Origami Axioms. O2 Given two marked points P and Q, we can fold a marked line that places P on top of Q.

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1 Origai Axios Given a piece of paper, it is possibe to fod ots of different ines on it. However, ony soe of those ines are constructibe ines, eaning that we can give precise rues for foding the without using a ruer or other too. Each fundaenta foding rue is caed an origai axio. When we start with a square piece of paper, we begin with four arked ines (the four edges) and four arked points (the four corners). Any crease created by appying an origai axio to existing arked points and ines is a new arked ine. Any pace where two arked ines cross is a new arked point. There are seven origai axios in a. O1 Given two arked points, we can fod a arked ine connecting the. O2 Given two arked points and, we can fod a arked ine that paces on top of.

2 O3 Given two arked ines and, we can fod a arked ine that paces on top of. O4 Given a arked point and a arked ine, we can fod a arked ine perpendicuar to passing through. O5 Given two arked points and and a arked ine, we can fod a arked ine passing through that paces on.

3 O6 Given two arked points and and two arked ines and, we can fod a arked ine that paces on and on. O7 Given a arked point and two arked ines and, we can fod a arked ine perpendicuar to that paces on. Restrictions on appying these axios O1 The fod exists and is unique for any two distinct points. O2 The fod exists and is unique for any two distinct points. O3 The fod exists and is unique for any two distinct ines. O4 The fod exists and is unique for any point and any ine. O5 The fod does not aways exist, and there can be up to two different fods that satisfy it. In this axio, the point is the focus for a paraboa and the ine is its directrix. The assertion is that we can find a tangent ine for the paraboa through. There are no tangent ines through points in the interior of the paraboa. Therefore, if ies inside the paraboa deterined by and, no tangent fod exists. If is any point outside of the paraboa, two tangent fods exist. If is on the boundary, there is exacty one tangent fod. If ies on, the paraboa is infinitey skinny and has no interior. Thus, in this case, can be anything and this axio becoes equivaent to O4. O6 The fod does not aways exist and it is not unique in genera. In this axio, is the focus for a paraboa with directrix and is the focus for a paraboa with directrix. Since foding a point to a ine aways gives us a tangent to the paraboa they deterine, the action of taking two points to

4 two ines akes the fod a tangent for both paraboas siutaneousy. There are at ost three such tangents for two paraboas, aking this probe a cubic in genera. There can aso be two tangents, one tangent, or no tangents. O7 This axio is equivaent to O6 in the case where one of the points is on one of the ines. O1 through O5 are sufficient to dupicate any straightedge and copass construction Here is how constructibiity works. We are given the points (0,0) and (1,0) in the pane. We want to know which points in the pane can be constructed by straight-edge and copass. These constructibe points are precisey those with coordinates which are soutions to soe equation ax 2 +bx+c = 0, where a, b, and c are integers. Using the quadratic forua, we know that the soutions of this equation are given by where a, b, and c are integers. x = b± b 2 4ac, 2a Thus, we need to verify that we can use the origai axios to add, subtract, utipy, and divide given engths. We aso need to be abe to take the square root of a given ength. Adding and subtracting engths To add two given engths, we need to be abe to copy a ength fro one ine segent to a particuar pace on another. One way to do this is to use O3 to fod the first ine onto the second. This wi pace the segent soewhere on the ine. We now need to ove one end of the segent to a particuar point, and ake the other end of the segent ie in our preferred direction on the ine. It is possibe that the point we are trying to hit ies in the idde of the segent. In this case, we first use O4 to ake a perpendicuar fod through one of the end points, we copy the segent to that part of the ine, and then unfod. Now we have a segent which does not touch the point we want to hit. We use O2 to fod the near endpoint of the segent to the target point. The segent ay or ay not be going in the desired direction. If it is not, we use O4 through the target point to fod the ine segent in the other direction. Notice that by copying one segent to the end point of another segent so that they both ie on the sae ine wi aow us to add the engths. To subtract engths, we need to copy the segent on top of the other one to find the difference in their engths. Mutipication and division To utipy two given segents of ength a and b, we first pace the so that they for an acute ange. We can do this using the copying engths ethods aready discussed. Next, we copy the unit ength segent onto the ine containing segent b so that one end of the unit ength segent ies at the ange vertex. We now use O1 to create a ine fro the end of a to the end of the unit segent. We now need to construct a parae ine through the point at the end of b. We can use O4 twice to do this, for exape. Now we ark the point on the ine containing a which intersects this parae ine. The ength fro the vertex to this point is ab by siiar trianges.

5 We use a siiar procedure to divide a by b. The set-up is the sae, but this tie, we use O1 to connect the end of a to the end of b. Now we construct a parae ine through the end of the unit ength. The point on the ine containing a which intersects this parae is the end point of a segent of ength a/b. Finding a square root A good way to take the square root of a ength n is to ask the paraboa y = x 2 to do it for you. We begin by copying n onto the y-axis. We construct the horizonta ine y = n at this height using O4 twice. Next we ark the focus at (0,1/4) and a point at (0, n) (which are both constructibe). We then use O5 to create a fod through the focus which takes the other endpoint to the horizonta ine. We know the iage point wi be on the paraboa y = x 2 because the distance fro the focus to the iage point is equa to n+1/4, which is aso the distance fro the iage point to the directrix at y = 1/4. The two points on the horizonta ine where the iage point can be are therefore n units away fro the vertica ine x = 0. We know that it is not aways possibe to use O5, so et us consider whether we have used it safey in this construction. When we use O5, we are using the horizonta ine at y = n as the directrix of the paraboa whose tangent ine we are constructing. We are using (O, n) as the focus of this paraboa. The paraboa therefore has vertex at (0,0) and opens downwards. Since the point we are finding a tangent through is at (0, 1/4) the desired tangent ine exists and so the construction is aways possibe. O2, O5, and O6 are the ony essentia axios We used a five of the origai axios in the constructions above, but we ony reay need O2 and O5 to accopish O1, O3, and O4. We have not anayzed the power axio O6 which aows us to trisect anges, doube cubes, and otherwise sove cube roots. However, O7 is reay O6 in disguise, as we pointed out earier. O4 is reay a specia case of O5 as we pointed out before. O1 can be repaced by O2 and O5 together. First use O2 to construct the perpendicuar bisector. Use O5 on the two origina points and the constructed ine to ark two points on the perpendicuar bisector. Now use O2 to bring these points together. This constructs the ine needed for O1. O3 is aso easiy repaced by O2 and O5. If the two ines are parae, we use the O4 version of O5 to construct a perpendicuar to both of the and then use O2 to bring one intersection point to the other. If the two ines are not parae, we first ark the intersection point and a different arbitrary point on one of the ines. We use O5 to ake a fod through the intersection point that brings point onto the other ine. This accopishes O3. O6 reates to a cubic equation See Thoas Hu s artice Soving Cubics with Creases: The Work of Beoch and Li in the MAA onthy, Apri 2011.

Sect 8.1 Lines and Angles

Sect 8.1 Lines and Angles 7 Sect 8. Lines and nges Objective a: asic efinitions. efinition Iustration Notation point is a ocation in space. It is indicated by aking a dot. Points are typicay abeed with capita etters next to the