From Euclid to Spigotry #7b Cubic Equations and Origami

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1 Last modified: August 26, 2011 From Euclid to Spigotry #7b Cubic Equations and Origami Reading Bold, pages Lecture topics 1. Menelaus proof of a lovely theorem I learned this theorem from Wendell Furry in Physics 22 in 196 but only recently learned how to prove it. Draw any triangle ABC. Trisect each side: F is 2 of the way from A to B, D is 2 of the way from B to C, E is 2 of the way from C to A. Construct lines CF, AD, and BE. Prove that the triangle that they bound has 1 of the area of ABC. 7 Tactic 1: Apply the theorem of Menelaus to triangle AF C cut by transversal BE. Tactic 2: If two triangles have bases on the same line and have a common vertex not on that line, their areas are in the same proportion as their base segments. Tactic : The area of the whole is the sum aof the areas of the disjoint parts. Let the area of triangle ABC be 21. Your construction has broken it into four triangles and three quadrilaterals. Show that the triangle in each corner has area 1 and that each quadrilateral has area 5. Then the triangle in the center must have area, and the theorem is proved! 1

2 2. How to prove that a construction is impossible - the key theorem (Bold, page 1) In a construction, we start with points whose coordinates are in the field F 0 of rational numbers. By using a compass once, we obtain points whose coordinates are in an extended field F 1, whose elements are of the form a + b k, where a and b are in F 0 and k is not the square of any element of F 0. By using a compass again, we obtain points whose coordinates are in an extended field F 2, whose elements are of the form a 1 + b 1 k, where a1 and b 1 are now in F 1 and k 1 is not the square of any element of F 1. By using a compass n times, we can obtain points whose coordinates are in an extended field F n, whose elements are of the form a n 1 + b n 1 k, where an 1 and b n 1 are in F n 1 and k n 1 is not the square of any element of F n 1. Theorem: If a cubic equation x + c 2 x 2 + c 1 x + c 0, where the coefficients c i are all in F n 1, has a root in F n, it also has a root in F n 1. Conclusion that will follow by induction (to be proved later) Theorem: If a cubic equation x + c 2 x 2 + c 1 x + c 0, where the coefficients c i are all in F n 1, has a root in F n, it also has a root in F 0 i.e. a rational root. First we must show that if a + b k = u is a root of polynomial p(u), then so is its conjugate a b k = u. In general we have that 0 = 0 = p(u) = p (u) For the case where the cubic has three distinct roots we are basically done. If one of them is in an extended field, then one of the other two roots must be its conjugate. The remaining distinct root must be in the same unextended field as c 1 and c 2. Otherwise its conjugate would also be a root, and the cubic would have more than three distinct roots. 2

3 . Example of finding a root in a less-extended field: Consider x x 2 + x + 1 = 0. You happen to notice that x = is a solution. (a) What happens when you try x = 1 2? (b) What is the sum of the three roots? (c) What is the third root? You are pondering x 2x 2 (1 + 2x (2 = 2) = 0 when a little bird tells you that x = is a solution. (d) What happens when you try x = 1 + 2? (e) What is the sum of the three roots? (f) What is the third root?

4 4. Proof of the key theorem (Bold pages 14 and 15): p, q, k and all the c i are in field F n 1, but x is in the extended field F n. We substitute x = p + q k into the cubic polynomial f(x) = x + c 2 x 2 + c 1 x + c 0 and discover that f(p + q k) = r + s k. (a) What is f(p q k)? (b) If x is a root of f(x) = 0, what is another root? (c) What is the sum of the three roots? (d) In what field does the third root lie? (e) Prove by induction that if p, q, k are in the field F n (i.e. x is constructible) and all the c i are in field F 0, the equation has a rational root. Equivalently, if a cubic equation has no rational root, none of its solutions is constructible. 4

5 5. Trisecting angles algebraically By repeated bisection of constructible angles like 45, 0, and 150, we can construct angles like 22.5, 15, 75 and 7.5. If an angle θ can be constructed, the angle θ can be trisected. That means that the cubic polynomial x x 2 cos θ can be factored as a linear factor times a quadratic factor. These factors may involve square roots, but those are constructible. Divide the class into three groups, and have each group try this approach for the following angles. In each case the group will find that one of the three solutions to the cubic involves only 2 while the other two are in an extended field. The solution to the cubic equation that involves only 2 may correspond to trisecting an angle that is greater than the original angle by either 60 or 720. (a) Since 15 is constructible by bisecting 0, we can trisect 45. (b) Since 45 is constructible by bisecting 90, we can trisect 15. (c) Since 75 is constructible by bisecting 150, we can trisect

6 6. Solving a cubic by angle trisection This the most famous contribution to mathematics of the French Renaissance mathematician Vieta, whom we will meet later. We want to solve the equation x px = q, where p and q are positive and q is not too large. We know that by using Archimedes marked ruler trisection, we can solve 4 cos θ cos θ = cos θ by determining θ given θ. So the goal is to convert the given equation into the one that we can solve. Assume that we can solve the problem by trisecting an angle and multiplying its cosine by a number that we shall call 2a. Set x = 2a cos θ in the original equation to get 8a cos θ 2pa cos θ = q. The best hope will be to cancel out 2a, so set p = a 2 and q = 2a cos θ. This is OK because p is positive and q is small enough that cos θ does not exceed 1. Now 8a cos θ 6a cos θ = 2a cos θ. Factor out 2a to obtain 4 cos θ cos θ = cos θ. To decide what angle to trisect, note that a = p cos θ = q = q. 2a So cos θ = 2( p ) 2 27q 2 4p and that The number whose square root we must take is positive (because p is) and we assume that q is small enough that 27q 2 < 4p. Now construct the angle θ and trisect it using a marked ruler. Alternatively, use a calculator or Mathematica to determine θ and divide the result by. Once we know θ we can determine cos θ, and then x = 2a cos θ = 2 p cos θ. To find all three solutions, we notice that there in addition to the acute angle θ, there are two other angles for which cos θ has the right value, namely θ + 2π and θ + 4π. These will both lead to negative values of cos θ. In principle every step of this solution can be done geometrically! So p and q could be specified as the lengths of line segments. 6

7 7. Axioms for origami A concise statement of the axioms can be found at axioms With a piece of paper, illustrate the seven axioms, and show that the first five correspond to operations that can be carried out with compass and unmarked straightedge. No proofs, please! 8. Pentagon construction by origami Start with a square of paper (there will be enough for the entire class). Show how, by using just the first five axioms of origami, you can create two folds with an angle of 2π between them and locate two adjacent vertices 5 of a regular pentagon. Just carry out the first few steps of Richmond s construction. 7

8 9. Trisecting an acute angle by using origami. The construction was invented by Japanese mathematician H. Abe in Start with two equidistant folds, both parallel to the bottom edge of the square. Now use axiom 6, Given two points p1 and p2 and two lines l1 and l2, there is a fold that places p1 onto l1 and p2 onto l2. Fold so that end d of the marked ruler along the left edge moves atop the lower fold at D, while end b moves atop one arm of the angle that is being trisected, at B. Finally, fold D on top of B to construct the perpendicular bisector of segment BD. The perpendicular bisector of bd passes through D and, by symmetry, the perpendicular bisector of BD will pass through d. The result, as shown in the top right diagram, is three congruent right triangles, each with an angle equal to one-third of the original angle. 8

9 10. Proof for origami angle trisection Notice that you have constructed an isosceles trapezoid (middle diagram), a pair of alternate interior angles (left diagram), and an isosceles triangle (right diagram). Let θ denote the complement of one of the angles that is shaded in the left two diagrams, and show that the two rays in the interior of the angle being trisected make an angle of θ with each other and with the arms of the angle being trisected. 9

10 11. Construct the cube root of 2 by using origami. This construction was published by Peter Messer in First fold a square horizontally into thirds. This is not trivial. Here is a diagram: and here are the instructions, attributed to Tom Hull of Merrimack College. Using a square piece of paper, make two creases by folding over from side to side and along a diagonal (left). Make a third crease that extends from one corner to the midpoint of one side (middle). Where the third crease meets the diagonal crease defines a point P that is on a vertical line dividing the sheet in thirds. A new crease that passes vertically through this point defines the boundary of one-third of the sheet. Hull claims that the proof is easy, and for once such a statement is correct. It is up to you to invent it the proof. One approach is to drop a perpendicular from P and show that it divides the triangle whose apex is P into two triangles whose areas are in the ration 2:1. Therefore their bases have the same ratio. The cube-root construction itself requires a single use of axiom 6, Given two points p1 and p2 and two lines l1 and l2, there is a fold that places p1 onto l1 and p2 onto l2. Make a single fold so that p 1 moves atop the left edge L 1 of the square while p 2 moves atop the horizontal fold L 2. The ratio of AC to BC is then 2 10

11 12. Proof for origami cube duplication (a) Because we started with a square, u+v = 1+z. Apply the Pythagorean theorem to triangle CBD to show that 2u(z + 1) = z 2 + 2z. (b) Right triangles CBD and SP C are similar, so BD CD = CP CS. The square was folded into thirds, so CS = 1+z. The length CP = z AP = 2z 1. The length CD = 1 + z u, and BD = u. Show that zu = (1 + z)(2z 1). (c) Eliminate U algebraically to show that z = 2. 11