How to best fold a triangle

We fold a triangle once along a straight line and study how small the area of the folded figure can be. It can always be as small as the fraction 2-2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$2-\sqrt{2}$$\end{document} of the area of the original triangle. This is best possible: For every positive number ε\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varepsilon$$\end{document} there are triangles that cannot be folded better than 2-2-ε\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$2-\sqrt{2}-\varepsilon$$\end{document}.

to students. But rectangles are quite special quadrilaterals. And then we should pass to polyhedra in n 3 dimensions ...

The answer
Theorem 1 Every triangle can be folded along a straight line so that the area of the doubly covered part is the fraction p 2 1 0.4142 of the area of the original triangle. This is best possible in the sense that, given any positive number ", there exist triangles that cannot be folded to yield an area of the doubly covered part larger than the fraction p 2 1 C ".
The area of the folded figure and the area of its doubly covered part sum up to the area of the original triangle. To ask for the smallest area of the folded figure is therefore equivalent to asking for the largest area of the doubly covered part. For the calculations to be made, it is easier to consider the latter quantity, which we shall do from now on. (See, e. g., Proposition 5. In the simple but important estimate in Proposition 7, the folded object has infinite area, so we can only work with the area of the doubly covered part of the strip.) Denote the side lengths of the triangle by A, B and C , with the notation chosen so that A Ä B Ä C .
We shall fold a triangle in two different ways: 1. If we fold along the bisector of the smallest angle, then the relative area of the doubly covered part is B=.B C C /. This quantity is p 2 1 if and only if B C= 2. If we fold along a line which is orthogonal to the longest side, then the best result to be obtained is C 2 =.B 2 A 2 C 2C 2 /. This quantity is p 2 1 if and only if One of (1) and (2) always holds. So to obtain the fraction p 2 1, we need two different kinds of folding: along lines that pass through a vertex and lines that are orthogonal to a side. Since we use only special foldings for these considerations, calculations are not very complicated.
To prove that it is not possible to do anything better, we must consider all foldings, so calculations become more involved. We shall study triangles with vertices at When the line avoids the direction of the longest side as well as a direction orthogonal to the longest side, a simple estimate shows that the area of the doubly covered part cannot be close to p 2 1 (Sect. 9.1 and Fig. 2). The situations when the line is almost parallel or almost orthogonal to the longest side of these thin triangles remain to be studied. This is done in Sects. 9.2 and 9.3, respectively.

The history
Sometime during the years 1997-1999, Bo Senje at Halmstad University College asked a question: How large can the doubly covered part of a folded triangle be? He had seen some student problems on the folding of rectangles but, to his surprise, never any on the folding of triangles. (Now we may suspect that the explanation was that triangles are more difficult than rectangles.) Georgi Mihailov Tchilikov , also at Halmstad University College, worked on the problem from time to time since December 2004. Some time later (according to Bo most probably in 2008) he found formula (19) for foldings along a bisector.
Bo also asked: Will the best folding always go through a vertex of the triangle?
The answer to this second question is in the negative as we shall see: for certain triangles, folding orthogonally to a side gives better results than folding along a bisector (see Remark 1). Also Georgi had found this. His results have not been published.
The questions posed by Bo reached me in April 2005. I did not know about the results proved by Georgi until February 2015.
Independently of what was known in Halmstad, I proved (in 2005 or 2006) the formula for foldings along lines through a vertex, and Martin Herschend then found a triangle where folding orthogonally to a side gives a better result than folding along a line through a vertex.
After having tried many different ways to approach the question -too many to be remembered -I could finally prove in February 2015 that p 2 1 is best possible. The proof was, however, so cumbersome that I did not dare to submit it to any journal. Only in the evening of 2016 October 13 did I find a proof that might convince more than one person. Of course I cannot exclude that there exist simpler proofs.
I thank Bo Senje for having asked these questions -a most important activity and the starting point of any kind of research. He also checked an earlier version of the manuscript and sent me comments.
I am grateful to two anonymous referees, whose comments have led to clarifications in the presentation.
For origamics in general, see [1] and [3]. More recently, [4,5], and [2] studied paper foldings. The first-mentioned article treats mostly folding of rectangles, then of right-angled triangles, and finally of general triangles. However, the best possible result is not obtained. Oswald's paper [5] is about results on paper foldings obtained by Adolf Hurwitz (1859-1919).

Best foldings and the worst best folding
Let T be a triangle, understood as the convex hull of three non-collinear points in the plane, and let area.T / be its area. We shall denote by T .a; b; c/ the triangle with vertices a, b, c.
We denote by Y.'; t/ the half plane defined by and by L.'; t/ its boundary, defined by forming an angle ' with the positive x 1 -axis. Obviously Y.' C 2 ; t/ D Y.'; t/ and L.' C ; t/ D L.'; t/.
Let Y be a half plane and let T OE@Y D T OEL be the reflection of T in its boundary @Y D L. We fold T along L and place the folded object in Y . Then the part which is doubly covered is T \ T OE@Y \ Y . We denote it by T Y . It is in general a quadrilateral. The folded object is .T [ T OE@Y / \ Y , to be denoted by T Y . It is in general a heptagon. See Figs. 1 and 4, respectively.
Depending of the shape of the triangle and the position of the line, some geometric elements can coincide: the doubly covered part can be a triangle; the folded object can have less than seven sides: six, five, four, or even three sides. See Fig. 3 for an example where T Y is a quadrilateral and T Y has five vertices. These special cases need not be treated separately since the calculations in the general case cover them all. Fig. 1 The original triangle T .a; b; c/ has its vertices at a, b and c. We fold along a straight line L.'; t /, which intersects OEa; b in a point p and OEc; a in a point q, and place the folded object to the right of the line. The doubly covered set T Y is then a quadrilateral Q.p; r; s; q/ with vertices at p, r, s, q. The folded object T Y is a heptagon with vertices at p, b, r, a , s, c, q, where a is the reflection of a in the line L.'; t /. In the figure, the parameters ' and t defining the line L.'; t / are chosen to satisfy 0 < ' <˛< 1 2 and a 1 < p 1 D t=sin' so that the relative areas satisfy The problem is to maximize the area of the set T Y which is doubly covered. We define the best folding of T as the largest relative area of the doubly covered object, where the supremum is taken over all possible half planes Y . For each triangle, the supremum is attained for at least one choice of Y . The quantity BF.T / satisfies As already noticed, the second inequality in (6) is an equality if and only if T is isoceles and is realized when L D @Y is the bisector between two equal sides.
How bad can BF.T / be? Let us define the worst best folding WBF as

Standard coordinates
It will often be convenient use the following coordinate system.
The angles at the vertices a; b; c are denoted by˛ÄˇÄ : We shall prove that, in standard coordinates, The right-hand side is larger than p 2 1 for all c 2 . It follows that BF.T / > p 2 1 for all triangles T , and that WBF p 2 1. Thus the first part of the main theorem follows.
The second part of the main theorem, to be proved in Sect. 9, says that the infimum of BF.T / over all triangles T is equal to p 2 1. It follows that WBF D p 2 1. The right-hand side in (9) can be extended to a continuous function of c 2 D [OE.1=2,0/; .1,0/, and this extension takes the value p 2 1 only at c D .1= p 2; 0/. Hence the bad triangles are only those with c close to .1= p 2; 0/. So, the infimum defining WBF is not attained: there is no worst triangle, but a family .T " / ">0 of bad triangles such that BF.T " / tends to WBF as " & 0; in standard coordinates, the vertex c 2 must tend to .1= p 2; 0/ 6 2 .

A calculus of points and lines
As already mentioned, the vertices of a triangle will be called a, b, c. Its angles will be denoted by˛,ˇ, ; its side lengths A Dk b ck 2 , B Dk c ak 2 , C Dk a bk 2 , where k xk 2 D q x 2 1 C x 2 2 denotes the Euclidean norm of a vector x. Often, but not always, we shall choose notation so that A Ä B Ä C . Then also˛ÄˇÄ , and sin˛Ä sinˇÄ sin D sin.˛Cˇ/. We have 0 <˛Ä =3, 0 <ˇ< =2; =3 Ä < : In addition to these inequalities we have˛ÄˇÄ 1 2 1 2˛, so that the admissible pairs .˛;ˇ/ belong to the closed triangle with vertices at .0,0/, . 1 3 ; 1 3 / and .0; 1 2 / with the segment OE.0,0/; .0; 1 2 / removed. We shall need some explicit calculations on lines and reflections in lines. We study half planes Y.'; t/ defined by (3) and lines L.'; t/ defined by (4).
Two different points p and q determine a line L.Â; s/ with K By reflection in the straight line L.'; t/, a point p is mapped to a point p , which is given by

A calculus of areas
We shall use the notation T .'; t/ for the relative area of T Y.';t/ , thus .';t/2 R R: We can take the supremum in two steps: the best that can be obtained if we fold only along lines parallel to a fixed line. If the vertices a, b, c are taken in the positive direction, the triangle T .a; b; c/ has the area where we think of a 2 R 2 as a column vector, a D .a 1 ; a 2 / T , and write A quadrilateral with vertices a, b, c, d will be denoted by Q.a; b; c; d /. If the vertices are taken in the positive direction, it has the area area.Q.a; b; c; d // D 1 2 .jabj C jbcj C jcd j C jdaj/ : (17)

Special foldings
We shall also briefly discuss quantities BF vertex .T /, BF ? .T /, WBF vertex , and WBF ? , defined in analogy with BF.T / and WBF, but when we fold only along lines that pass through a vertex or are orthogonal to a side, respectively. Clearly BF vertex .T /; BF ? .T / Ä BF.T / for every triangle T; yielding WBF vertex ; WBF ? Ä WBF. Actually we have strict inequality here: We shall find families .T vertex " / ">0 and .T ? " / ">0 of bad triangles for these two special foldings.
In the next sections, we shall start by studying some special foldings. We first fold along a straight line which is parallel to a side; see Sect. 5. This kind of folding can never give a better result than 1 3 < p 2 1. To prove that a triangle can always be folded to yield the relative area p 2 1 of the doubly convered set we shall study two special ways of folding a triangle.
We fold along a line which passes through a vertex; see Sect. 6 and Corollary 2. In this case where˛ÄˇÄ are the angles of T . If we fold along a line which is parallel to a bisector, the result is the same, i. e., the supremum of area.T Y / over all half planes with L D @Y parallel to a certain bisector is the same as the supremum of the area over all half planes Y with L D @Y passing through the vertex of that bisector.
The bad triangles for this kind of folding are those with vertices in .0,0/, .0,1/, .c 1 ; c 2 /, where c 1 D Au and c 2 is a small positive number. Here Au is the Golden Ratio, equal to 1 2 .
Then we fold along a straight line which is orthogonal to a side (see Sect. 7). It turns out that the worst best folding WBF ? for these foldings is 1 3 . It can happen that the supremum when we keep L parallel to a given direction which is not a bisector is larger than when L passes through a vertex (see Remark 1).
Combining two of the methods, viz. folding either along lines through a vertex or orthogonally to a side (a choice depending on the triangle), we find that the universal constant WBF is at least equal to p 2 1 0.4142; see (9).

Folding along a line parallel to a side
When folding along a line parallel to a side, it matters whether the angles adjacent to the side are at most 1 2 or whether one is larger then 1 2 . In the first case, the best result is always 1=3, thus smaller than p 2 1; in the second case it is even smaller than 1=3. Thus folding along a line parallel to a side never yields a good result.
Its maximum is attained for It can be observed that by performing a shear, the general case 0 Ä c 1 Ä 1 can be reduced to the special case when c 1 D 1, which means that the line is orthogonal to the side OEa; b. In turn this is equivalent to the special case when c 1 D 1 in Proposition 5. Anyway, the proof is very simple even if we assume only that 0 Ä c 1 Ä 1, and is therefore omitted.
If one of the adjacent angles is larger than 1 2 , the best result is smaller than 1=3: where the half plane Y.0; t/ is defined by x 2 Ä t, is Its maximum is attained for t D c 2 2c 1 C 1 and is T .0/ D T .0; c 2 =.2c 1 C 1// D 1 2c 1 C 1 : Thus, in both cases, The proof of this proposition is also easy and is omitted here. When c 1 % C1, equivalentlyˇ% , the best folding becomes very bad: the relative area of the doubly covered set tends to 0 Corollary 1 If no angle in a triangle is larger than 1 2 , then the best result for the area of the doubly covered object when folding along a line parallel to a side is 1 3 in all three cases.
If one angle, sayˇ, is larger than 1 2 , then folding along a line parallel to OEa; b, OEb; c, and OEc; a yields, respectively, the best result Its maximum is attained for It is clear that, for c 1 Ä t Ä 0, the value of the doubly covered area is proportional to the the square of t c 1 , and similarly proportional to .b 1 t/ 2 for 0 Ä t Ä b 1 . To determine K 1 and K 2 we only have to calculate We see that the side OEa; b reflected in the line L. =2; t/ is parallel to the side OEa; c, and this accounts for the discontinuity in the derivative of T . =2; t/ for t D 0.
It follows that and that the universal constant WBF vertex is equal to WBF vertex D 3 2 1 2 p 5 0.3820.
Since the constant WBF vertex is 3 2 1 2 p 5, the constant WBF is larger than or equal to this number.
Proof. There are three bisectors, and Proposition 4 shows that the best we can do is the maximum of the three quantities is B=.B C C /, A=.A C B/ and A=.A C C /. However, the last quantity is always majorized by each of the other two.  p 5 C 1/, which in turn implies that A C B Ä C , contradicting the strict triangle inequality C < A C B.
The function t 7 ! T . =2; t/ is a spline function of degree 2. Its derivative at the nodes t D 0, 1=2, c 1 , 1 has the values 0, 1=c 1 , 2, and 0, respectively. Its maximum is attained for t D 2c 1 2c 1 C 1 and is T . =2/ D T . =2/ D T . =2,2c 1 =.2c 1 C1// D 1 2c 1 C 1 : By choosing c 1 < 1 close to 1 we can obtain that the best relative area is just a little bit larger than 1 3 , similarly to the result when we fold along a line parallel to a side; see Proposition 2. In fact, the bad triangles for this kind of folding are T ?
" D T .a; b; c/ with c 1 D 1 " and c 2 D " 2 . Then folding along a normal to OEa; b yields 1=.2c 1 C 1/ D 1=.3 2"/ & 1 3 as " & 0. Foldings along normals to OEb; c or OEc; a give worse results since > 1 2 . Therefore WBF ? D 1 3 . If we choose c 1 D 1 2 p 2, the largest relative area is We get the following values of T . =2; t/ for three values of t: By (17), twice the area of Q.p; r; s; q/ is given by the formula which divided by c 2 , twice the area of the triangle, gives the second-degree polynomial in the statement of the proposition. The derivative is zero at t D 2c 1 =.2c 1 C 1/ and this yields the maximum area.
When c 1 1=2 is a small positive number, the second derivative of T with respect to t in the interval OE1=2; c 1 is a large negative number. When c 1 D 1=2, we pass directly from the expression 2t 2 for 0 Ä t Ä 1=2 to 2.1 t/ 2 for 1=2 Ä t Ä 1; there is a jump in the first derivative.
We note that, when c 1 > 1=2, the maximum is not attained when the line passes through the vertex (t D c 1 ) but at a point t to the left of c 1 : 1=2 < t D 2c 1 =.2c 1 C 1/ < c 1 .
If we want to find only the maximum of T . =2; t/, we can observe that, using the notation in the proof, it is attained when c 2 .s 1 t/=c 1 is half of q 2 D c 2 t=c 1 .

Combining two methods of folding
We have seen that folding through a vertex and folding orthogonally against a side sometimes yield good results, but for different shapes. We now combine the two methods.

Theorem 2 Any triangle can be folded along a straight line such that the area of the doubly covered part is at least
p 2 1 0.4142. This bound can be obtained by using one of two methods: folding along the bisector of the smallest angle and folding orthogonally to the largest side. This proves the first part of the main result.
In standard coordinates, the best relative area that can be obtained by these two methods iŝ The values ofˆsatisfy p 2 1 <ˆ.c/ Ä 1 2 . The infimum ofˆis p 2 1. It is not attained, but is the limit as c approaches the boundary point c D . 2, then c 1 < B < 1 2 p 2 and folding along a normal to the side OEa; b gives 1=.2c 1 C 1/ > p 2 1 while folding along the bisector at a yields B=.B C C / < p 2 1. Proof. Let the notation be chosen so that A Ä B Ä C . We shall see that it is enough to fold along a line through a or orthogonally against the side opposite to c.
Consider the two inequalities If we choose coordinates as in (8), these inequalities are equivalent to The first covers the permitted values of c except those in the set with c 2 1 C c 2 2 < 1 2 and c 2 > 0. These points are then taken care of by the second case. If (21) holds, we fold along the bisector through a and obtain from Corollary 2 If (21) does not hold, then (22) and (24) hold, and we can fold along a line orthogonal to the side OEa; b. We obtain 1=.2c 1 C 1/ p 2 1.

K 9 The constant found is best possible
We shall now prove the second part of Theorem 1. This will be done in three steps.
First we shall prove that if the angle ' of the line avoids the set 1 2 Z, then an easy estimate shows that the doubly covered set cannot have a relative area larger than p 2 1 (Sect. 9.1). Next we shall study the behavior of the best folding when the angle is in a neighborhood of zero (Sect. 9.2); finally when the angle is in a 2 and the angle˛is very small. The smaller " in the statement of Theorem 1 is, the smaller we have to take˛. This has the advantage that we can approximate to simplify many expressions, for example all of the quantities c 2 =c 1 D tan˛, sin˛, 1 2 sin2˛can be approximated by˛C O.˛3/ as˛& 0, which means that the error is not larger than a constant times˛3 when˛is positive and sufficently small; in fact for all˛satisfying 0 Ä˛Ä =4. Similarly, cos˛D 1CO.˛2/ as˛& 0. In the sequel we shall use these approximations many times, however without repeating the expression "as˛& 0." When implying that these two special methods of folding give approximately the same result when˛is small. However, this is not sufficient: we now have to consider all kinds of folding.

Folding along a line with angle avoiding 1 2 Z
Proposition 7 Let T be a triangle in standard coordinates (8). With a half plane Y.'; t/ and its boundary L.'; t/ as defined in (3) and (4), assume that ' 6 2 1 2 Z. Then the relative area of the doubly covered set when folding along L.'; t/ is less than c 2 =jsin2'j.
In the inequality for the relative area, we can either fix an angle ' 6 2 1 2 Z and choose a small c 2 to make the inequality area.T Y /=area.T / < c 2 =jsin2'j Ä p 2 1 valid, or fix c 2 and consider the set of angles ' which satisfy the inequality.
This means that when jsin2'j . p 2 C 1/c 2 , the relative area is smaller than p 2 1, implying that only a certain neighborhood of 1 2 Z remains to be studied. This neighborhood is where Á, which depends on c 2 , is given by the conditions 0 < Á < 1 4 and sin2Á D c 2 . p 2 C 1/. When˛is small, the angle Á is approximately equal to . We find that p 1 < q 1 < r 1 and that sin2' D c 2 k q rk 2 D c 2 r 1 s 1 ; where s D .s 1 ; c 2 / is the reflection of q in L.'; t/. The area of T Y.';t/ is therefore less than 1 2 c 2 .r 1 s 1 / D 1 2 c 2 2 =sin2' . The area of T is 1 2 c 2 , so that the relative area is less than c 2 =sin2' , a finite positive number.

Folding with an angle near zero
Consider a triangle T .a; b; c/ with standard coordinates (8) and intersect it with a line L.'; t/. In view of Proposition 7 it is enough to study the lines L.'; t/ with ' either close to 0 or close to 1 2 , more precisely, when ' belongs to the intervals OE Á; Á and OE =2 Á; =2 C Á, where Á is defined after formula (25). To do so, it is enough to use the approximations that follow.
K For ' near 0 and in the neighborhood W .c 2 / defined by (25), we shall study angles ' 2 OE Á; Á. When ˛Ä ' Ä 0, we can use the results on folding along a line parallel to the bisector of the angle at b. Then the best result is see Corollary 2. So this is much worse than the results for positive ' to be discussed now. We study separately the cases 0 Ä ' Ä 1 2˛a nd 1 2˛Ä ' Ä˛. so also t is of the order˛when we study lines with angle ' close to zero that actually cut the triangle.
The relative area, which is a polynomial in t 2 R with coefficients that are rational functions of sines and cosines of˛,ˇand ', can therefore be approximated by a polynomial in t with coefficients that are rational functions of˛and '.
We obtain the following values for the coordinates of the points introduced above.
This is a polynomial in t of order 2, the coefficients of which are rational functions in the trigonometric functions of the angles indicated. Our task is to prove that this expression cannot exceed 2. The points expressed in these constants and variables are: Twice the area of the quadrilateral Q.p; r; s; q/ is 2area.Q.p; r; s; q// D jprj C jrsj C jsqj C jqpj D p 2 .q 1 r 1 / C q 2 .s 1 p 1 / D˛ 1 K so that the relative area is clearly an increasing function on OE0; 1 2 , which implies that on this interval, the supremum of S is equal to the value S. 1 2 / D p 2 1. Since T . ˛/ and S. / differ at most by a constant times˛, this result implies that, for any given positive number ", the largest relative area of T Y is smaller than p 2 1 C " when˛is sufficiently small.
A similar study can be made for 1 2˛Ä ' Ä˛.

Folding with an angle near =2
We shall now study an angle ' 2 W .c 2 / close to 1 2 .
Proposition 9 Let us, in a triangle with standard coordinates (Sect. 4.2), fold along a line L.'; t/ with angle ' satisfying j' 1 2 j Ä˛, and let " be any positive number. Then, for˛small enough, the relative area of the doubly covered set T Y is never larger than p 2 1 C ". Proof. It turns out that we can make a direct comparison of the folding along a line L.'; t/ with the folding along the line L. 1 2 ; q 1 / D L. 1 2 ; q 1 /, where q D .q 1 ; q 2 / is the point where L.'; t/ intersects OEa; c. We already know that the relative area T . 1 2 ; q 1 / is at most equal to T . 1 2 ; 2 p 2/ D p 2 1. The only values of q 1 that are of interest are those that satisfy 1 2 Ä q 1 Ä 1 2 p 2. We introduce D ' 1 2 , a small angle to which q 1 and t are related by the equation the angle at p also being 2 . We define V .˛; ; t/ D .1 p 1 / k r pk 2 k s ? qk 2 k s qk 2 : Then the area of the quadrilateral Q.p; r; s; q/ can be estimated from above as C slightly larger than a right angle and comparing it with folding along the line L. =2; q 1 / with angle equal to =2 and intersecting the side OEc; a in the same point q as L.'; t /. We compare Q.p; r; s; q/ with Q.p ? ; b; s ? ; q/ K