In this chapter we will search for a way to describe numerically locations of points in two-dimensional space and study the properties of space itself.
Planes
To begin with, let’s try to construct a new structure that goes beyond straight lines. To do that we will use a flat table as a model.
If we choose any two points on a table and stretch a string in such a way that it touches both, it’s clear that all points along the string are on the table too. We will use this observation as a defining feature of a plane.
To construct a plane, we simply choose three points that aren’t members of the same straight line and declare them to be on the same plane. By passing straight lines through any two points that already belong to the plane, we identify other members of this plane.
Note that the shortest distance between two points A and B on the surface of a globe measured along the globe’s surface also constitutes a “straight” segment according to our definition (if we ignore the postulate 2). This segment is depicted in the picture below with a black line linking A with B:
You may point out that the actual shortest distance between A and B corresponds to the red line instead. Flat creatures inhabiting the surface may not be aware of the third dimension though. They can perform measurements only along the curved surface. Can we be sure that our space is flat? Maybe it’s an illusion. Note that postulate 2 is not good enough to eliminate this possibility because curved space may be infinite and undulated.
Let’s try to find some features that would define flatness. If we draw any straight line on a table, it divides all locations on this table into two subsets located on the opposite sides of the demarcating line. To get from one side to the other without losing touch with the table, we must cross the boundary line at least in one location.
Do the planes constructed according to the recipe laid out in this section have the property described? Observation suggests they do. Let’s postulate the following:
Postulate 3: If a straight segment linking point P with Q intersects some straight line k that is on the same plane as P and Q, then P and Q are on the opposite sides of k i.e. every route on the plane that links P and Q must also intersect k. If on the other hand the straight segment PQ doesn’t intersect k, both points are on the same side of k.
What does it have to do with flatness though? Imagine that the black lines in the picture below are perceived to be straight (they correspond to curved lines projected on a flat plane of the picture). Even though PQ intersects line k, there exists a route (the one in red) that doesn’t intersect k so P and Q are on the same side of k. Postulate 3 eliminates this possibility and is in agreement with observations so we will assume it to be true.
Coordinates of points on a flat plane
Let’s pick three points P0, P1, P2 constituting a plane and some other point P that is on this plane too. We want to answer the question: what is the location of P with respect to P0, P1 and P2.
Of course, we can measure the distance from P0 towards P. That would be our first coordinate.
If we could somehow measure the inclination between the segments P0P1 and P0P, that would give us the second coordinate and the ability to pinpoint any location on the plane with two numbers.
Inclinations
Let’s examine every possible inclination between two segments. To do that, grab a compass and draw a circle with the radius equal to the length of a straight segment linking two arbitrary points P0 and P1. If we connect the centre of the circle P0 with some point P on the circle, we get a specific inclination between the segments P0P1 and P0P:
Every point P on the circle corresponds to a different inclination and every possible inclination is covered. Now, we have to find a way to express this somehow in terms of numbers. Before doing that, let’s discuss the approach we are going to adopt moving forward.
The Method
We could make myriads of observations just by playing with geometrical shapes on a plane. Instead of cataloging them all, we will aim to reduce everything down to a handful of definitions and postulates. The rest must simply follow as logical consequences. This way we reduce complexity and may even stumble on some underlining structure we didn’t expect.
Every claim either relies on the validity of some other claims or is assumed or verified to be true.
In pure mathematics, we assume a handful of assertions as self-evident and study their consequences. Here we are interested in models of reality, so we must verify our claims against the physical world. We have already adopted a handful of definitions and postulates, and we will be adding more but only when it is absolutely necessary. The goal is to have as few of them as possible.
Initially, deriving obvious results from definitions and postulates is going to feel overly pedantic and awkward, yet once we establish the method for simple geometry, the same method will work in the areas where intuition fails completely, and the approach presented here will be the only way to move forward.
Circles
A circle is defined as a collection of points on the surface with the shortest distance to some point O being equal to some arbitrary length r. Circles will be denoted with the symbol C(O,r) where O is circle’s centre and r is the radius.
Let’s derive results of a handful of observations using previously accepted definitions and postulates.
If on a straight line we grab some point O, there are only two points on this line that are r units away from the point O. One in the positive, the other in the negative direction.
In other words, a circle with its centre located on a straight line must intersect this line in two points only.
If two lines intersect, they must intersect in only one point because otherwise there would be two shortest distances between points of intersection and according to the postulate 1, that is not possible.
Postulate 1 is not enough though, because of the situations like this:
If the shortest straight line between A and C goes through ABC, and the shortest line between A and D happens to contain ABD (both measured in space with constraints indicated in black), then half of both lines coincide entirely. We want to exclude curved spaces, so we have to add another postulate:
Postulate 4: If two straight lines “a” and “b” pass through the same two points A and B (A≠B), then “a” must be identical to “b”.
With this in place we are certain that if two straight lines intersect, they must intersect in one point only.
Note that, if the line k intersects some other line h on the same plane, then according to the postulates 3 and 4, all points of k cannot be on the same side of h. Half of k must be on one side and the other half on the opposite side of h. By the same token, if between k and h we draw some line f, it must occupy two quarters that are exactly opposite to each other (quarters 1 and 3 in the picture):
It’s all dead obvious but instead of trusting our senses, we have derived everything from postulates and definitions.
Let’s keep adding (on the same plane) more and more lines that intersect point O. If on each line we mark points that are r units away from O, the resulting points will belong to the same circle C(O,r). It looks like the inclination between straight lines passing through: OP0 and OP1 could be described by the length of the arc P0P1 of the circle C(O,|OP0|):
Measuring lengths of arcs is somewhat cumbersome. Let’s find some more user-friendly way to pinpoint inclinations.
Inclinations, part II
In this section we will start with experiments. Grab a piece of paper, pencil and circle. Draw a straight line and two circles with their centres on the line. Depending on the radii, there are three possible scenarios: circles do not intersect at all, they intersect in one point or we have two intersections, one below and one above the line:
If we change the radius of one circle, the intersections are in different places. From this it follows that if the radius of the first circle is fixed |P0P1|=1, inclinations between line k and other lines going through P0 could be pinpoint with the radius of the second circle. There is a problem though. In the situation in the picture |P1P2|=|P1P3|=r so according to the measure of inclination we just adopted, inclination between P0P2 and P0P1 is the same as the inclination between P0P3 and P0P1, and our aim was to pinpoint inclinations uniquely. Perhaps one could be positive and the other negative. How to distinguish between + and – in this case though? Plane is constructed around initial three points: P0, P1 and P2 which can be used to define a positive direction:
Depending on which side of the plane we look at, the sequence P0,P1,P2 looks clockwise or counter-clockwise. In agreement with a dominant convention, we will consider counter-clockwise to be the front side, and positive rotation will be counter-clockwise as seen on the front side of the plane.
What’s left is figuring out whether all of that follows from the postulates we already have.
Distance between a circle and a point
In agreement with the method adopted, please remember to suspend the judgement of your eyes. We will use them only to confirm the facts established by reasoning based on the accepted postulates. Let’s start with figuring out what is the shortest distance between a point and a circle.
Draw a circle C(O,r) and pick some point P in such a way that |PO|>r where |PO| denotes the shortest distance between P and O. Now we can measure the shortest distance d=|PP’| between P and point P’ on the circle C(O,r).
C is the point where the straight segments PO intersects the circle. Lower case letters in the picture represent the shortest distances between points: r=|P’O|=|CO|, s=|CP|. Distance r+d≥r+s because r+s is the shortest route between O and P. From this it follows that d≥s, so the shortest distance between P and C(O,r) must be equal to s=|CP|. If P was on the circle, then the distance between P and C(O,r) would be obviously zero.
What if |OP|<r?
The shortest distance between O and P’ is r, so r≤|OP|+d. We also know that r=|OP|+s, so |OP|+s≤|OP|+d, thus s≤d. Conclusion: the shortest distance between P and C(O,r) must be equal to s=r-|OP|=|PC| where C, P and O are on the same straight segment.
Intersection between two circles.
To begin with, let’s figure out when circles can intersect. Consider the situation when segment AB is longer than the radius R:
As proven in the previous section, for every point P of the circle C(A,R), the segment PB is not shorter than the segment CB: |CB|≤|PB|. If r<|CB| then the shortest segment between point P and the circle C(B,r) must be on the segment linking P with B. Because r<|CB|≤|PB| the circles C(A,R) and C(B,r) cannot intersect in this scenario: |PP’|=|PB|-r>0.
If on the other hand r=|CB|, then C(A,R) and C(B,r) intersects only in point C because for every other point P on C(A,R), its distance to C(B,r) is: |PP’|=|PB|-|CB|>0.
Also, note that If r≤|CB|, then r+R≤|AB| and the shortest segment linking C(A,R) and C(B,r) must coincide with AB and be equal to |AB|-R-r.
Next, let’s take a look at the situation when |AB|<R:
In this case the reasoning is almost the same as in the previous, so I will just summarize the findings. |BC|≤|BP| so if r<|BC|, the circles cannot intersect:
r<|BP|, |PP’|=|BP|-r>0 and the shortest segment linking them coincides with AC and its length is equal to |BC|-r. If on the other hand r=|BC|, then the circles intersect only in the point C.
What’s left to analyze are the situations when: R-r<|AB|<R+r (assuming r≤R). Let’s focus on the upper halves of the circles:
|AB|<R+r so |AI1|<R and |AI2|>R thus I1 is inside and I2 outside of the circle C(A,R). This means that every route linking I1 with I2 including C(B,r) must either cross or go underneath C(A,R) (postulate 3 applies to straight lines, not circles). How do we know it does cross? Let’s zoom in on the area in question:
Points C1, C2 on C(A,R) and points C’1, C’2 on C(B,r) are chosen in such a way that |AC’1|<R and |AC’2|>R so the points C’1, C’2 cannot be on the same side of the line passing through C1 and C2. If that’s the case, then from the postulate 3 it follows that the segment C’1C’2 must intersect the line going through C1 and C2. Finally, the segment C’1C’2 is between the straight lines coinciding with AC1 and AC2 respectively so in fact the segment C’1C’2 must intersect the segment C1C2.
The shorter C1C2 and C’1C’2 are, the better they approximate corresponding arcs. On the arc C’1C’2 we can add another point C’3 and find C3 where C(A,R) intersects the line linking A with C’3. After that we can repeat the reasoning to establish that the segments C1C3 and C’1C’3 must intersect. By continuing this procedure, we can prove our thesis to any arbitrary accuracy thus the upper arcs of C(A,R) and C(B,r) must intersect at least once. Next, we need to figure out how many times they do intersect.
If there exists some other intersection point between the upper halves of the circles, it must be in one of the four sectors indicated in red:
Let’s assume the additional intersection point P’ is somewhere in the sector 1 (see the picture below). To study the consequences of its occurrence, first, we will pass a straight line a’ linking points A and P’. Line a’ is between lines a and c, so points P and B must be on the opposite sides of a’. From this and the postulate 3, it follows that the segment PB must intersect line a’ in some point that we will denote with B’. By the same token, the segment AP intersect line b’ in some point A’ and line p’ that is between a’ and b’ intersects segments BB’ and AA’ in points B” and A” respectively.
Because the segment AP’ is the shortest possible route between A and P’ the route going through A” must be longer:
|AP’| < |AA”|+|A”P’|
For the same reason, we also have:
|BP’| < |BB”|+|B”P’|
After combining these two inequalities, we get:
|AP’|+|BP’|< |AA”|+|BB”|+|A”P’|+|B”P’|
A”, P’ and B” are on the same straight segment A”B” so:
|AP’|+|BP’|< |AA”|+|BB”|+|A”B”|
A”B” is the shortest route between A” and B” thus:
|A”B”|<|A”P|+|B”P|
If that’s the case, we must also have:
|AP’|+|BP’|<|AA”|+|A”P|+|BB”|+|B”P|
Note that: |AA”|+|A”P|=|AP| and |BB”|+|B”P|=|BP| so:
|AP’|+|BP’|<|AP|+|BP|=R+r
Conclusion: |AP’|=R and |BP’|=r is not possible. In other words, any point in the sector 1 cannot be a second intersection point between upper halves of C(A,R) and C(B,r). In sector 3 the situation is similar. Using almost the same reasoning, it can be proven that: |AP’|+|BP’|>R+r, thus sector 3 is excluded as well.
Let’s assume P’ is in the sector 2:
Now we have:
|AP|<|AB’|+|B’P|=|AP’|-|B’P’|+|B’P|
We have assumed: |AP|=|AP’|=R so:
R<R-|B’P’|+|B’P|
and finally the above can be transformed into:
|B’P’|<|B’P|
By the same token, we can write:
|BP’|<|BB’|+|B’P’|=|BP|-|B’P|+|B’P’|
We have assumed: |BP’|=|BP|=r so:
r<r-|B’P|+|B’P’|
and finally:
|B’P|<|B’P’|
Conclusion: if |AP|=|AP’|=R and |BP’|=|BP|=r then |B’P’| is smaller and greater than |B’P| at the same time. This cannot be true, so sector 2 also cannot contain the second intersection. In the sector 4 the situation is identical, so we must conclude that if the upper halves of C(A,R) and C(B,r) intersect at all, they intersect in one point only. Obviously, the same must be true about the lower halves as well.
Let’s summarize. C(A,R) and C(B,r) either:
– don’t intersect
– intersect in a single point located on the line passing through AB
– intersect in two points: one above (P+) and one below (P–) the line AB
If we choose |AB|=R=1 in some units, then for different r, the intersections must happen in different points P+, P-. In other words, we have a unique correspondence between r and inclinations.
Coordinates
Let’s locate the position of a random point P on the plane defined by the points: P0, P1, P2.
The line linking P with P0 intersects C(P0, |P0P1|) in some point P’, so if we know lengths of the segments: |P0P|, |P1P’| and are told that the inclination between P0P1 and P0P’ is measured in the clockwise direction, we can find where the point P is with respect to P0, P1 and P2.
Triangles
Now that we have a problem of coordinates out of the way, it’s time to take a closer look at space. It appears to be nothing, an empty canvas, where everything else happens, yet interestingly enough, this nothing has properties.
We will start with a question: how two-dimensional creatures could tell whether their space is flat or curved? In an attempt to answer this, we will construct a triangle on a flat plane and another one on the surface of a globe, both produced using the same procedure. In the next step, we will try to spot a difference.
How to construct a triangle though? A triangle is constructed out of three segments with some arbitrary lengths: a, b, c. Let’s draw a segment “a” units long anywhere on a given surface, then on each end of the segment we will draw a circle. One with radius “b” and the other with radius “c”. If these circles intersect in two different points, both of them constitute a third vertex of a triangle with the properties in question.
Triangles just like planes have a front and back side. If we walk along each triangle’s edges following them in the order: a, b, c, we move either clock-wise or counter-clockwise. For the counter-clockwise version denoted with +T(a,b,c), we will assume the triangle’s front side to be up. The clock-wise version -T(a,b,c) will be considered to be back-side up.
Now let’s use the procedure described above to construct two triangles, each built with 3 segments “a” units long. The first will be on the flat table, the second on the surface of a globe with a circumference equal to 4a.
The triangle of the left is on the flat surface. The segment between the top vertex and the point halfway through the base is shorter than “a” (a>h).
On the right of the picture, there is a globe seen from the north pole denoted with the point C. The points A, B are on the equator, and the segment AB (coinciding with the equator) constitutes a base of the triangle ABC. It is easy to verify that a “straight line” (the upper part of one of the meridians) that links C with the point in the middle of the segment AB has the length “a”. In other words, the triangle on the left is not the same as the one on the right despite the same construction method and equal lengths of the sides. From this it follows that two-dimension creatures can tell the difference between flat and undulated surface – the same triangle has a different shape.
Moving objects in space
How do we know that triangles constructed in different places of the same surface have the same size and shape? When we construct a triangle in a new place, we simply move there a ruler and/or compass used to construct the first one. From this it follows that to check whether two triangles are identical, we can transport one of them in space to check whether it would coincide with the second. What exactly does it mean though to move objects in space?
Imagine we have the following two triangles:
Let’s transport the triangle on the left towards the one on the right:
We can break this operation apart into three separate moves: translation, rotation and reflection. Translation changes only the position of the object. The orientation and the side that is up stay the same. Rotation changes the orientation only and reflection is an equivalent of flipping the triangle along one of its edges (AB in the picture).
If it is possible to make one triangle to coincide with the other using these operations, we conclude that they are identical.
In what follows, we will take a closer look at each of the operations.
Reflection
Construction of a triangle out of some arbitrary segments: a, b, c yields the result as shown in the picture:
The lower triangle looks like a mirror image of the upper one, so to define a reflection let’s try to use the triangle construction procedure.
Consider some point P and a straight line k:
To find a reflection of P against k, we pick any two points A and B on k, and construct two circles C(A,|AP|), C(B,|BP|). One of the intersection points must be in P. The other intersection point P’ constitutes the reflection of P.
Let’s reflect some arbitrary triangle against a straight line to check whether the procedure actually produces a mirror image:
If we reflect each point of the upper triangle separately, it is easy to verify experimentally that the resulting triangle has exactly the same size and shape. The only difference is which side is up (counter-clockwise/clockwise). We can encapsulate this fact using the following postulate:
Postulate 5: Reflection symmetries:
- reflection preserves distances (the distance between any two points A and B is the same as between their mirror image)
- reflection flips a triangle on the other side (a mirror reflection of a clockwise triangle is counter-clockwise)
An important consequence of the postulate 5 is preservation of collinearity. If points: A, B, C are on the same straight segment:
we have the relationship: |AC|=|AB|+|BC|. Reflection of these points produces some points: A’, B’, C’. Thanks to the point 1 of the postulate 5, we must have:
|A’C’|=|A’B’|+|B’C’| so A’, B’, C’ must be on the same straight segment with B’ between A’ and C’. From this it follows that reflection of a straight line produces a straight line. In other words: reflection preserves collinearity.
Note that thanks to the postulate 5, the result of reflection does not depend on the choice of circles’ centres:
If we reflect CP and DP using reflection based on circles in A and B, then both segments CP’ and DP’ must be straight and |CP’|=|CP| and |DP’|=|DP|.
Before we move on to discuss rotations, let’s take a look at some consequences of the reflection symmetry (postulate 5). They will turn out to be useful later.
Bisection of an angle
Bisection of an angle simply means division in half. Imagine we have two segments at some inclination with respect to each other (the left side of the picture):
On each segment we choose a point r units away from A, let’s call them B and C. Next, we construct two circles C(B,r) and C(C,r). They intersect in A and D. Note that the triangle ADC is a reflection of ADB against line going through A and D. From this and the postulate 5 it follows that the angle DAB must be equal to the angle CAD. The segment AD divides the angle CAB in half.
Bisection of a segment
Bisection divides a segment exactly in half. To pull it off, we will use the properties of reflection. Let’s take some arbitrary segment AB:
If the radius r is big enough, the circles C(A,r) and C(B,r) will intersect in points P and P’. The segment AB must intersect the segment PP’. B is a mirror image of A against PP’, and O is its own mirror image so according to the postulate 5 |AO|=|BO| i.e. we have divided AB in half.
Right angles
A quarter of a rotation that brings any object back to its starting orientation, corresponds to the angle that is called right. To construct it in some point O on the line k, first, we choose points A and B that are “a” units away from O to the left and right respectively. Then we find P and P’ as intersection points of the circles C(A,r) and C(B,r) for some r>a. As we know from the previous section, the segment PP’ must cross AB in O. It is also apparent that P and P’ are each others mirror image against AB so |OP|=|OP’|.
Now it is clear that the angles ∢BOP, ∢POA, ∢AOP’, ∢P’OB must be identical because they are constructed out of the same segments: a, b and r. This in turn means that all four must be right, as indicated in the picture with small squares.
Rotations
We all know intuitively what a rotation is, however, right now we are after a formal definition of this operation on a flat plane. Let’s say we want to rotate counter-clockwise some point P around the pivot O by the angle constructed with the segment “r”.
First, we draw a line passing through O and P. On this line there is a point (S) that is one unit away from O towards P. Next, in points O and S we construct the circles C(O, 1) and C(S, r). Out of two possible intersection points, the one in the counter-clockwise direction is the one we want (N in the picture). Finally, we pass a line through O and N and find on it such point P’ that |OP’|=|OP|. That’s it.
To discover properties of rotation, we will simply rotate a triangle on a flat surface and examine what happened to it:
The properties in question can be summarized as follows:
Postulate 6: Rotation symmetries:
- rotation preserves distances (the distance between any two points A and B stays the same after both of them were rotated)
- rotation leaves a triangle on the same side (a clockwise triangle stays clockwise after it was rotated)
Next, we have to take a look at some of the consequences of this postulate to check whether they agree with what we observe experimentally. If points are on the same segment, after the rotation they still constitute the same segment (the reasoning is the same as in the section about reflection). The triangle B’OC’ is constructed out of the segments having the same length as the edges of the triangle BOC. Both triangles are counter-clockwise so ∢B’OC’=∢BOC. By the same token we must also have: ∢C’OA’=∢COA and ∢B’OA’=∢BOA.
The last thing to mention is another way we can express angles. After constructing a line perpendicular to given, we can keep dividing each angle in half ad infinitum as demonstrated in the picture:
If we accept 360 degrees to represent a rotation taking object to its starting point, then 180 degrees means half of this rotation, 90 a quarter etc. In other words, we can express every angle by providing a single real number between 0 and 360. To designate what this number represents, we will use degree (°) as a unit e.g. 90° is the right angle.
Finally, note that for two intersecting straight lines, the angles opposite to each other must be the same because rotation of both lines by 180 degrees transforms each of them into itself:
Translations
Let’s start with moving a single point along some straight line h by the distance “a”. If the point is on the line h (point A), the task is straightforward. The more interesting case is when our point is not on h (see point B).
We are going to demand that B after the translation corresponding to the translation of A, must move “a” units just like A, and at the same time |A’B’|=|AB|. To pull it off, we will construct two circles C(A’, |AB|) and C(B,|AA’|). The intersection point that is not on the same side of A’B with A, will represent B after the translation (B’ in the picture).
As usually let’s translate a triangle to discover experimentally the properties of this operation.
The properties in question can be summarized as follows:
Postulate 7: Translation symmetries:
- translation preserves distances (the distance between any two points A and B does not change after both are translated)
- translation leaves a triangle on the same side (a clockwise triangle stays clockwise after translation)
Let’s take a look at some of the consequences of the symmetries postulated.
Imagine a straight line with equally spaced points on it: A, B, C. Let’s translate all three along the line h to new positions: A’, B’, C’. According to the postulate 7, A’, B’, C’ must be on the same straight line (reasoning the same as for reflections). It is apparent that triangles: ABC, B’A’B, BCB’ and C’B’C are identical because we can make them coincide by means of rotation and translation. If that is the case, the internal angles of these triangles: α, β, φ must add up to 180 degrees: α+β+φ=180°. This result is independent of triangles’ size and shape. Neither it depends on where in space a triangle is drawn. It is true for every triangle.
The last thing to mention is that postulate 7 is different than postulates 5 and 6, even though they look similar. If we perform a reflection or rotation on the surface of a globe, postulates 5 & 6 hold. In fact, for a globe we don’t need translations at all because rotations are enough to move objects on its surface. Nevertheless let’s take a look how translation would work in this case.
We are going to translate a triangle: A, B, C. Point C is on the northern pole, and points: A, B are on the equator, with a distance between them equal to the quarter of the equator’s length (q in the picture). A moves to A’, and B moves to B’, both along the equator and both covering the distance q. To find a new position of C, we will follow a procedure described in this section. The circles: C(A’, q) and C(C, q) intersect in two points: A and C’. C’ is opposite to A, so this is the location of C after the translation. Note that |BC|=q and |B’C’|=0, so the postulate 7 is violated! In other words, if postulate 7 holds, we are on a flat surface.
Parallel lines
Consider some point P’ and a straight line h.
Let’s translate h in such a way that it coincides with the point P’. To do that we simply choose any point P on the line h and pass a straight line k between P and P’. Translating P to P’ is straightforward, it simply moves along k. Any other point Q on the line h can be translated according to the procedure described in the previous section. Angle α<180° because P’ is above h. Angle ∢Q’QP is equal to α so Q’ must be on the same side of h as P’. The same reasoning is applicable to every point on h’ so h and h’ cannot intersect. Lines like that will be called parallel. Note that k and k’ are also parallel.
Perpendicular lines.
Imagine the same situation as in the previous section (picture below on the left), this time round tough we want to find line f that passes through the point P’ and is inclined to h at the right angle.
For this purpose we choose any points A and B on h and find a reflection of P’ on the other side of h. The line linking P’ with its reflection P must cross h in some point O. The triangle POA is a mirror reflection of AOP so ∢AOP’=∢POA and because ∢AOP’+∢POA=180°, we must have ∢AOP’=∢POA=90°, that is the angle between f and h is right. If any two lines have this property, we say they are perpendicular.
Inclinations, part III
An inclination can be constructed once we know direction of the rotation and length of the segment r corresponding to the angle in mind. Let’s construct our inclined line in some point O:
To do that we draw two circles C(O,1) and C(S, r). If the angle in question is supposed to be positive, the line we search for must go through the points O and N where N is the intersection point between the circles that is above the line h.
Let’s scale both segments used for the construction i.e. 1 becomes m=|OP| and r is replaced with m∙r=r’. C(O,m) and C(P, m∙r) will intersect in some point P’. It would be interesting to know whether P’ is on the same straight line as O and N? We can easily confirm this experimentally. Does it follow from our postulates though or do we have to add another one? Let’s find out.
Out of the triangle OSN, we can construct a triangle AS’N’ that is twice bigger by translating OSN along OS and then along ON as well.
|OS’|=2|OS|, |N’S’|=2|S’S”| and N’ is on the same line as O and N, so we have proved that if scaling factor m=2, the angle constructed with scaled segments must be the same. In the same fashion we can prove our thesis for any natural m. What if m is a real number? We need to adopt some strategy to find out.
Note that the area of the triangle OS’N’ is four times greater than OSN. That suggest that area of a triangle depends on the lengths of its arms. Obviously it must depend on the angle between them as well:
A(OSN) = F(∢SON)∙|OS|∙|ON|
F(∢SON) in the formula above represents some number that changes with angle ∢SON.
If this formula is correct than the area of the triangles: SS’N’ and SS’O (see the picture below) must be the same.
Let’s replace ON’ with any MM” parallel to SS”:
If we could revert our reasoning and prove that the areas of triangles SS’M” and MS’S” are identical then we would have:
which is equivalent to:
This in turn can be transformed to:
and that’s exactly what we are trying to prove. Is our formula for the area correct though? We will find out in the next section.