While we won t be figuring out how to fold any kinds of models, we will see what points we can find using the folds of origami.

Transcription

1 Origami Geometry While we won t be figuring out how to fold any kinds of models, we will see what points we can find using the folds of origami. Origami is the art of folding paper into interesting shapes. We re not actually going to try to figure out what shapes we can get, but we re going to unfold the paper and see where all the lines cross. What points in the plane can we actually identify from folding? First, we need to know what we can do with folding. As usual we will be given some starting points, and identify new points as the intersections between lines that we have folded. But what kinds of folds make sense. Making the following folds should be clear. 1. Given two points, you can fold the paper so that the fold passes through the two points. This basically means that, given two points, you can draw the line through them. We have a straightedge! 2. Given two points, you can fold the paper in such a way that one ends up on top of the other. The fold will have to fall along the perpendicular bisector between these two points. Now it is clear that there are other ways we can sensibly fold the paper according to some geometry rule, but it is amazing just how much we can already do with just these two rules! Theorem Given point P and line segment AB we can translate the line segment to P Q so that P Q AB and P Q = AB. Proof: Using the second rule, find the perpendicular bisector of AB. It intersects AB in point C which is the midpoint between A and B. So our rules allow us to find midpoints! Now assume P is not on the line through A and B. By what we know from the Poncelet- Steiner work, since C is the midpoint between A and B, we can draw the line parallel to AB through P. (Remember, to draw parallels we only needed three evenly spaced points!). Now find the midpoints between A and P and draw the parallel to AP through B. Mark Q at the intersection of these two parallels, and since ABQP is a parallelogram, we have accomplished our task. If P is on the line through A and B, first translate to some random point off the line, and then translate again back to P. Theorem to l. Given a line l and a point P we can construct the line through P perpendicular Proof: Take any two points on l and construct their perpendicular bisector. Now use the previous theorem to construct the parallel to this through P. 1

2 Theorem Given collinear points A, B, and C, and point P, we can construct Q so that A, P, and Q are collinear and AP/P Q = AB/BC. Proof: First, if P is not on the line, construct ABP. Construct the line through C parallel to BQ, which will mee AP at a point Q that satisfies the requirements because of similar triangles. If P is on the line, first choose an arbitrary point P not on the line and find Q, and then transfer back to the line. Theorem Given line l. If we are given a point P, we can reflect it across l. If we are given a line m, we can reflect it across l. Proof: These are equivalent. If we can reflect points, then given a line, pick two points on the line and reflect them, and the ilne through the reflected points is the reflected line. Conversely, if we can reflect lines, then given a point choosse two lines that pass through it and reflect them; the reflection of the point is at the intersection of the reflected lines. Now to reflect a point P across l, simply find the perpendicular to l through P. This meets l at Q. Now translate P Q to start at Q. Now let s say we start with the numbers 0 and 1 in the complex plane. What s constructible? So far, all we can really do is add and subtract numbers. This is because our translation is parallel we have no way to rotate anything. We can t even construct (0, 1) on the y-axis. Now there are certainly other kinds of folds that we ll get to soon that will allow us to do that, but they are much more powerful in terms of what they bring us algebraically, so we don t want to invoke them yet. If we consider our plane to be the complex numbers, then given any two constructible complex numbers as we add and subtract them, by parallel translation and/or reflection across lines. We can project them onlo the real and complex axes, so that if a + bi is constructible, so is just a and just bi. But b might not be constructible, nor ai. That would require rotation! Certainly all the integers are constuctible, but nothing else that can be identified is yet. Now pick any point on the y-axis (the imaginary axis, if we re thinking of the complex plane). Call this point bi. From what we know, we can now construct any lattice point x+yi where x is an integer, and y is an integer multiple of b. Theorem We can multiply and divide by integers. That is, if z is constructible, then so is pz/q where p and q are integers, q 0 of course. Proof: If z is not on the real axis, construct the line through 0 and z, and through 1 and z. Then construct the line through p parallel to this last line; it meets the first line at point pz. Construct the line through this and q on the x-axis. Construct the line parallel to it through 1 on the x-axis; where this last meets the line through 0 and z is pz/q. If z is on the real axis, replace 1, p and q on the real axis with bi, pbi and qbi on the imaginary axis and the proof is complete. 2

3 So now we can add, subtract, multiply, and divide. We can identify all of the real rational numbers, but note that there is still no single complex number we can identify yet since we have no way to get a scale in the imaginary direction. Let s change that. One thing that it should be pretty clear we can fold is that given two lines, we can fold the paper so that they end up on top of each other. This leads to two cases, depending on whether the lines are parallel or not: 3. Given two lines, we can create a fold so that one is placed on top of the other: (a) If the two lines intersect, we can fold either or both bisectors. (b) If the lines are parallel, this means that we can fold the midline between them. Has anybody noticed yet that these are the rules that I gave for constructions in the last project? Did anyone notice that the rule 7) given there has actually been rendered redundant by our work here, as we proved it follows from the other rules? And did anyone notice that the parallel thing above is actually redundant. We could already have created the midline first, make a perpendicular to both parallel lines, then find the perpendicular bisector of the segment that runs between the two lines. This new axiom allows us to move distances into any (constructible) direction. So, for instance, we can now mark a unit distance in the y-direction. In fact: Theorem Axiom 3 is equivalent to being able to take a given length and translate it in any direction. Specifically, given segment AB and ray P Q we can mark point R on the ray so that P R = AB. Proof: First, by our earlier work, parallel translate AB to P C. Now bisect angle QP C. Construct the perpendicular to P C at C and let S be the point at which this perpendicular meets the angle bisector. Draw the perpendicular to P Q through S which meets P Q at R. Then RP S and CP S are both right triangles with common hypotenuse and one equal angle, so P R = P S = AB. Conversely, If we can construct the line segment as required, we can bisect the angle as follows. Again, parallel translate AB to P C and construct the perpendicular through C. Now translate the distance to P R and draw the perpendicular through R. These two perpendiculars meet at S, so are right triangles with the same hypotenuse and one equal side P C = P R, so the corresponding angles RP S and CP S are equal and we have bisected angle CP Q. With this ability, we can now construct all rational complex numbers, a + bi where a and b are rational. These can be added, subtracted, multiplied, and divided. But there are also some more lengths that can be constructed. For instance, we can construct 2 as the segment from 0 to 1 + i. In fact, if x and y are constructible distances, we can also now construct x 2 + y 2 by making a right triangle. Theorem The first three axioms allow us to generate all lengths that are Pythagorean numbers, that is, the smallest set of numbers that can be generated by adding, subtracting, multiplying, dividing, and taking all numbers of the form 1 + a 2 where a is a Pythagorean number. 3

4 Proof: We have already shown we can add, subtract, multiply, and divide. And if a is constructible, then we can construct the points (1, 0) and (0, a), and the distance between these it 1 + a 2. So all these distances are constructible, as advertised. We don t need the full power of the Pythagorean theorem, because a 2 + b 2 = a 1 + (b/a) 2 and if a and b are constructible we can certainly divide b by a and multiply the entire result by a, so we don t get anything we couldn t already get with 1 + a 2. We don t have to worry about the angle bisection giving us a number that is not of this form, either. For let θ be a constructible acute angle, and let a = tan(θ). Then, by drawing a line at angle θ to the x-axis and drawing the vertical line through (1, 0), we see these lines intersect at the point (1, a). One of the versions of the half-angle formula for tangent is tan( θ) = 1+tan 2 (θ) 1. Since these operations are all already things we are considering, we 2 tan(θ) don t need anything new. It can be a little difficult to determine if a number is a Pythagorean number. For example, it is clear 2 is Pythagorean. So is And thus 1 + (1 + 2) 2 = is Pythagorean. Even so, if we had a right triangle with this length as hypotenuse and 2 as one leg, the other leg would be which is not Pythagorean. Try convincing yourself of this fact! So even though we can get a 2 + b 2 we can t always get a 2 b 2. Let s put the next axiom into action. We ve seen it before in the project, and it allowed us to get all the same things that we could get with compass and straightedge. It was: 4. Given a line l and points P and Q, we may fold any line through Q that places P on l. The way this works in terms of folding is to fold the paper over so that P is on l, and slide P along l until the fold passes through Q. Then crease the paper at that place. There are either zero, one, or two possibilities, depending on the relative positions of P and Q to the line. Basically, we consider the parabola with focus P and directrix l. The folds that place P onto l are exactly tangents to this parabola. There are two tangents to the parabola through Q if Q is outside the parabola, one (the tangent line at Q itself) if Q is on it, and none if Q is inside the parabola. The proof that this new axiom gets you exactly the same points as compass and straightedge was the content of the project. Basically, we prove we can construct square roots. Let P be at (0, 1) and l be the line y = 1. Then the parabola with directrix l and focus P has equation y = 1 4 x2. The tangent line to the parabola at the point (a, a 2 /4) has slope a/2 and thus equation y = a(x a)/2 + a 2 /4. Substituting x = 0 we get y = a 2 /4. Working backward, if we want to find the square root of r, place Q at the point (0, r/4). Then the tangent line to the parabola will meet the parabola at x = r. We fold this tangent line and the line y = r/4, which meet on the parabola at ( r, r/4), so we can locate square roots. It is easy to prove we can t get anything else because intersections of lines and parabolas, or of two parabolas, only require solving quadratic equations, so no intersections of constructible lines at this point need anything other than addition, subtraction, multiplication, division, and square roots. Time to introduce the last three axioms of origami the last three ways we can naturally fold paper. Let s see what they get us: 4

5 5. Given a point P and a line l, we can fold a line through P perpendicular to l. 6. Given a point P and lines l 1 and l 2, we can fold a line perpendicular to l 2 that places P on l Given points P and Q, and lines l and m, we may make any single fold that places P on l and simultaneously place Q on m. Now we ve already proven that we can do the first of these, so it doesn t get us anything that we didn t already have. It just makes it more convenient to make certain folds. This is also true of the sixth axiom. Basically, we fold the paper so that the two halves of l 2 overlap, and slide the fold back and forth until P lands on l 1, and then crease the paper. We could have achieved this result by folding a line through P parallel to l 2, finding where it intersects l 1 and marking that point Q, and then finding the perpendicular bisector of P Q. This just allows us to do it faster. The seventh axiom is something entirely new, though. In order to do this, fold the paper over so that P is on l, and slide it along l until Q lands on m. This may or may not actually happen. For instance, if P and Q are unit distance from each other, and the lines l and m are parallel and distance 2 apart, there is no way to do this. On the other hand, there may be as many as three such line, as we will see. If this sliding business reminds you of a neusis, that s because it is! We are essentially verging from line l onto line m with a fixed length ruler. So it will not be surprising that we can now do exactly the same things we could do in neusis constructions basically we can solve all third and fourth degree equations, allowing trisection, construction of the regular heptagon, enneagon, and triskasidecagon, for example. What we are really doing by placing P on l is finding a tangent line to the parabola with focus P and directrix l. So this axiom allows us to find any simultaneous tangents of two parabolas. That s why there might be zero, one, two, or three solutions. These cases might be: y = x 2 and y = x (no common tangent lines); y = x 2 and y = x 2 (the x-axis is the only common tangent); y = x and y = x 2 1 with the two common tangent lines y = 2x and y = 2x; and y = x and x = y with one common external tangent and two common internal tangents. To prove that we can now solve cubics, let a and b be constructible numbers and consider the two parabolas (y a/2) 2 = 2bx and y = x 2 /2. While we can t actually construct these parabolas, we can construct their foci and directrices (they can be found from these equations by arithmetic and square roots). That means we can construct common tangents to them. So let y = µx + β be a simultaneous tangent, to the first parabola at (x 1, y 1 ) and to the second at (x 2, y 2 ). Differentiate (implicitly) the first equation to get 2(y a/2)y = 2b. At the point (x 1, y 1 ) the slope y must equal µ so we find µ = b y 1. Differentiate the second equation to find a/2 y = x. Inserting (x 2, y 2 ) also gives slope µ, so µ = b/(y 1 a/2) = x 2. Also, y 2 = x 2 2/2 = µ 2 /2 and x 1 = (y 1 a/2) 2 /(2b) = bµ 2 /2. Of course, since µ = y 2 y 1 x 2 x 1 we get µ = µ 2 a b 2 2 µ. µ b 2µ 2 5

6 This simplifies to µ 2 + aµ + b = 0. This is the depressed cubic, and we know we can depress any cubic with just arithmetic. So we can solve any cubic equation whose coefficients are constructible (and hence also any fourth degree equation). The axioms we have given for origami constructions are (in a slightly different order) called the Huzita-Hatori or axioms. Sometimes, they are also called the Huzita-Justin axioms. They can be shown to be complete there are no other lines constructible by a single fold by matching up points and lines already found. The proof of this is quite beyond us. It can also be shown that our last axiom contains all the others, that is, all we ever need to be able to do is to fold as in the final axiom, and we can obtain the same results as using all the axioms. For fun, if we allow two or more simultaneous folds, we can get much more! For instance, if you allow yourself to align combinations of points and lines that can t be done with a single fold but can be done with two folds if done at the same time, you can quintisect an angle. This will allow construction of a regular 11-gon, as well as the solution of many (though not quite all) fifth-degree equations. To see how this is done, consult Robert Lang s construction at 6