Cracking the Cubic: Cardano, Controversy, and Creasing Alissa S. Crans Loyola Marymount University
MAA MD-DC-VA Spring Meeting Stevenson University April 14, 2012
These images are from the Wikipedia articles on Niccolò Fontana Tartaglia and Gerolamo Cardano. Both images belong to the public domain.
A brief history...
• 400 BC Babylonians • 300 BC Euclid
323 - 283 BC
This image is from the website entry for Euclid from the MacTutor History of Mathematics. It belongs to the public domain.
3 Quadratic Equation • 600 AD Brahmagupta
To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value. Brahmasphutasiddhanta Colebook translation, 1817, pg 346 598 - 668 BC
This image is from the website entry for Brahmagupta from the The Story of Mathematics. It belongs to the public domain.
4 Quadratic Equation • 600 AD Brahmagupta
To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value. Brahmasphutasiddhanta Colebook translation, 1817, pg 346 598 - 668 BC ax2 + bx = c
This image is from the website entry for Brahmagupta from the The Story of Mathematics. It belongs to the public domain.
4 Quadratic Equation • 600 AD Brahmagupta
To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value. Brahmasphutasiddhanta Colebook translation, 1817, pg 346 598 - 668 BC √4ac + b2 b 2 x = − ax + bx = c 2a
This image is from the website entry for Brahmagupta from the The Story of Mathematics. It belongs to the public domain.
4
7 Quadratic Equation
• 800 AD al-Khwarizmi
• 12th cent bar Hiyya (Savasorda) Liber embadorum
• 13th cent Yang Hui 780 - 850
This image is from the website entry for al-Khwarizmi from The Story of Mathematics. It belongs to the public domain.
5 Luca Pacioli
1445 - 1509 Summa de arithmetica, geometrica, proportioni et proportionalita (1494)
This image is from the Wikipedia article on Luca Pacioli. It belongs to the public domain.
Challenge: Solve the equation ax3 + bx2 + cx + d = 0
7 Cubic Equation
Challenge: Solve the equation ax3 + bx2 + cx + d = 0
The quest for the solution to the cubic begins!
7 Cubic Equation
Challenge: Solve the equation ax3 + bx2 + cx + d = 0
The quest for the solution to the cubic begins!
Enter Scipione del Ferro...
7 Scipione del Ferro
• 1465 - 1526, Italian • Chair of math dept at University of Bologna
8 Scipione del Ferro
• 1465 - 1526, Italian • Chair of math dept at University of Bologna
3 • First to solve depressed cubic: x + mx = n
8 Scipione del Ferro
• 1465 - 1526, Italian • Chair of math dept at University of Bologna
3 • First to solve depressed cubic: x + mx = n
• Kept formula secret!
8 Scipione del Ferro
• 1465 - 1526, Italian • Chair of math dept at University of Bologna
3 • First to solve depressed cubic: x + mx = n
• Kept formula secret! • Revealed method to student Antonio Fior on deathbed
8 Nicolo of Brescia (Tartaglia)
This image is from the Wikipedia article on Niccolò Fontana Tartaglia. It belongs to the public domain.
9 Nicolo of Brescia (Tartaglia)
• 1500 - 1557, Italian
This image is from the Wikipedia article on Niccolò Fontana Tartaglia. It belongs to the public domain.
9 Nicolo of Brescia (Tartaglia)
• 1500 - 1557, Italian • Feb 13, 1535 solved x3 + mx2 = n
This image is from the Wikipedia article on Niccolò Fontana Tartaglia. It belongs to the public domain.
9 Nicolo of Brescia (Tartaglia)
• 1500 - 1557, Italian • Feb 13, 1535 solved x3 + mx2 = n • Won challenge!
This image is from the Wikipedia article on Niccolò Fontana Tartaglia. It belongs to the public domain.
9 Girolamo Cardano
• 1501 - 1576, Italian
This image is from the Wikipedia article on Gerolamo Cardano. It belongs to the public domain.
10 Girolamo Cardano
• 1501 - 1576, Italian • Numerous ailments when young
This image is from the Wikipedia article on Gerolamo Cardano. It belongs to the public domain.
10 Girolamo Cardano
• 1501 - 1576, Italian • Numerous ailments when young
• Became a physician
This image is from the Wikipedia article on Gerolamo Cardano. It belongs to the public domain.
10 Girolamo Cardano
• 1501 - 1576, Italian • Numerous ailments when young
• Became a physician • Wrote treatise on probability
This image is from the Wikipedia article on Gerolamo Cardano. It belongs to the public domain.
10 Girolamo Cardano
• 1501 - 1576, Italian • Numerous ailments when young
• Became a physician • Wrote treatise on probability • Brought Tartaglia to Milan to learn secret of the cubic
This image is from the Wikipedia article on Gerolamo Cardano. It belongs to the public domain.
10 The (encoded) solution!
When the cube and things together Are equal to some discreet number, Find two other numbers differing in this one. Then you will keep this as a habit That their product should always be equal Exactly to the cube of a third of the things. The remainder then as a general rule Of their cube roots subtracted Will be equal to your principal thing
11 The (encoded) solution!
In the second of these acts, When the cube remains alone, You will observe these other agreements: You will at once divide the number into two parts So that the one times the other produces clearly The cube of the third of the things exactly. Then of these two parts, as a habitual rule, You will take the cube roots added together, And this sum will be your thought.
12 The (encoded) solution!
The third of these calculations of ours Is solved with the second if you take good care, As in their nature they are almost matched. These things I found, and not with sluggish steps, In the year one thousand five hundred, four and thirty. With foundations strong and sturdy In the city girdled by the sea.
13 The (encoded) solution!
This verse speaks so clearly that, without any other example, I believe that your Excellency will understand everything. - Tartaglia
14 The (encoded) solution!
This verse speaks so clearly that, without any other example, I believe that your Excellency will understand everything. - Tartaglia
I swear to you, by God's holy Gospels, and as a true man of honour, not only never to publish your discoveries, if you teach me them, but I also promise you, and I pledge my faith as a true Christian, to note them down in code, so that after my death no one will be able to understand them. - Cardano
14 The (encoded) solution!
This verse speaks so clearly that, without any other example, I believe that your Excellency will understand everything. - Tartaglia
I swear to you, by God's holy Gospels, and as a true man of honour, not only never to publish your discoveries, if you teach me them, but I also promise you, and I pledge my faith as a true Christian, to note them down in code, so that after my death no one will be able to understand them. - Cardano
15 The (encoded) solution!
This verse speaks so clearly that, without any other example, I believe that your Excellency will understand everything. - Tartaglia
I swear to you, by God's holy Gospels, and as a true man of honour, not only never to publish your discoveries, if you teach me them, but I also promise you, and I pledge my faith as a true Christian, to note them down in code, so that after my death no one will be able to understand them. - Cardano
16 Lodovico Ferrari
• 1522 - 1565, Italian
17 Lodovico Ferrari
• 1522 - 1565, Italian • Started out as Cardano’s servant
17 Lodovico Ferrari
• 1522 - 1565, Italian • Started out as Cardano’s servant
• Quickly became colleagues
17 Lodovico Ferrari
• 1522 - 1565, Italian • Started out as Cardano’s servant
• Quickly became colleagues
• Cardano reveals Tartaglia’s secret solution
17 Lodovico Ferrari
• 1522 - 1565, Italian • Started out as Cardano’s servant
• Quickly became colleagues
• Cardano reveals Tartaglia’s secret solution • Together solved general cubic and quartic!
17 Cardano and Ferrari • Due to oath, could not publish their work!
18 Cardano and Ferrari • Due to oath, could not publish their work! • Traveled to Bologna seeking del Ferro’s original work (1543)
18 Cardano and Ferrari • Due to oath, could not publish their work! • Traveled to Bologna seeking del Ferro’s original work (1543)
• Found solution to depressed cubic!
18 Cardano and Ferrari • Due to oath, could not publish their work! • Traveled to Bologna seeking del Ferro’s original work (1543)
• Found solution to depressed cubic! • Cardano publishes Ars Magna in 1545
18 Cardano and Ferrari • Due to oath, could not publish their work! • Traveled to Bologna seeking del Ferro’s original work (1543)
• Found solution to depressed cubic! • Cardano publishes Ars Magna in 1545 • Chapter XI “On the Cube and First Power
Equal to the Number”
18 Ars Magna
In our own days Scipione del Ferro of Bologna has solved the case of the cube and first power equal to a constant, a very elegant and admirable accomplishment...In emulation of him, my friend Niccolo Tartaglia of Brescia, wanting not to be outdone, solved the same case when he got into a contest with his [Scipione’s] pupil, Antonio Maria Fior, and, moved by my many entreaties, gave it to me.
19 Ars Magna
In our own days Scipione del Ferro of Bologna has solved the case of the cube and first power equal to a constant, a very elegant and admirable accomplishment...In emulation of him, my friend Niccolo Tartaglia of Brescia, wanting not to be outdone, solved the same case when he got into a contest with his [Scipione’s] pupil, Antonio Maria Fior, and, moved by my many entreaties, gave it to me.
20 Ars Magna
For I had been deceived by the world of Luca Paccioli, who denied that any more general rule could be discovered than his own. Notwithstanding the many things which I had already discovered, as is well known, I had despaired and had not attempted to look any further. Then, however, having received Tartaglia’s solution and seeking for the proof of it, I came to understand that there were a great many other things that could also be had. Pursuing this thought and with increased confidence, I discovered these others, partly by myself and partly through Lodovico Ferrari, formerly my pupil.
21 Ars Magna
For I had been deceived by the world of Luca Paccioli, who denied that any more general rule could be discovered than his own. Notwithstanding the many things which I had already discovered, as is well known, I had despaired and had not attempted to look any further. Then, however, having received Tartaglia’s solution and seeking for the proof of it, I came to understand that there were a great many other things that could also be had. Pursuing this thought and with increased confidence, I discovered these others, partly by myself and partly through Lodovico Ferrari, formerly my pupil.
22 Ferrari vs. Tartaglia
• Public debate on August 10, 1548
23 Ferrari vs. Tartaglia
• Public debate on August 10, 1548 • Refereed by Governor of Milan
23 Ferrari vs. Tartaglia
• Public debate on August 10, 1548 • Refereed by Governor of Milan
• Each posed 62 problems
23 Ferrari vs. Tartaglia
• Public debate on August 10, 1548 • Refereed by Governor of Milan
• Each posed 62 problems
• Ferrari wins
23 Ferrari vs. Tartaglia
• Public debate on August 10, 1548 • Refereed by Governor of Milan
• Each posed 62 problems
• Ferrari wins There is a right-angled triangle, such that when the perpendicular is drawn, one of the sides with the opposite part of the base makes 30, and the other side with the other part makes 28. What is the length of one of the sides?
23 Cardano’s Solution Method to solve x3 + mx = n:
Cube one-third the coefficient of x; add to it the square of one-half the constant of the equation; and take the square root of the whole. You will duplicate [repeat] this, and to one of the two you add one-half the number you have already squared and from the other you subtract one-half the same. Then, subtracting the cube root of the first from the cube root of the second, the remainder which is left is the value of x.
24 Cardano’s Solution Method to solve x3 + mx = n:
Cube one-third the coefficient of x; add to it the square of one-half the constant of the equation; and take the square root of the whole. You will duplicate [repeat] this, and to one of the two you add one-half the number you have already squared and from the other you subtract one-half the same. Then, subtracting the cube root of the first from the cube root of the second, the remainder which is left is the value of x.
25 Cardano’s Solution Method to solve x3 + mx = n:
Cube one-third the coefficient of x; add to it the square of one-half the constant of the equation; and take the square root of the whole. You will duplicate [repeat] this, and to one of the two you add one-half the number you have already squared and from the other you subtract one-half the same. Then, subtracting the cube root of the first from the cube root of the second, the remainder which is left is the value of x.
26 Cardano’s Solution
3 n n2 m3 3 n n2 m3 x = + + − + + 2 4 27 − 2 4 27
27
7 Cardano’s Solution
28 Cardano’s Solution
3 Vol of Pink Cube = u
28 Cardano’s Solution
3 Vol of Pink Cube = u 3 Vol of Green Cube = (t - u)
28 Cardano’s Solution
3 Vol of Pink Cube = u 3 Vol of Green Cube = (t - u) Vol of Clear and Blue Slabs
= 2tu(t - u)
28 Cardano’s Solution
3 Vol of Pink Cube = u 3 Vol of Green Cube = (t - u) Vol of Clear and Blue Slabs
= 2tu(t - u) 2 Vol of Yellow Block = u (t - u)
28 Cardano’s Solution
3 Vol of Pink Cube = u 3 Vol of Green Cube = (t - u) Vol of Clear and Blue Slabs
= 2tu(t - u) 2 Vol of Yellow Block = u (t - u) Vol of Red Block = u(t - u)2
28 Cardano’s Solution
3 Vol of Pink Cube = u 3 Vol of Green Cube = (t - u) Vol of Clear and Blue Slabs
= 2tu(t - u) 2 Vol of Yellow Block = u (t - u) Vol of Red Block = u(t - u)2
Total Volume (simplified): t3 - u3 = (t - u)3 + 3tu(t - u)
28 Cardano’s Solution Make a clever substitution in: t3 - u3 = (t - u)3 + 3tu(t - u)
29 Cardano’s Solution Make a clever substitution in: t3 - u3 = (t - u)3 + 3tu(t - u) Let x = t - u to obtain: x3 + 3tux = t3 - u3
29 Cardano’s Solution Make a clever substitution in: t3 - u3 = (t - u)3 + 3tu(t - u) Let x = t - u to obtain: x3 + 3tux = t3 - u3 This is depressed where m = 3tu and n = t3 - u3.
29 Cardano’s Solution Make a clever substitution in: t3 - u3 = (t - u)3 + 3tu(t - u) Let x = t - u to obtain: x3 + 3tux = t3 - u3 This is depressed where m = 3tu and n = t3 - u3.
Solving for u in the first gives u = m/3t and substituting this into the second gives: n = t3 - m3/27t3
29 Cardano’s Solution
Multiplying n = t3 - m3/27t3 by t3 produces: t6 - nt3 - m3/27 = 0
30 Cardano’s Solution
Multiplying n = t3 - m3/27t3 by t3 produces: t6 - nt3 - m3/27 = 0 which we can rewrite as: (t3)2- n(t3) - m3/27 = 0
30 Cardano’s Solution
Multiplying n = t3 - m3/27t3 by t3 produces: t6 - nt3 - m3/27 = 0 which we can rewrite as: (t3)2- n(t3) - m3/27 = 0
This is a quadratic!!
30 Cardano’s Solution
(t3)2- n(t3) - m3/27 = 0
31 Cardano’s Solution
(t3)2- n(t3) - m3/27 = 0
The quadratic formula gives solutions for t. Then we use n = t3 - u3 to solve for u and finally use x = t - u to solve for x. Thus, we have:
31 Cardano’s Solution
(t3)2- n(t3) - m3/27 = 0
The quadratic formula gives solutions for t. Then we use n = t3 - u3 to solve for u and finally use x = t - u to solve for x. Thus, we have:
3 n n2 m3 3 n n2 m3 x = + + − + + 2 4 27 − 2 4 27
32
7 Ars Magna Chapter XI: example illustrating technique for x3 + 6x = 20
33 Ars Magna Chapter XI: example illustrating technique for x3 + 6x = 20
Chapter XII: solved x3 = mx + n
33 Ars Magna Chapter XI: example illustrating technique for x3 + 6x = 20
Chapter XII: solved x3 = mx + n
3 Chapter XIII: solved x + n = mx
33 Ars Magna Chapter XI: example illustrating technique for x3 + 6x = 20
Chapter XII: solved x3 = mx + n
3 Chapter XIII: solved x + n = mx
But what about the general cubic: ax3 + bx2 + cx + d = 0
33 General Cubic ax3 + bx2 + cx + d = 0
34 General Cubic
ax3 + bx2 + cx + d = 0
The key is to make a clever substitution:
x = y - b/3a
34 General Cubic
ax3 + bx2 + cx + d = 0
The key is to make a clever substitution:
x = y - b/3a
This results in a depressed equation
y3 + py = q
34 Negative Roots
Puzzle: But what about negative roots?
35 Negative Roots
Puzzle: But what about negative roots?
Example: Find the roots of x3 - 15x = 4
3 3 x = 2+√ 121 2+√ 121 − − − −
35
7 Negative Roots
(2 + √ 1)3 =8+12√ 1 6 √ 1 − − − − − =2+11√ 1 − =2+√ 121 −
Rafael Bombelli 1526 - 1573
This image is from the website entry for Rafael Bombelli from the MacTutor History of Mathematics. It belongs to the public domain.
36
7 Negative Roots
(2 + √ 1)3 =8+12√ 1 6 √ 1 − − − − − =2+11√ 1 − =2+√ 121 −
Rafael Bombelli 3 3 1526 - 1573 x = 2+√ 121 2+√ 121 − − − − This image is from the website entry for Rafael Bombelli from the MacTutor History of Mathematics. It belongs to the public domain.
36
7
7 Negative Roots
Plus times plus makes plus Minus times minus makes plus Plus times minus makes minus Minus times plus makes minus Plus 8 times plus 8 makes plus 64 Minus 5 times minus 6 makes plus 30 Minus 4 times plus 5 makes minus 20 Rafael Bombelli Plus 5 times minus 4 makes minus 20 1526 - 1573
37 Negative Roots
Plus by plus of minus, makes plus of minus. Minus by plus of minus, makes minus of minus. Plus by minus of minus, makes minus of minus. Minus by minus of minus, makes plus of minus. Plus of minus by plus of minus, makes minus. Plus of minus by minus of minus, makes plus. Minus of minus by plus of minus, makes plus. Minus of minus by minus of minus makes minus. Rafael Bombelli 1526 - 1573 √ x = “plus of minus” −
38
7 Quartic Equation
Puzzle: What about the quartic? ax4 + bx3 + cx2 + dx + e = 0
39 Quartic Equation
Puzzle: What about the quartic? ax4 + bx3 + cx2 + dx + e = 0
Step One: Divide by a and make a substitution to obtain a depressed equation:
y4 + my2 + ny = p
39 Quartic Equation
Puzzle: What about the quartic? ax4 + bx3 + cx2 + dx + e = 0
Step One: Divide by a and make a substitution to obtain a depressed equation:
y4 + my2 + ny = p
Step Two: Replace this by a related cubic, then use previous techniques.
39 Origami Solution
Elementary Moves:
40 Origami Solution
Elementary Moves:
Given two points P and Q, we can make a crease line that places P onto Q when folded.
40 Origami Solution
Elementary Moves:
Given two points P and Q, we can make a crease line that places P onto Q when folded.
Given a line l and point P not on l, we can make a crease line that passes through P and is perpendicular to l.
40 O1: Given two points P1 and P2, we can make a crease line that places P1 onto P2 when folded. O2: Given a line l and a point P not on l, we can make a crease line that passes through P and is perpendicular to l. For more information on these basic moves see [13] and [17]. Note, however, that the above two basic moves can also be done by a straightedge and compass. The one basic folding move which sets origami apart from straightedge and compass constructions is the following: Beloch Fold
The Beloch Fold. GivenGiven two two points points PP1andand PP 1andand two two lines lines l l 1andand l l2 we can, when- ever possible, make a single fold that places1 P2 onto l and P1 onto l2 simultaneously. we can, whenever possible,1 make a1 single 2fold that2 (See Figure 1.) places P1 onto l1 and P2 onto l2 simultaneously.
P1 P2 l2
l1
Figure 1. The Beloch origami fold. 41
One way to see what this fold is doing is to consider one of the point-line pairs. If we fold a point P to a line l, the resulting crease line will be tangent to the parabola with focus P and directrix l (the equidistant set from P and l). This can be demonstrated by the following activity: Take a piece of paper, draw a point P on it, and let the bottom edge of the paper be the line l. Then fold P to l over and over again. An easy way to do this is to pick a point on l and fold it up to P, unfold, then pick a new point on l and fold it to P, and repeat. After a diverse sampling of creases are made, the outline of a parabola seems to emerge. Or, more precisely, the envelope of the crease lines seems to be a parabola. (See Figure 2(a).) A proof of this can be established as follows: After folding a point P on l to P, draw a line perpendicular to the folded image of l, on the folded flap of paper from P to the crease line, as in Figure 2(b). If X is the point where this drawn line intersects the crease line, then we see when unfolding the paper that the point X is equidistant from the point P and the line l. (See Figure 2(c).) Any other point on the crease line will be equidistant from P and P and thus will not have the same distance to the line l. Therefore the crease line is tangent to the parabola with focus P and directrix l.
X X P P P
l llP (a) (b) (c)
Figure 2. Folding a point to a line creates tangents to a parabola.
308 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 118 Beloch Fold What is this fold accomplishing?
P
42 Beloch Fold What is this fold accomplishing?
P
43 Beloch Fold
This image was created using the java applet from the website Cut The Knot.
44 Beloch Fold
This image was created using the java applet from the website Cut The Knot.
45 Beloch Fold
These crease lines are tangent to the parabola with focus P and directrix l. This image was created using the java applet from the website Cut The Knot.
46 Review of Parabolas
focus
directrix
This image is from the Wikipedia article on parabola. It belongs to the public domain.
47 Beloch Fold Picture Proof:
P
48 Beloch Fold Picture Proof:
P
The Beloch fold finds a common tangent to two parabolas!
48 Beloch Fold Morals:
Folding a point to a line is equivalent to solving a quadratic equation.
49 Beloch Fold Morals:
Folding a point to a line is equivalent to solving a quadratic equation.
The Beloch fold, then, is equivalent to solving a cubic equation.
49 Margherita Piazzolla Beloch
• 1879 - 1976, Italian • Algebraic geometer, Chair at Univ. of Ferrara
50 Margherita Piazzolla Beloch
• 1879 - 1976, Italian • Algebraic geometer, Chair at Univ. of Ferrara • First to discover origami can find common
tangents to two parabolas!
50 Margherita Piazzolla Beloch
• 1879 - 1976, Italian • Algebraic geometer, Chair at Univ. of Ferrara • First to discover origami can find common
tangents to two parabolas! • Beloch fold is most complicated paper-folding move possible
50 Margherita Piazzolla Beloch
51 Beloch Square Beloch Square: Given two points A and B and two lines r and s in the plane, construct a square WXYZ with two adjacent corners X and Y lying on r and s, respectively, and the sides WX and YZ, or their extensions, passing through A and B, respectively.
52 Beloch Square Beloch Square: Given two points A and B and two lines ar lineand ss .in (See the Figureplane, 5,construct left.) Note a square that these lines r and s can be constructed easily WXYZ withvia two paper adjacent folding corners by, say, X folding and Y along lyingr , marking where A lands under this fold, and on r and sthen, respectively, making a sequenceand the sides of perpendicular WX and folds O2 described above. (The details of YZ, or theirthis extensions, are left as anpassing exercise.) through A and B, respectively.
A r A r A X r r r r B B A B A s s s s Y s s B B 53 Figure 5. Constructing the Beloch Square using origami.
We then perform the Beloch fold, folding A onto r and B onto s simultaneously. (See Figure 5, center.) This will fold A to a point A on r and B onto a point B on s. The crease made from this fold will be the perpendicular bisector of the segments AA and BB. Therefore, if we let X and Y be the midpoints of AA and BB, respectively, we have that X lies on r and Y lies on s because of the way in which r and s were constructed. The segment XY can then be one side of our Beloch square, and since AX and BY are perpendicular to XY, we have that A and B are on opposite sides, or extensions of sides, of this square.
3. CONSTRUCTING √3 2. Next we will see how Beloch’s square allowed her to construct the cube root of two. (Actually, what follows is her construction set on co- ordinate axes.) Let us take r to be the y-axis and s to be the x-axis of the plane. Let A ( 1, 0) and B (0, 2). Then we construct the lines r to be x 1 and s to be = − = − = y 2. Folding A onto r and B onto s using the Beloch fold will make a crease which crosses= r at a point X and s at a point Y . Consulting Figure 6, if we let O be the origin, then notice that OAX, OXY, and OBY are all similar right triangles. This follows from the fact that XY is perpendicular to AA and BB.
rr 3 B A B s –2 X 1 A Y s –4 –3 –2 –1 O 1234 –1
–2 B –3
Figure 6. Beloch’s origami construction of the cube root of two.
Therefore, we have OX / OA OY / OX OB / OY , where denotes the length of the segment. Filling| | | in OA|=| 1| and| OB|=| 2| gives| | us OX |·|OY / OX | |= | |= | |=| | | |=
310 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 118 Beloch Square Beloch Square: Given two points A and B and a line s. (See Figuretwo lines 5, left.) r and Note s in that the these plane, lines constructr and s a cansquare be constructed easily via paper folding by, say, folding along r, marking where A lands under this fold, and WXYZ with two adjacent corners X and Y lying then making a sequence of perpendicular folds O2 described above. (The details of this are left as anon exercise.) r and s, respectively, and the sides WX and YZ, or their extensions, passing through A and B, respectively. A r A r A X r r r r B B A B A s s s s Y s s B B
54 Figure 5. Constructing the Beloch Square using origami.
We then perform the Beloch fold, folding A onto r and B onto s simultaneously. (See Figure 5, center.) This will fold A to a point A on r and B onto a point B on s. The crease made from this fold will be the perpendicular bisector of the segments AA and BB. Therefore, if we let X and Y be the midpoints of AA and BB, respectively, we have that X lies on r and Y lies on s because of the way in which r and s were constructed. The segment XY can then be one side of our Beloch square, and since AX and BY are perpendicular to XY, we have that A and B are on opposite sides, or extensions of sides, of this square.
3. CONSTRUCTING √3 2. Next we will see how Beloch’s square allowed her to construct the cube root of two. (Actually, what follows is her construction set on co- ordinate axes.) Let us take r to be the y-axis and s to be the x-axis of the plane. Let A ( 1, 0) and B (0, 2). Then we construct the lines r to be x 1 and s to be = − = − = y 2. Folding A onto r and B onto s using the Beloch fold will make a crease which crosses= r at a point X and s at a point Y . Consulting Figure 6, if we let O be the origin, then notice that OAX, OXY, and OBY are all similar right triangles. This follows from the fact that XY is perpendicular to AA and BB.
rr 3 B A B s –2 X 1 A Y s –4 –3 –2 –1 O 1234 –1
–2 B –3
Figure 6. Beloch’s origami construction of the cube root of two.
Therefore, we have OX / OA OY / OX OB / OY , where denotes the length of the segment. Filling| | | in OA|=| 1| and| OB|=| 2| gives| | us OX |·|OY / OX | |= | |= | |=| | | |=
310 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 118 Beloch Square a line s. (See Figure 5, left.) Note that theseBeloch lines r Square:and s can Given be constructed two points easily A and B and via paper folding by, say, folding along r, markingtwo lines where r andA lands s in the under plane, this construct fold, and a square then making a sequence of perpendicular foldsWXYZ O2 with described two adjacent above. (Thecorners details X and of Y lying this are left as an exercise.) on r and s, respectively, and the sides WX and YZ, or their extensions, passing through A and B, respectively. A r A r A X r r r r B B A B A s s s s Y s s B B 55 Figure 5. Constructing the Beloch Square using origami.
We then perform the Beloch fold, folding A onto r and B onto s simultaneously. (See Figure 5, center.) This will fold A to a point A on r and B onto a point B on s. The crease made from this fold will be the perpendicular bisector of the segments AA and BB. Therefore, if we let X and Y be the midpoints of AA and BB, respectively, we have that X lies on r and Y lies on s because of the way in which r and s were constructed. The segment XY can then be one side of our Beloch square, and since AX and BY are perpendicular to XY, we have that A and B are on opposite sides, or extensions of sides, of this square.
3. CONSTRUCTING √3 2. Next we will see how Beloch’s square allowed her to construct the cube root of two. (Actually, what follows is her construction set on co- ordinate axes.) Let us take r to be the y-axis and s to be the x-axis of the plane. Let A ( 1, 0) and B (0, 2). Then we construct the lines r to be x 1 and s to be = − = − = y 2. Folding A onto r and B onto s using the Beloch fold will make a crease which crosses= r at a point X and s at a point Y . Consulting Figure 6, if we let O be the origin, then notice that OAX, OXY, and OBY are all similar right triangles. This follows from the fact that XY is perpendicular to AA and BB.
rr 3 B A B s –2 X 1 A Y s –4 –3 –2 –1 O 1234 –1
–2 B –3
Figure 6. Beloch’s origami construction of the cube root of two.
Therefore, we have OX / OA OY / OX OB / OY , where denotes the length of the segment. Filling| | | in OA|=| 1| and| OB|=| 2| gives| | us OX |·|OY / OX | |= | |= | |=| | | |=
310 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 118 Constructing the Cube Root of 2
r
A s
B
56 Constructing the Cube Root of 2
r r’
s’
A s
B
57 Constructing the Cube Root of 2
r r’
s’
A s
B
58 Constructing the Cube Root of 2
r r’
A’ B’ s’
A s
B
58 Constructing the Cube Root of 2
r r’
A’ B’ s’ X A Y s
B
59 Constructing the Cube Root of 2
r r’
A’ B’ s’ X A Y s O
B
61 Constructing the Cube Root of 2
r r’
A’ B’ s’ X A Y s O
OAX B
61 Constructing the Cube Root of 2
r r’
A’ B’ s’ X A Y s O
OAX B OXY
61 Constructing the Cube Root of 2
r r’
A’ B’ s’ X A Y s O
OAX B OXY OBY
61 Constructing the Cube Root of 2
r r’
A’ B’ s’ X A Y s O
OAX OX OY OB B OXY OA =OX =OY OBY
62 Constructing the Cube Root of 2
r r’
A’ B’ s’ X A Y s O
OAX OX OY 2 = = B OXY 1 OX OY OBY
63 Constructing the Cube Root of 2
r r’
A’ B’ s’ X A Y s O
3 OY 2 (OX) =OX . . OX OY B = 2
64 Epilogue Puzzle: What about the quintic?
65 Epilogue Puzzle: What about the quintic?
65 Epilogue Puzzle: What about the quintic?
Does there exist a “solution by radicals,” that is, a formula for its roots that involves only the original coefficients and the algebraic operations of addition, subtraction, multiplication and division?
65 Epilogue
• 250 years since quartic solved
Paolo Ruffini 1765 - 1822
This image is from the Wikipedia article on Paolo Ruffini. It belongs to the public domain.
66 Epilogue
• 250 years since quartic solved • 1790’s sends work to Lagrange
Paolo Ruffini 1765 - 1822
This image is from the Wikipedia article on Paolo Ruffini. It belongs to the public domain.
66 Epilogue
• 250 years since quartic solved • 1790’s sends work to Lagrange The algebraic solution of general equations of degree greater than four is always impossible. Behold a very important theorem which I believe I am able to assert (if I do not err): to Paolo Ruffini present the proof of it is the main reason for 1765 - 1822 publishing this volume. The immortal Lagrange, with his sublime reflections, has provided the basis of my proof.
This image is from the Wikipedia article on Paolo Ruffini. It belongs to the public domain.
66 Epilogue
• 250 years since quartic solved • 1790’s sends work to Lagrange • Sends work to Institute of Paris
and Royal Society
Paolo Ruffini 1765 - 1822
This image is from the Wikipedia article on Paolo Ruffini. It belongs to the public domain.
67 Epilogue
• 250 years since quartic solved • 1790’s sends work to Lagrange • Sends work to Institute of Paris
and Royal Society
... if a thing is not of importance, no Paolo Ruffini notice is taken of it and Lagrange 1765 - 1822 himself, “with his coolness” found little in it worthy of attention.
This image is from the Wikipedia article on Paolo Ruffini. It belongs to the public domain.
67 Epilogue
Geometers have occupied themselves a great deal with the general solution of algebraic equations and several among them have sought to prove the impossibility. But, if I am not mistaken, they have not Niels Abel succeeded up to the present. (1824) 1802 - 1829
This image is from the Wikipedia article on Niels Henrik Abel. It belongs to the public domain.
68 Epilogue
Geometers have occupied themselves a great deal with the general solution of algebraic equations and several among them have sought to prove the impossibility. But, if I am not mistaken, they have not Niels Abel succeeded up to the present. (1824) 1802 - 1829
Why does Abel get the credit?
This image is from the Wikipedia article on Niels Henrik Abel. It belongs to the public domain.
68 Epilogue
... the mathematical community was not ready to accept so revolutionary an idea: that a polynomial could not be solved in radicals. Then, too, the method of permutations was too exotic and, it must be conceded, Ruffini's early account is not easy to follow. ... between 1800 and 1820 say, the mood of the mathematical community ... changed from one attempting to solve the quintic to one proving its impossibility...
69 References • Dunham, W. Journey through Genius: The Great Theorems of Mathematics, John Wiley & Sons: New York, 1990, 133 - 154
• Hull, T. “Solving Cubics with Creases: The Work of Beloch and Lill,” American Mathematical Monthly Vol. 118, No. 4 (April 2011), 307 - 315
• MacTutor History of Mathematics
70