

a computer to study the behaviour of ∆(θ) directly, plotting it in the isosceles

case a/b = 1, where we already intuit the answer by symmetry, and in the case

a/b = 1/2, where an answer is less obvious.

Is the neatness of (3) something special to Nahin’s problem? To ﬁnd out, we

investigate some related minimisation problems. For example, in Section 3, we

consider minimising variously the displacement CC

of C and the width UV of the

resulting crease. The former is nice enough, as the displacement is minimised by

folding parallel to the hypotenuse, as in Figure 1(ii), when the displacement is the

altitude and the width is half the hypotenuse. But the latter is of the same order

of diﬃculty as Nahin’s problem, in that it, too, requires the solution of a cubic

equation, but without an answer as neat as (3). Curiously enough, if we switch

to folding A onto BC, then minimising either the area of folded triangle or the

width of the crease require only the solution of quadratic equations, as we show in

Section 4. In terms of the functions involved in these problems, some might have

served Nahin’s interests in computer-aided studies as well as, if not better than,

minimising the area ∆ in the envelope-folding problem. However, Nahin’s choice

seems inspired in the combination it oﬀers of arresting challenge and attractive

outcome.

So, we are left with the puzzle: what is the meaning of the angle trisection in

(3) for minimisation? To point up this question, in Section 5, we leave the reader

to consider what is in eﬀect a limiting case of Nahin’s problem where, for ﬁxed a,

the right triangle becomes a strip of width a — but now, following [5], in doing

without calculus.

Normat 2/2009 Trond Steihaug and D. G. Rogers 81

2 Minimising the area

Our ﬁrst task is to provide a straightforward derivation of (2). Naturally, we should

keep in mind the key property of folding, that the crease UV is the perpendicular

bisector of the displacement CC

. Thus, ∠C

UV = ∠CUV = θ, so that ∠CUC

2θ (see Figure 3).

b-yy

a-x

2"-#

Figure 3: Setting for Sine rule

But ∠CU C

, as an exterior angle of 4AC

U, is the sum of the two opposite angles

∠C

AU = ∠BAC = α and ∠AC

U. Turning this equality around, we deduce that

∠AC

U = ∠CUC

− ∠C

AU = 2θ − α.

This means that, with reference to Figure 3, application of the Sine rule yields

sin α

sin ∠C

sin ∠AC

b − y

sin(2θ − α)

Collecting terms in y gives

b sin α = y(sin α + sin(2θ −α)) = y(sin α + sin 2θ cos α − cos 2θ sin α).

Hence, on recalling that cot α = b/a and deploying the double angle formulae, we

ﬁnd that

y =

2 sin θ(a sin θ + b cos θ)

. (4)

Turning now to the analogous triangle 4BC

V , we see that

∠C

V B = π − ∠C

V C = ∠C

UC = 2θ,

because ∠UC

V = ∠U CV = π/2, while

∠BC

V = π − ∠UC

V − ∠AC

U =

+ α − 2θ.

So, we are also in a position to apply the Sine rule in this second triangle, deriving

thereby an expression for x matching (4):

x =

2 cos θ(a sin θ + b cos θ)

. (5)

82 Trond Steihaug and D. G. Rogers Normat 2/2009

But, in view of Figure 1(i), ∆ = xy/2, so (4) and (5) delivers (2). Thus we are

able to dispense with Nahin’s much longer derivation, if at the price of knowing

the double angle formulae. However, these are also suggestive on turning to the

minimisation problem.

Now, as ab is a constant, (2) is really just a reciprocal. So, minimising ∆(θ) is

equivalent to maximising a multiple of the denominator, say,

f(θ) = 2 sin θ cos θ(a sin θ + b cos θ)

taking care to avoid f (θ) vanishing. Diﬀerentiation will be simpliﬁed if we revert

to double angles and write

f(θ) = sin 2θ(a sin θ + b cos θ)

It follows that

(θ) = 2 cos 2θ(a sin θ + b cos θ)

+ 2 sin 2θ(a sin θ + b cos θ)(a cos θ − b sin θ)

= 2[a(sin θ cos 2θ + cos θ sin 2θ) + b(cos θ cos 2θ − sin θ sin 2θ)](a sin θ + b cos θ)

= 2(a sin 3θ + b cos 3θ)(a sin θ + b cos θ), (6)

making use of the treble angle formulae to achieve the ﬁnal simpliﬁcation in (6).

Consequently, the only turning points of f (θ), with f (θ) non-zero, are given by

tan 3θ = −

. (7)

But since tan α = a/b, we also know that

tan(π/2 + α) = −

Hence, the solution to (7) in the range (1) is given by (3); and this solution does

give a maximun for f(θ), as expected.

What if we have (2), but do not (think to) make use of double and treble angles?

Then there is some choice as to what might stand in place of the term in 3θ in the

product in (6). But if we leave it as

a(3 cos

θ − sin

θ) sin θ + b(cos

θ − 3 sin

θ) cos θ,

then instead of (7), the location of the minimum of ∆(θ) is given by a cubic in

t = tan θ:

+ 3bt

− 3at − b = 0. (8)

Clearly, (8) disguises the angle trisection in (3). Nevertheless, it turns out to be

tractable.

In order to recast (1) in terms of t, it is helpful to recall that the tangents of

the half-angles of a right triangle can be expressed by means of the sides together

with the radius r of the inscribed circle:

tan

b − r

, tan(

−

) =

a − r

Normat 2/2009 Trond Steihaug and D. G. Rogers 83

where r = (a + b − c)/2. It follows that (1) translates to

a + b −c

b + c −a

≤ t ≤

a + c −b

a + b −c

. (9)

On the other hand, since a ≤ b, it is straightforward to check that (8) has a root,

t = t

∆

, say, in the interval [

,1], so satisfying (9). The sign of the constant term in

(8) implies there are either one or three positive real roots. But with three positive

real roots, the cubic would have both its own turning points positive, which is not

the case. Hence the root t

∆

in [

, 1] is, in fact, the unique positive root of (8).

Thus, comparing (8) with (6), at this turning point for f(θ), f

(θ) changes from

positive to negative, ensuring that f(θ) assumes a maximum.

3 Minimising the crease

In obtaining (2) in the previous section, we made use of the fact that the area ∆

of the folded triangle 4UV C is given by ∆ = xy/2. Of course, this depends on the

angle at C being right. But, whenever we fold, the displacement l = CC

and the

width w = UV of the crease are orthogonal. Thus, with reference to Figure 1(i),

we have an alternative expression for ∆:

∆ =

. (10)

Now, since the angle at C is right, w

= x

+ y

. So, from (4) and (5), we ﬁnd

that

w =

2 sin θ cos θ(a sin θ + b cos θ)

, (11)

and then comparison of (2), (10) and (11) gives

l =

a sin θ + b cos θ

. (12)

In (11) and (12), we have two further examples where, as with (2), minimisation

can be eﬀected through maximising a denominator. Of these, the minimisation of

the displacement l in (12) is the simpler and neater. Indeed, diﬀerentiation of the

denominator a sin θ + b cos θ quickly reveals that l attains a minimum when θ = α.

that is, when the crease is parallel to the hypotenuse. Consequently, in this case, C

is the foot of the altitude of 4ABC from C, while U and V are the midpoints of AC

and BC (compare Figure 1(ii)). As sin α = a/c and cos α = b/c, with c

= a

+ b

we infer from (11) and (12) that w = ab/c and l = c/2.

On the other hand, minimisation of w itself proceeds along the lines of our

discussion in the ﬁnal paragraph of the previous section. The upshot is that the

minimum is located by another cubic in t = tan θ:

+ 2bt

− 2at − b = 0. (13)

84 Trond Steihaug and D. G. Rogers Normat 2/2009

Without being alert to the quirk of angle trisection, there would seem little to tell

(8) and (13) apart. In particular, arguing as for (8), (13) has a unique positive root

t = t

, say, which is located in the interval [

,1], so satisfying (9), and gives a

minimum for w in (11). Moreover, these unique positive roots of (8) and (13) are

related by

≤ t

∆

≤ t

≤ 1.

Any one of these inequalities holds with equality only in the case a = b. that is,

the right triangle 4ABC is isosceles, and equality holds throughout. But Nahin’s

computer-aided study might come in more handy with (13), where we miss out on

such a neat relation as (3).

4 Minimising another area

Envelopes make a handy source of right triangles ready for folding. But there is

nothing in the nature of folding to restrict attention to right triangles and readers

might be encouraged to experiment with folding vertices of triangles onto opposite

sides for triangles and vertices of their choice. Entering into this spirit, we stay

with Nahin’s right triangle 4ABC, but now consider folding A onto BC at, say,

, by a crease UV with U on AC and V on AB.

zb-z

Figure 4: Another folding

If A

is on the line segment BC, rather than the line BC produced, the internal

angle bisectors at B and C are inaccessible as folds for a < b, and the crease varies

instead from the perpendicular bisector of AB, when A is folded onto B, to the

perpendicular bisector of AC, when A is folded onto C. Hence, if we let φ = ∠AUV

be the angle that the crease UV makes with CA, as shown in Figure 4, then

− α ≤ φ ≤

. (14)

Further, let O be the point of intersection of the crease U V with the displacement

. Since the crease and the displacement intersect at right angles, the right

triangles 4AA

C and 4AUO are similar, so that

∠AA

C = ∠AU O = ∠AU V = φ. (15)

Normat 2/2009 Trond Steihaug and D. G. Rogers 85

We convert our notation to this new setting, writing ∆ for the area of the folded

triangle 4AUV , l for the displacement AA

, w for the width of the crease UV

and z = AU = A

U. With this understanding, ∆ is given by (10), as before. From

Figure 4, we see that

l sin φ = b, 2z sin φ = l.

Thus, z comes out as:

z =

2 sin

. (16)

Now, arguing on the lines of our derivation of (4) in Section 2, application of the

Sine rule to 4AUV yields, with reference to Figure 4,

sin α

sin ∠UAV

sin ∠AV U

sin(π − α − φ)

that is,

z sin α = w sin(α + φ) = w(sin α cos φ + cos α sin φ).

Noting once more that cot α = b/a and eliminating z by means of (16) leads to

w =

a cos φ + b sin φ

2 sin

φ(a cos φ + b sin φ)

. (17)

Hence,

∆ =

8 sin

φ(a cos φ + b sin φ)

. (18)

It is immediate that l is minimised by folding A onto C. But the functions

in (17) and (18) provide further candidates for computer-aided study of the sort

favoured by Nahin. However, as with the counterpart expressions in the previous

two sections, there is really no problem in locating minimums in (17) and (18) by

maximising the denominators. Perhaps what is most interesting here is that, in

contrast with the cubic equations (12) and (8), these new minimums are located

as the positive roots of a quadratic, meaning that, in some sense, the problems in

this section are simpler — for example, we can construct their solutions by means

of straightedge and compasses, which is not possible for the solution of the cubics.

Thus the minimum of w in (17) is located at the positive root of

− 4bt − 3a = 0,

while that of ∆ in (18) is located at the positive root of

− 3bt − 2a = 0,

where now t = tan φ. It is easy the check that these positive roots are greater than

b/a, in keeping with (14), although it is less easy to recognise the angles that have

these tangents, even in the isosceles case.

86 Trond Steihaug and D. G. Rogers Normat 2/2009

There may be somewhat more transparency here in using as the parameter h =

C. From (15), ht = h tan φ = b. Moreover, if L is the foot of the perpendicular

from O onto AC, then L is the midpoint of AC while OL =

h. Thus, the discussion

in this section can be reformulated in the language of coordinate geometry by means

of h, if so wished. At all events, in terms of h, we ﬁnd that

l =

+ h

, w =

a(b

+ h

)

√

+ h

2b(ah + b

)

, ∆ =

a(b

+ h

)

8b(ah + b

)

Of course, the minimum of l is b. The minimum of w occurs for

= b(

+ 9b

− 3b)/4a;

that for ∆ occurs for

∆

= b(

+ 4b

− 2b)/3a.

For what it is, worth 0 < h

∆

< h

< a.

5 A stripped-down minimisation

Nahin has not been alone in combining folding with minimisation. For example, the

questions of minimising the width of the crease when a corner C of a strip of width

a is folded onto the opposite edge, as in Figure 5, has already achieved a certain

veneration, perhaps because the answer is surprisingly neat: the minimum occurs

when x = 3a/4, that is when tan θ =

√

2/2 (compare (13)). For the folded triangle

in Figure 5, the minimum area occurs when x = 2a/3, that is, when θ = π/6, the

limiting case of (3) as α tends to zero. Clearly, Nahin’s envelope-folding problem

can be seen as extending this setting to allow the line through B onto which C is

folded to be at an acute angle to BC, rather than a right angle.

a-x

Figure 5: The folded strip

The well-known puzzlist Henry Ernest Dudeney (1857–1930) posed this minimum

crease problem in the issue of The Strand Magazine for August, 1919, as part of his

long-running series of Perplexities [3, (a), Prob. 469] that had appeared there since

Normat 2/2009 Trond Steihaug and D. G. Rogers 87

May 1910 (and was to run until his death in 1930). He included it in Modern Puzzles

and How to Solve Them [3, (b) Prob. 139, pp. 55, 144], a collection drawn mainly

from this series that came out in 1926 (for a contemporary review, see [3, (c)]).

However, in this instance no justiﬁcation is given for the answer — the pages from

Dudeney’s book are reproduced in Figure 6 on p. 89. Similarly, Martin Gardner

(1914–) gave the problem a further airing as an aside when discussing paper-folding

in an early column on Mathematical Games [6] in the issue of Scientiﬁc American

[6] for July, 1959. Although the problem has done long service as a staple in calculus

textbooks, a fresh look has been taken recently in [4, 9].

But the questions of minimising the folded triangle and the width of the crease

in Figure 5 are also among the examples in an article [5, Ex. 17 and Fig. 7] by

Trevor James Fletcher (1921–) in 1971. Fletcher’s intent is to show how to dispense

with calculus in working a variety of maximisation and minimisation problems —

an aim of which Dudeney and Gardner would no doubt approve. The “fundamental

principle”, as he puts it, running through his solutions is the inequality between

the arithmetic and geometric means. But he concedes that the folding problems

are “a little tricky” managed this way, and opts to leave them for “enthusiasts to

solve for themselves”. We are happy to pass on this encouragement to readers of

Normat, now with Nahin’s generalisation.

Indeed, there is a further puzzle we should like to include. Folding the corner of

a paper strip over onto the opposite edge has been developed for classroom activity,

not only as a computer-aided exercise in optimisation, for example in [1], but also

as a construction for Pythagorean triples, for example in [7]. However, the triangle

of interest in [1, 7] is not the folded triangle 4UV C, but rather the right triangle

V B. Now, the maximum area of this triangle can be obtained by a simple

direct argument. Let V

be the reﬂection of V in the line through B perpendicular

to BC. Then 4C

B is congruent to 4C

V B. Hence, 4C

V V

is isosceles, with

equal sides x and base 2(a − x), and so with constant perimeter 2a. But among

all (isosceles) triangles with given perimeter, the equilateral triangle has largest

area. Thus, 4C

V B attains its maximum area when 4C

V V

is equilateral, that

is, when x = 2(a − x) or x = 2a/3. So, our further puzzle is to explain why this

maximum area is achieved where the folded triangle 4UV C attains its minimum

area.

Moreover, this agreement is not peculiar to the folded strip, but holds more

generally in Nahin’s envelope-folding problem. For, let Γ = Γ(θ) denote the area of

triangle 4C

V B in Figure 3, so that

Γ = Γ(θ) =

(a − x)x sin 2θ.

From (5), after a little algebra with trigonometric expansions, we ﬁnd that

Γ(θ) =

b sin θ(a sin 2θ + b cos 2θ)

4 cos θ(a sin θ + b cos θ)

Now, if Nahin thought ∆(θ) in (2) was messy to diﬀerentiate, then Γ(θ) looks even

more formidable. Yet, the derivative comes out comparatively cleanly:

(θ) =

(ab)

(a sin 3θ + b cos 3θ)

4 cos

θ(a sin θ + b cos θ)

88 Trond Steihaug and D. G. Rogers Normat 2/2009

Hence, we see that turning points for Γ(θ) are also given by (7), as for ∆(θ), and

we can then verify that Γ(θ) attains a maximum at (3). But this working only

quickens our interest in having some more transparent explanation.

References

[1] S. M. Arnold, Online mathematics resources, available at

hhttp://www.compasstech.com.au/ARNOLD/maths.htmi.

[2] D. A. Cox, (a) Galois Theory (Wiley Interscience, Hoboken, NJ, 2004). MR2119052;

(b) with J. Shurman, Geometry and number theory on clovers, Amer. Math. Monthly,

112 (2005), 682–704. MR2167769; (c) Why Eisenstein proved the Eisenstein crite-

rion and why Schönemann discovered it ﬁrst, Normat, 57 (2009), this issue.

[3] H. E. Dudeney, (a) Perplexities, The Strand Magazine, 58 (August, 1919), 200; (b)

Modern Puzzles and How to Solve Them (C. A. Pearson, London, 1926; 2nd ed.,

1936) (c) review of [3, (b)] by W. Hope-Jones, Math. Gaz., 13 (1927), 337–338.

[4] S. E. Ellermeyer, A closer look at the crease length problem, Math. Mag., 81 (2008),

138–145.

[5] T. J. Fletcher, Doing without calculus, Math. Gaz., 55 (1971), 4–17.

[6] M. Gardner, Mathematical Games: Origami, Scientiﬁc American, 201 (July 1959);

reprinted as Chap. 16 in M. Gardner, The Second Scientiﬁc American Book of Math-

ematical Puzzles and Diversions (Simon and Schuster, New York, NY, 1961), esp.

pp. 144–145; updated as Origami, Eleusis, and the Soma Cube. Martin Gardner’s

Mathematical Diversions. New Martin Gardner Mathematical Library (Cambridge

University Press, Cambridge, UK; Math. Assoc. Amer., Washington, DC, 2008).

MR2441570.

[7] G. Hatch, Note 80.40: Still more about the (20, 21, 29) triangle, Math. Gaz., 80

(1996), 548–550.

[8] K. Holing, På gjengrodde stiger I, Normat, 45 (1997), 62–78; II: tillegg og rettelser,

ibid, 46 (1998), 45; III: geometriske løsninger, ibid, 48 (2000), 83–90; IV: epilog,

ibid, 50 (2002), 92–95. MR1780823.

[9] R. B. Kirchner, The crease length problem revisited, Spring Meeting, MAA North

Central Section, 25 April, 2009; demonstrations available at

hhttp://public.me.com/rkirchne/CreaseLengthProblemi;

hhttp://demonstrations.wolfram.com/ExploringTheCreaseLengthProblemi

[10] P. J. Nahin, When Least Is Best: How mathematicians discovered many clever ways to

make things as small (or as large) as possible (Princeton University Press, Princeton,

NJ, 2004). MR2022170.

[11] B. Scimemi, Algebra og geometri ved hjelp av papirbretting, Normat, 46 (1998),

170–185, 188. MR1682447.

Normat 2/2009 Trond Steihaug and D. G. Rogers 89

Figure 6: Facscimiles from Dudeney’s 1926 book Modern Puzzles and How to Solve

Them.