– Note: This is the condensed English version of the preceding three posts on that topic.
Some time ago – at the end of January 2010, as we were waiting for her train on a platform at Berlin Hauptbahnhof – my daughter Pamela obtained the rectangular wrapper of a small chocolate bar—the brand is called Kinderschokolade and is made by Ferrero—and said, a friend of hers, Florian, had shown her how to fold that wrapper into an equilateral triangle. We proceeded to deduce the condition for the length-to-width ratio of the paper as well as the folding sequence that makes it possible to turn a rectangular piece of paper into a highly symmetrical triangular shape. Florian also surmised that the dimensions of the wrapper corresponded to the Golden Ratio.
The following gives a condensed version of the three preceding post in this blog which record sequentially the insights gained into the features and properties of the folding process. The analysis shows that
(i) the ratio of length b and width a of a piece of paper from which to fold an equilateral triangle as described below has to be p = b/a = √3 = 1.732...; the dimensions of the wrapper were found to be 99.5 and 57.5 mm, giving a length ratio of the sides of the paper of 1.73 so that the folding would indeed be successful, and
(ii) the conjecture that the paper's length ratio p would be that of the Golden Ratio (or Golden Section) b/a = φ = (1 + √5)/2 = 1.618... is justified but does actually not apply to the case of folding an equilateral triangle.
A) Geometric Analysis
It follows from its three sides being equal that the three angles of an equilateral triangle are also equal, each measuring 60º. If the side ratio of the paper equals √3, that is, if
b = a √3 . . . . . [1]
where AD = a and AB = b, then that angle can be obtained by folding one corner of a rectangular piece of paper diagonally onto the opposite corner as illustrated in the diagram at right: Folding corner D onto corner B (where it is labeled D'), makes a fold XY which forms an angle of 60º with the paper edge XB.
This follows from applying Pythagoras' theorem to the diagonal d = DB to give d² = a² +b² which, for b = a√3 equals a² + 3a² = 4a² or d = 2a. Thus, the diagonal of a paper with side ratio p = √3 is twice the length of side a and therefore, since the sine of the angle ∠ABD is a/d, one has sin∠ABD = 1/2 which makes ∠ABD = 30º. For symmetry reasons, the diagonal DB makes a right angle with the fold XY at point M, therefore ∠YXB = 60º and this is also the value of ∠DYX. Then the apex angle at corner B, ∠XBY must also be 60º.
It is further seen from the diagram, that the short side a of the paper forms the height of the equilateral triangle XBY. One also notes that the new corners X and Y of the triangle divide the sides AB and CD into the sections XB = DY = u and AX = YC = v, respectively. Between these lengths the proportionality equation
b/a = a/v . . . . . [2]
holds because the triangles ABD and MBX (or ADX, etc.) are similar. Between the quantities a and b, however, exists the relationship given in equ.[1] as required for folding the equilateral triangle. From equ.[2] the lengths AX = YC = v can be calculated from a² = v·b = v·a√3 or
v = b/3 . . . . . [3]
It follows that u = 2b/3 so that the parts into which the folding corners X and Y divide the long side b of the rectangle have a length ratio u:v = 2:1.
The height h of the folded triangle is equal to the short side a of the rectangle. All of this also follows from symmetry or directly from the folding method described below in Section C.
B) Comparison to Golden RectangleIts sides are obtained by starting with a line of length AB = b and finding a point X that divides the line into two segments AX = v and XB = u in such a way that
b / u = u / v . . . . . . [4]
and then using the length XB = u as the shorter side BC = a in constructing the Golden Rectangle as is illustrated in the drawing on the right. For u = a, this expression is identical to equ.[2] above, namely b/a = a/v.
The analogy between the two cases is obvious; in particular it justifies Florian's conjecture that the sides of the rectangle with which one can fold an equilateral triangle would have lenghts in the Golden Ratio. This, however, is not actually the case because there exist relationships between the quantities a (or u) and v in the two equations [2] and [4], which are different from each other: For folding the equilateral triangle, the relationship is only between sides a and b of the rectangle as given by equation [1] while for the Golden Rectangle the relationship is between u = a and v which add up to the quantity b as
b = a + v. . . . . . [5]
Inserting this expression into equ.[4] to eliminate v one obtains a² = b·(b – a) which leads to the quadratic equation b² – a·b – a² = 0 that can solved to express b relative to a as
b = a (1 + √5) / 2 = 1.618...·a . . . . [6]
This is different from equ.[1] also numerically. The ratio b/a in equ.[6] is the golden ratio φ.
The relative length of the small section v can be calculated as a fraction of the long side b as
v/b = (3 – √5) / 2 = 0.382... . . . . [7]
and from equ.[6] one has a/b = v/b = 0.618..., both adding up to 1.
C) Folding Methods
1. After folding corner D onto corner B as described above, the apex angle at B is formed when fold YB is made along the partial edge YD' (folding upwards) with corner C falling onto the paper's center M which is located on the middle of the first fold XY. Similarly, folding now corner A also onto M along edge XB (folding downwards) forms the fold DX. This finishes the folding of the equilateral triangle YBX. Both triangle surfaces look identical, each showing one short side of the rectangle as its height.
2. Instead of corner A and C going around the paper edges XB and YD' the folds DX and BY, respectively, can also be made on their own, after opening the paper on fold XY again. Then the corner A and C are on the inside of the triangle, directly touching each other as they both fall onto center M. Closing the paper by folding D onto C again results in an equilateral triangle with both flaps inside.
Any combination of these folds can be used to give the triangle different appearances, especially if folded from a piece of paper with two different surface colors or patterns. Incidentally, the corners X, Y, and B=D' of the equilateral triangle can be further folded to its center so that a regular hexagon is formed.
3. The analysis of the folding of an equilateral triangle from a rectangle with a side ratio of p = √3 has also suggested a way of achieving this with any paper rectangle for which p ≥ (2/3)√3 = 1.1547... . It is based on the fact that either of the short parts AX or YC fall onto the fold XY to form an angle of 60°, and that the corners A and C both lie on the center. Thus, folding any corner of a rectangular piece of paper onto the center line (parallel to the longer edge) in such a way that the fold goes exactly through the adjacent corner creates an angle of 60° in that corner between the fold and the long edge. This is illustrated with the red lines in the diagram at right where corner D is folded as D' onto M on the center line so that the fold goes through corner A. The green and blue lines show the effect of folding too little or too much material, respectively, as the corner D' is moved along the center line. With the correct position of M established, the base fold XY is made by folding the rest of the paper along the edge YD' so that the top edge aligns with the first fold AY, and finishing with folding the remainder of the paper around edge AX, forming again an equilateral triangle. Depending on how long the paper is, the folding needs to continue until the corners B and C (not shown) fall within the area of the triangle. Alternatively, the paper is cut along the line AX after the second fold.
In this way, the equilateral triangle can be folded from any piece of rectangular paper provided it has the minimal side ratio. Of course, this is a somewhat imprecise method since one has to move one corner along the center line while at the same time nudging the fold through the other corners. Also, the center line needs to be established first, for instance by folding the paper parallel to the two long edges or marking it with a pencil. But with some skill, the result will be convincing.
4. The smallest rectangular piece of paper for which a complete equilateral triangle can be obtained is one for which the long edge, b = AB, equals just AY and YB, that is, the other sides of the triangle; here the corner B coincides with X as compared to the preceding diagram. This case is depicted in the graphic at right. From the analysis in part 3 above, and X coinciding here with B, one has YX = 2·YM = 2v which equals b here so that v² + a² = (2v)² = b². This gives v = (a/3)√3 and with b = 2v the side ratio p = b/a = (2/3)√3. It also follows directly from the cosine of ∠YAD = 30° which equals a/b and has a value of √3/2. One obtains b = 2a/√3 which is the same result.
The fact that for this paper format, p = 1.15..., 2v = b also means that point Y divides DC in half so that its position is more easily found by folding corners C and D together and then make folds AY and BY directly, without first finding the center line.
D) General Case: Folding an Isosceles Triangle
The folding scheme discussed in section A) above and illustrated with the first diagram there is also applicable to a rectangular piece of paper with an arbitrary side ratio and will generally result in an isosceles triangle. This is illustrated in the following diagram which depicts the situation after the first fold has been made by bringing the corner D over to corner B creating the fold XY. Broken lines indicate the movement of the corners; after a fold is made, the new position of a corner is indicated by a singly primed letter, and after a second fold by a doubly primed one. Thus, corner A now lies at A' and the quadrangle A'D'YX shows the (blue shaded) underside of the rectangular paper while its red-shaded front face can still be seen in the triangle BCX.
The two equivalent second steps consist of folding corner C over edge YD' creating fold BY and then corner A' over edge XB creating fold XD'. The corners A' and C then fall on positions A" and C", respectively, either within the area of the triangle XBY as shown, or outside of it. In the latter case one continues folding the corners in until they do. The edges XB and YB of triangle XBY are of equal length but in general different in length from edge XY, therefore, XBY is an isosceles triangle.
The relationships between some of the characteristic quantities which appear in the folding process will be analyzed in the following. As before, the ratio of the lengths of the long side, b, and the short side, a, of the folding paper rectangle is designated as the parameter p = b/a.
The first fold—corner D falls as D' onto corner B—establishes the fold angle α at Y as ∠XYD' and also at X as ∠YXB. Since ΔYMD' and ΔDAB are similar, ∠BDA = α and, therefore, tan α = b/a = p. The apex angle β is then formed as ∠XBY by the edge parts XB and YD'. These edges, along which the second folds are made, together with the short edges a = A'D' and BC, respectively, form the edge angle γ = ∠A'D'X = ∠CBY.
Between these angles and lengths the following relations then hold:
2α + β = π and β + γ = π/2. It follows that α = β/2 + γ.
From the diagram one sees further that ∠BYC = β so that cos β = sin γ = v/u, see also below.
The length f of the first fold XY follows from the similarity of the triangles created in the folding process, as expressed by f/d = a/b = 1/p, which can be written as
f = d/p = (a/p) √(p² + 1)
where BD = AC = d = √(a² + b²) = a √(p² + 1) is the diagonal of the paper rectangle. The length u of the two isosceles legs XB and YD' can also be expressed in terms of the side ratio p as
u = (a/2p) (p² + 1) = (a/2) (p + 1/p)
and similarly for the length v of the remaining part of the long edge b, namely v = (a/2)(p – 1/p). The ratio of those two parts into which points X and Y divide the long edge of the rectangle is, therefore,
v/u = (p² – 1) / (p² + 1).
With these expressions the three angles can be directly obtained from the paper format. The fold angle is simply given by α = arctan p. The apex angle β and edge angle γ are derived from β = arccos(v/u) and γ = arcsin(v/u), respectively. Also, since cot β/2 = b/a = p, the apex angle can be calculated directly from β = 2·arctan(1/p).
E) Other Properties of Interest
Some particular values of the parameter p, which describes the shape of the rectangular paper, are briefly discussed here.
• Folding a paper with the trivial value of p = 1, i.e. a square with b = a, gives a diagonal for the first fold XY and doesn't allow the second folds to be made. The apex angle is 90°.
• The paper formats in the DIN A ... C series for which p = √2 require a third folding to reach a triangular shape because after the second folding the corners A" and C" fall beyond the first fold XY and need to be folded around that edge in order to come within the area of the triangle. Only for p ≥ √3 (provided p ≤ 1+√2 = 2.141... as discussed below) two folding steps suffice. The apex angle is 70.2°.
• Folding an isosceles from a rectangle with sides in the Golden Ratio p = φ = (√5 + 1)/2 = 1.618... will also require these third folding steps; its apex angle has a value of 63.4°, only slightly larger than that for the equilateral triangle.
• In this context one can also ask what side length ratio p will make the first fold's endpoint X or Y divide the long edge of the rectangle in the Golden Ratio, that is such that u/v = φ = (√5 + 1)/2 which is equivalent to having v/u = (√5 – 1)/2 = 0.618... . This occurs for p = √[(φ+1)/(φ–1)] or p = √[√5+2] = 2.058... . Here the apex angle is 51.8°.
• The transition at p = √3 = 1.732... is also characterized by the apex angle being equal to the fold angle, β = α, with the triangle being equilateral and the corners C" and A" meeting at the middle of the first fold XY.
• Another interesting transition is reached when the edge angle γ becomes equal to the apex angle β so that γ = β = 45°; this happens for a length ratio of p = (√2 + 1) = 2.414... . Then the corner C" (second folding) falls exactly on the lower long edge (and A" on the upper one). For values of p > (√2 + 1) the corners will extend beyond the triangle area and need to be folded a third time, now around the edges XB and YD', as mentioned above. The larger p gets, the more often another folding sequence must be added to obtain increasingly pointed isosceles triangles.
In addition to the three angles α, β, and γ, the lengths u and v into which the long edges are partitioned, and the length of the first fold XY = f , there are two other quantities of interest: (i) the area A∆ of the folded isosceles triangle, and (ii) the position of corner A" (and C" as its mirror image with respect to the diagonal BD) after the second folding operation.
(i) The area of the triangle is half of that of the rhombus XBYD from which it was obtained by folding D onto B; the area of the rhombus A◊ = f·d/2 = a·u from which
A∆ = (a²/4p) (p² + 1) = (a/2)² (p + 1 / p) .
A simpler expression for this is obtained when the length d of the rectangle diagonal is introduced again, giving A∆ = (d/2)²/p. The areas of the triangles BCY and D'A'X folded over in the second folding sequence are the complements of the rhombus to form the initial rectangle with area Arctg = a·b = a²·p. Relative to this area, the area of the triangle is
Arel = A∆ / Arctg = (p² + 1) / (4p²) = (1 + 1/p²) / 4 = (d / 2b)².
For the case of the equilateral triangle, that is, for p = √3, these values are A∆ = a²/√3 and Arel = 1/3.
(ii) In the second folding step corner C comes to lie at position C" (and A at A" symmetrically with respect to the diagonal). In a coordinate system centered at the mid-point M of the rectangular paper (in the middle of fold XY), and with axes parallel to the rectangle's edges, each having a unit length of a/2, the location of corner C is described by coordinates (p,1). After the second folding point C is at C" where its coordinates are (xs,ys), that is C(p,1) → C"(xs,ys). Corners C" and A" are mirror images of C and A with respect to the folds YD' and XB. Using analytical geometry, the following expressions are found for the coordinates of C" as a function of the shape parameter p of the rectangle:
xs = p (p² – 3)² / (p² + 1)² and ys = [ 4p² – 3(p² – 1)² ] / (p² + 1)² . . . . . [8]
For p = 1, i.e., a square, C" stays on C, that is xs = ys = 1. For the rectangle with p = √3 which gives the equilateral triangle, C" and A" coincide on the center point M, that is xs = ys = 0. For p > √3 the corners C" and A" approach the long sides of the (isosceles) triangle and reach them for p = (√2 + 1) = 2.414... as mentioned above. Point C" then has ys = –1 which is at the lower long edge of the rectangle, coinciding with that of the triangle; its other coordinate at that point is xs = √2 – 1 = 0.414... . For larger values of p the folding of the triangle requires additional folding sequences in order to place the corners within the area of the triangle. The diagram at right shows the position of C" in the xy–coordinate system introduced above. The values of p increase along the curve from the top right to the bottom. The first point C"(1,1) corresponds to p = 1; point C"(0,0) is associated with p = √3; the last point at C"(0.414...,–1) is reached for p = 2.414... as discussed above.
The discussion above assumes that the folding starts with a rectangular piece of paper with the long edge oriented horizontally, as described by a value for the ratio of its sides p = b/a > 1 where b is the length of the horizontal edge. Extending the range of this parameter to smaller values, i.e., p < 1 essentially means to fold a rectangle with its long edge oriented vertically. The folding sequence then gives a mirror image of the shapes obtained with p > 1. This can be also described by saying that folding corner D onto corner B for the first fold when p < 1 is equivalent to folding corner A to corner C when p > 1 (i.e., with a horizontally oriented paper). In any case, the expressions for the coordinates of corner C" given above in equ.[8] are not valid for p < 1 unless one first performs the coordinate transformation of mirroring the x- and y-axes along the +45° direction, a process which essentially restores the horizontal orientation of the longer edge of the rectangular paper.
It would be more useful and much more intuitive to calculate the distances of points C" and A" to the first fold XY and to the symmetry line of the isosceles triangle, i.e., its height. That means setting up the coordinate system parallel to the fold XY with length f and the diagonal DB of length d, using as unit f/2 for both axes. Thus, the corners X and Y of the isosceles triangle will have the coordinates (0, –1) and (0, 1), respectively, and the apex B(=C') lies at (p, 0). This approach is explored in a different blog called Isosceles Acute Triangles.
Another parameter of interest in this coordinate system is the distance of the points A" and C" from the long edges of the isosceles triangle: it determines whether another folding sequence will be needed to finish the triangle, depending on whether those corners fall already inside or still outside of the triangle's area.
F) Epilog
Finally, the question should be asked of the engineer who designed the wrapper for the Kinderschokolade bars why the rectangular piece of material has a side ratio of √3. Is there a standard format for that purpose? Or did s/he consciously select the shape so that an equilateral triangle could actually be folded? I doubt that it happened by pure chance. An inquiry to Ferrero, sent by e-mail from their web site, has so far not resulted in an answer.
Comment: The earlier posts to this blog, see below, constitute the German version of the material above, presented in three installments written as insights into the folding process developed and its formal description evolved. Let me know if there is something else that could be said about this topic ... or if you find any errors.
