Geometry in two dimensions

In this chapter we will search for a way to numerically describe 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 this we will use a table as a model.

If we choose any two points on a table and we stretch a string in such a way that it touches both points, 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 start with choosing three points that aren’t on the same straight line and declare them to be on the same plane. By passing straight lines through any two points known to belong to our plane, we identify other members of this plane.

In three dimensional space we can always fly above or below any given location. Planes on the other hand impose some limits. 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 in the previous paragraph? 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.

Note that shortest route between points A and B on the surface of a globe is different than the very shortest route. The black line in the picture below corresponds to the equator of the globe. The shortest segment linking A with B along the globe’s surface runs along the equator:

The very shortest route on the other hand 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 three dimensional space is flat? Maybe it’s an illusion. We will get back to this problem later.

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 just two numbers.

Inclinations, part I

Let’s examine every possible inclination between two segments starting in the same point. To do that, choose any arbitrary two points P₀ and P₁ and draw a circle with the radius equal to the length of a straight segment linking P₀ with P₁.

For every point P on the circle the inclination between P₀P and P₀P₁ is different, and the points on the circle give us every possible inclination. Next, we have to find a way to express the inclination with 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 cataloguing 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 underlying 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 all points on the surface with a fixed distance to some point O. Circles will be denoted with the symbol C(O,r) where O is the circle’s centre and r is the distance of every point on this circle to the centre O. “r” is called radius. By this logic, C(A,R) is a circle with origin in the point A and with radius R, and C(B,r) is a circle with origin in B and with radius r.

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 point 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.

Imagine a situation depicted below where flat creatures draw straight lines on the surface that is not a flat plane. In their world the lines ABC and ABD are straight and they coincide in multiple points. This is in agreement with our postulates, yet straight lines and segments in our world do not behave like this. We need yet another postulate to capture this.

Postulate 4: If straight lines or segments 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 our postulates and definitions.

Let’s keep adding (on the same plane) more and more lines that intersect point O. If on each of them we mark the points that are r units away from O, the resulting points will belong to the same circle C(O,r). It looks like the length of the arc linking P₀ with P₁ is shorter than the length of the arc P₀P₂. In other words, we could use the length of arcs to describe inclinations.

Measuring lengths of arcs is somewhat cumbersome though. It would be good to find something better.

Inclinations, part II

Let’s get back to experiments. Grab a piece of paper, pencil and a compass. Draw a straight line and two circles with their centres on this line. Depending on the radii, there are three possible scenarios: the two circles do not intersect at all, they intersect in one point or we have two intersections, one below and one above the line:

Let’s focus on the circles: C(P₀,R) and C(P₁,r) that intersect in two points.

Let’s take a look at the inclination between the segments P₀P₁ and P₀P₂. For R kept fixed, different r correspond to different inclinations. The problem is that by this token the inclination between P₀P₁ and P₀P₃ is the same as between P₀P₁ and P₀P₂. How can we tell the difference between the two? Perhaps one could be positive and the other negative.

The points: P₀, P₁ and Q are on the same plane. This can be used to define a positive direction:

Depending on which side of the plane we look at, the sequence P₀,P₁,Q 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 establishing whether all of that follows from our postulates.

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 only 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| and s=|CP| so for P’≠C it must be true that 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 for P’≠C. 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 of two circles

To begin with, let’s figure out when circles can intersect. Consider the situation when the segment AB is longer than R+r.

Please bear in mind that in the reasoning presented below, the points A, C, C’, B are fixed and P can be anywhere on C(A,R). If P is elsewhere than in the picture, P’ will be elsewhere as well.

For every P≠C, |PB|>|CB|>r because the shortest route between C(A,R) and B is CB and |AB|>R+r so |CB|>r.

For every point P on C(A,R), the shortest segment between P and C(B,r) must be PP’. From the previous paragraph it follows that PP’ is the shortest when P=C, so the shortest route between C(A,R) and C(B,r) must be CC’. Its length is |CC’|=|AB|-R-r so for |AB|>R+r the circles cannot intersect. If on the other hand |AB|=R+r, then the circles intersect in one point only C=C’.

Next, let’s take a look at the situation when |AB|+r<R:

In the previous section, we have established that: |BC|<|BP| for every point P on C(A,R) apart from C. Also, |AB|+r<R=|AB|+|BC| so r<|BC|. Altogether we have r<|BC|<|BP| for P≠C.

For every P on C(A,R), the shortest route between P and C(B,r) is PP’ and from the previous paragraph it follows that PP’ is the shortest for P=C when its length must be |BC|-r. |BC|>r so the circles won’t intersect. Of course if |BC|=r then the circles intersect only in point C.

Above we have analysed the situations: R+r≤|AB| and |AB|+r≤R. What’s left to analyse then is the situation in the picture below where: R+r>|AB|>R-r (assuming r≤R). Let’s focus on the upper halves of the circles:

R+r>|AB| 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). Underneath? Remember! It doesn’t follow directly from any of our postulates that a closed loop route divides the plane into inside and outside of the loop. Do we need another postulate? 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 and C’₂ cannot be on the same side of the line passing through C₁ and C₂. If that’s the case, then from postulate 3 it follows that the segment C’₁C’₂ must intersect the line going through C₁ and C₂. Finally, segment C’₁C’₂ is between the straight lines coinciding with segments AC₁ and AC₂ respectively, so in fact segment C’₁C’₂ must intersect segment C₁C₂ because C₁C₂ is between the same straight lines as C’₁C’₂.

Next, on arc C’₁C’₂ we can add another point C’₃ such that |AC’₃|>R. If that’s the case then we can use the same argument as above to prove that the segments C₁C₃ and C’₁C’₃ must intersect.

With more points added between C’₁ and C’₃, the segments we operate on get shorter and as a result they approximate the corresponding arcs better. By continuing this procedure, we can prove our thesis to any arbitrary accuracy, thus we conclude that the upper arcs of C(A,R) and C(B,r) do intersect at least once. Next, we need to figure out how many times they do intersect.

Let’s assume there are two intersection points and see where it will take us. Remember that we suspend the judgment of our eyes, and derive everything from the postulates and their consequences.

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’ (P indicates the first intersection) is somewhere in 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 postulate 3, it follows that segment PB must intersect line a’ in some point that we will denote with B’. By the same token, segment AP intersects line b’ in some point A’, and any line between a’ and b’ that goes through P’ (line p’ in the picture) must intersects the segments BB’ and AA’. For line p’ intersections are in points B” and A” respectively.

Because the segment AP’ is the shortest possible route between A and P’, the alternative 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, it must be true that:

|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: For the second intersection point P’ we must have |AP’|=R and |BP’|=r, however, according to the inequality above it cannot be true. In other words, any point in the sector 1 cannot be a second intersection point between the 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 for P’ in sector 3 that: |AP’|+|BP’|>R+r, thus second intersection cannot be in sector 3 either.

Let’s assume the second intersection point P’ is in 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 the lengths of 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 and space

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. 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 this 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.

These two triangles are not identical though. 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 on the left is on a 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 (AC 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 these 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 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 coincide with 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 the 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. This operation is called a bisection of angle and the line passing through A and D is called: bisector.

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 a right angle 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 b:=|OP|=|OP’|.

Now it is clear that the angles ∠BOP, ∠POA, ∠AOP’, ∠P’OB (∠ is a shorthand for angles) must be identical because they are constructed out of the same segments: a, b and r. This in turn means that all four angles must be right. From now on, the right angle will be indicated with a small square as indicated in the picture.

Rotations

We all know intuitively what 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 and ∠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.

Note that all of this gives us another way to describe a reflection. To reflect P’ against h, first, we construct a line f perpendicular to h that passes through P’. f intersect h in some point O. Reflection of P’ against h is on the other side of h on the line f in such a position that |P’O|=|PO|.

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, an inclined 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.

Using the triangle OSN, we can construct a triangle AS’N’ that is twice bigger:

|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?

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 OS’S” (see the picture below) must be the same.

$\frac{A(SS'N')}{A(OS'S'')}=\frac{F(\angle S''S'S)\cdot |SS'|\cdot |S'N'|}{F(\angle S''S'S)\cdot |OS'|\cdot |S'S''|}=$
$=\frac{F(\angle S''S'C)\cdot |SS'|\cdot 2\cdot |S'S''|}{F(\angle S''S'C)\cdot 2\cdot |SS'| \cdot |S'S''|}=1$

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:

$\frac{A(SS'M'')}{A(MS'S'')}=\frac{F(\angle S''S'C)\cdot |SS'|\cdot |S'M''|}{F(\angle S''S'C)\cdot |MS'|\cdot |S'S''|}=1$

which is equivalent to:

$\frac{|S'M''|}{|S'S''|}=\frac{|MS'|}{|SS'|}=m$

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.

Area

Let’s start with area of some arbitrary flat object. If the object can be covered with a bunch of tiles that are identical, then its area can be specified with the number of tiles used. The question is what tile shape is the most appropriate. Certainly they must cover the area tightly i.e. without gaps between tiles. Any triangle has this property but we cannot use a triangle because if we divide it, it’s not clear what the area of each fragment is. A square seems to be a better choice. Let’s take a look.

Squares

A square is a polygon that has four sides that are equally long and each is perpendicular to its neighbouring sides. Let’s cut a square with the sides one unit long with a line that divides in half the sides that are opposite to each other:

The resulting upper and lower polygons must have the same area because we can make them coincide with a translation. A half can be divided again, and the resulting half as well. This process can be continued indefinitely.

We will be denoting a single tile with U² where U is a single unit. In the picture “1” is used instead to demonstrate a unit independence of the reasoning presented below. U can be any unit we wish, an inch, meter or anything else. Superscript ² next to U will be used to distinguish U² from units of length. From this it follows that the areas of the binary subdivisions of a tile described in the previous paragraph are equal to: 1/2 U², 1/4 U², 1/8 U² etc.

Let’s divide a square U² with some arbitrary line parallel to the square’s base:

The area of the polygon indicated in red can be obtained by stacking the binary subdivisions of the unit tile one on top of the other just like we did for a segment in this section. The result can be expressed as:

$\sum_{i=1}^{\infty} b_i \cdot \frac{U^2}{2^i}$

where each b_i is equal to 0 or 1. This series:

$\sum_{i=1}^{\infty} b_i \cdot \frac{1}{2^i} = v$

must be convergent to the real number “v” (see picture above) so the area of our polygon must be:

$\sum_{i=1}^{\infty} b_i \cdot \frac{U^2}{2^i}=(\sum_{i=1}^{\infty} b_i \cdot \frac{1}{2^i})\cdot U^2 = v U^2$

Let’s pick some other real number 0<u<1 and cut from U² the following polygon:

Repeating the reasoning laid out above, this time for u, leads to the conclusion that area of the resulting polygon must be equal to u∙v U².

Rectangles

A rectangle has four sides just like squares. Each side is perpendicular to its neighbours, yet not all of them must be equal to each other.

Let’s grab a rectangle built with segments u and v units long. This time u and v are any positive real numbers.

Using the same argument as in the previous section, it is straightforward to prove that the area of this rectangle must be u∙v. If for example v=3.5cm and u=4.8cm, then the are of the rectangle is:
3.5 ∙ cm ∙ 4.8 ∙ cm = 3.5 ∙ 4.8 ∙ cm ∙ cm = 16.8 cm²

where cm² represents a single tile.

Parallelograms

Will the area of a rectangle change if we tilt it as depicted below?

The upper and lower sides are equally long and parallel. The same can be said about left and right sides. A polygon like this is called a parallelogram.

Let’s construct a segment h perpendicular to the base and passing through the upper left corner. The upper side is parallel to the lower so the upper side must be perpendicular to the segment h as well. The left side is parallel to the right, so if we cut the parallelogram along h and move the triangle obtained this way to the right (see the picture below), we get a rectangle.

The area of this rectangle and of the initial parallelogram must be u∙h where u is the length of the base and h represents the height.

Triangles

Let’s draw two triangles both constructed out of three segments: u, v and q. They share the side q and the upper triangle can be transformed into the lower by 180 degrees rotation and translation.

All of this means that the polygon in the picture is a parallelogram and from this it follows that the area of both triangles must be u∙h/2.

Let’s see if we can use this formula to solve the problem formulated in section about inclinations, part III.

The Thales theorem

Consider two straight lines a and b intersecting in point O and the other two lines c and d that are parallel to each other and also intersect the lines a and b.

Using the formula for area of triangles, we can prove that the area of triangles A’BO and AB’O must be identical because the area of both triangles A’BA and AB’B is equal to |AB|∙h/2. If that’s the case, we can write:

$\frac{A(A'BO)}{A(ABO)}=\frac{A(AB'O)}{A(ABO)}$

where A(triangle) represents the area of a triangle. After inserting formulas for the areas in question, the formula above reads:

$\frac{1/2\cdot |A'O|\cdot h_a}{1/2\cdot |AO|\cdot h_a}=\frac{1/2\cdot |B'O|\cdot h_b}{1/2\cdot |BO|\cdot h_b}$

$\frac{|A'O|}{|AO|}=\frac{|B'O|}{|BO|}$

Note that the theorem covers more than one situation. By means of translation or rotation we can transform the configuration in the picture above into three other scenarios:

In all four situations the following equality must be true:

$\frac{m'}{m}=\frac{n'}{n}$

The reason why we delved into all of that stems from the investigation of inclinations. It’s time to get back to it then.

Inclinations, part IV

Thanks to the Thales theorem we know that in the situation depicted in the picture below the following must be true: |NS|/|OS|=|PP’|/|OP|.

This means that the inclination constructed using circles C(O, |OS|) and C(S, |NS|) is exactly the same as the inclination obtained with: C(O, |OP|) and C(P,|PP’|). Inclination is given by a ratio |NS|/|OS|=|PP’|/|OP| so contrary to distance, it doesn’t require a unit. Note that direction of inclination can be incorporated into this number with a sign i.e. positive – counterclockwise, negative – clockwise.

But wait, there is even greater discovery staring in our face! Imagine two circles C(O,r) and C(O,R) and two lines a₁ and a₂ crossing in the point O. Let’s focus on the arcs of these circles that are between a₁ and a₂.

We can divide the angle between a₁ and a₂ using a bisector b₁ and then divide each of the slices again with bisectors b₂ and b₃. The segments d and D linking intersections of the lines: a₁, a₂, b₁, b₂, b₃ with circles C(O,r) and C(O,R) approximate the arcs of these circles. Obviously we can easily increase the accuracy of this approximation by bisecting resulting slices time and again. After every step, the combined length of segments will be getting closer and closer to the length of the corresponding arcs. In the picture above, the arcs’ lengths are approximated with: 4∙d and 4∙D respectively.

Thanks to the Thales theorem we have:

$\frac{d}{r}=\frac{D}{R}$

so it follows that the length of an arc stretched between a₁ and a₂ must be equal to its corresponding radius multiplied by some constant.

Note that this reasoning can be applied to a full circumference too. Resulting formula for a circumference must be: S∙r where S is some number independent of radius and r is radius.

Rotation about the centre transforms a circle into itself so we must also have:

$Arc(\alpha, r)=\frac{\alpha}{360}\cdot S \cdot r$

where Arc(α, r) is the length of arc cut from the circle as in the picture above with α denoting the angle between lines a₁ and a₂. This gives us yet another method to measure inclinations: Arc(α, r)/r because this ratio is independent of radius.

Let’s try to estimate S. Before we do, we have to establish a few things. Take a look at the picture below.

We have: |OP|=|OP’|=|OP”|=r. The segment OP” is between OP and OP’. It is easy to confirm experimentally that the segment OP” must intersect PP’. Does it follow from our postulates though?

Let’s reformulate the question a bit. What is the shortest distance between a point and line? It is easy to verify experimentally that the shortest route between a point and line (marked red in the picture below) is perpendicular to this line i.e. c>b. In the next section it will be derived from the postulates.

The Pythagoras theorem

Let’s find a relationship between the sides of a right triangle i.e. a triangle with one of the internal angles equal to 90 degrees.

The first observation to make is that α:=∠BCO=∠CAO and β:=∠OCA=∠OBC. This must be so because α+β+90^o=180^o. In other words, the triangles OBC and OCA have the same internal angles as ABC. Imagine we create an exact copy of OBC and OCA. Using reflection, rotation and translation the first one can be transformed into A”B”C in the picture below, and the second into A’B’C.

Now using the Thales theorem we can link the heights with the lengths of bases:

$\frac{a_1}{h_a}=\frac{b_1}{h_b}=\frac{c_1}{h_c}$

$\frac{a_2}{h_a}=\frac{b_2}{h_b}=\frac{c_2}{h_c}$

Let’s combine the above two into:

$\frac{a_1+a_2}{h_a}=\frac{b_1+b_2}{h_b}=\frac{c_1+c_2}{h_c}$

According to the first picture: a₁+a₂=|BC|, b₁+b₂=|AC| and c₁+c₂=|AB|. After shortening the notation into a:=|BC|, b:=|AC| and c:=|AB|, we can rewrite the equation above into:

$\frac{a}{h_a}=\frac{b}{h_b}=\frac{c}{h_c}=P$

Variable P will be useful later.

The areas of triangles OBC and OCA add up to the area of ABC:

$\frac{1}{2}\cdot a \cdot h_a + \frac{1}{2}\cdot b \cdot h_b = \frac{1}{2}\cdot c \cdot h_c$

Using the previous equation, we can replace h_a with a/P, h_b with b/P and h_c with c/P. After that, we get:

$\frac{1}{2\cdot P}\cdot a^2 + \frac{1}{2\cdot P}\cdot b^2 = \frac{1}{2\cdot P}\cdot c^2$

Finally, after multiplying both sides of the equality above times 2P, we get a relation called in mathematics the Pythagorean theorem:

$a^2 + b^2 = c^2$

As a consequence of this theorem, the shortest segment linking a line with a point must be perpendicular to this line i.e. c>b.

Irrational numbers are real

In one of the previous chapters it was proven that 2^1/2 is not a rational number.
The way 2^1/2 was introduced might feel like it is an artificial construct, a useless mathematical extravaganza. Thanks to the Pythagoras theorem we know that it is quite the opposite. If we construct a triangle with two sides one unit long, and at the right angle to each other, according to the Pythagoras theorem the third one must be 2^1/2 units long.

Circumference of a circle

The circumference of every circle is equal to its radius multiplied by some number. To establish what this number is equal to, we will estimate the length of the circumference for a circle with radius r=1.

Take a look at the picture above. A and B are each others mirror image against the line going through OC that is the bisector of the angle ∠BOA. The segment AD is perpendicular to AO so its mirror image DB must be perpendicular to OB.

|OP’|=1 and according to the Pythagoras theorem: |OP|<1 and |OP”|>1. From this it follows that the arc AB of the circle C(O,1) must be within the triangle ABD. Of course, |AD|+|BD| is greater than |AB| yet as we will see the difference between them shrinks down to 0 with decreasing angle ∠BOA.

Let’s construct a segment EG perpendicular to OD in the point F:

By the same reasoning as before the triangle AEO is a mirror image of EFO against the segment EO, the bisector of the angle ∠FOA. △FGO is a mirror image of △BGO against the bisector GO, △EFO is a mirror image of △FGO against the bisector OF, and △AFO is a reflection of △BFO against FO. From this it follows that |AF|=|BF|, |AE|=|EF|=|FG|=|BG|.

Also, it is apparent that: |AD|+|BD|>|AE|+|EG|+|BG|>Arc(60^o, 1)>|AF|+|BF|>|AB|, so the approximation of the length of the circle’s arc Arc(60^o, 1) (the arc linking points AFB) with segments is getting better and better with increasing number of segments.

Let’s actually calculate what the circumference is equal to. Our first upper estimate is 12|BD| and lower 12|BC| because six triangles identical to ABO can be arranged around the circle C(O,1).

From the Thales theorem it follows that:

$\frac{|BD|}{1}= \frac{1/2}{|CO|}$

and thanks to the Pythagoras theorem we have:

$|CO|^2+(1/2)^2=1^2$

$|CO|=\sqrt{1-1/4}=\sqrt{3}/2$

thus

$|BD|=1/\sqrt{3}$

All of that leads to the conclusion that the circumference of C(O,1) is between $12/\sqrt{3}$ and 6. Using the technique to approximate $\sqrt{3}$ with rational number, we can estimate the circumference to be between 6.92821 and 6.0.

To increase the accuracy, we double the number of segments. Let’s start with the upper limit i.e. the estimate that overshoots past the actual length. The length of |BD| was calculated above. Now we will use it to calculate |BG|=|FG|.

Thanks to the Thales theorem with have:

$\frac{x_1}{k}= \frac{1}{|DO|}$

Also: $k=x_0-x_1$ and $|DO|= \sqrt{1+x_0^2}$ so:

$\frac{x_1}{x_0-x_1}= \frac{1}{\sqrt{1+x_0^2}}$

and from this we can unentangle x₁:

$x_1=\frac{x_0}{1+\sqrt{1+x_0^2}}$

We know that $x_0=|BD|=1/\sqrt{3}$ so after inserting this in the formula above, and after a few transformations the result reads:

$x_1=\frac{1}{\sqrt{3}+2}$

We doubled the number of segments, so the upper estimation for the circumference must be 24∙x₁, that is approximately 6.4308.

If we double the number of segments again, the formula for the upper approximation of the circumference is:

$48 \cdot x_2=48 \cdot \frac{x_1}{1+\sqrt{1+x_1^2}}$

where x₁ was calculated in the previous step. Of course, we can continue this process ad infinitum.

The lower approximation for the circumference after i-th round of segments doubling can be obtained from the upper estimate:

x_i represents a single segment of the upper approximation and y_i represents the lower. Using the Thales and Pythagoras theorem, we find that:

$\frac{y_i}{1}=\frac{x_i}{\sqrt{1+x_i^2}}=\frac{1}{\sqrt{1/x_i^2+1}}$

so after i-th round of segments doubling the circumference must be between 6∙2^ⁱ⁺¹∙y_i and 6∙2^ⁱ⁺¹∙x_i.

For i=1 we have $x_1=\frac{1}{\sqrt{3}+2}$ thus 24∙y₁ must be equal to 6.2116 approximately so the circumference is between 6.2116 and 6.4308.

With increasing number of segments the upper and lower approximations get closer to each other. Is the final result rational or irrational though? Also, is there a better way of calculating it? We will get back to these question later. For now we will follow the accepted convention, and denote the ratio between a circumference and diameter (double of radius) of a circle with the symbol π.

Radians

By now we have established that the length of an arc cut out of a circle is given by the formula:

$Arc(\alpha, r)=\frac{\alpha}{360}\cdot 2 \pi \cdot r$

so we can write:

$\frac{Arc(\alpha, r)}{r}=\frac{\alpha}{360}\cdot 2 \pi$

The fraction of 2π on the right is yet another popular way for measuring angles where 2π corresponds to 360^o, π to 180^o and π/2 to 90^o. Number on the right is not given in any units because on the left side the length is divided by length. If some number is supposed to represent Arc(α, r)/r, then we add to it the word radians so its clear what we mean. Note that a degree (^o) that was introduced earlier, is defined as 1/360 of the rotation taking object to its initial orientation.

Radial coordinates

The location of any point P on the plane constructed around the points P₀, P₁ and P₂ can be defined by radius r and angle α:

The radius tells us how far we are from the reference point P₀ and the angle specifies the inclination between the segment P₀P and P₀P₁. This type of coordinates is called radial.

Cartesian coordinates

The shortest route between a straight line and a point goes along a segment perpendicular to this line. We can use this fact to introduce another way to specify a location. Imagine a plane build around the points P₀, P₁, P₂ and some point P on its surface.

The segment |PO| is the shortest route between P and the line linking P₀ with P₁. The length of |P₀O| will be our first coordinate. It will be measured in the direction from P₀ towards P₁ thus for the situation in the picture, it will be positive. The second coordinate will be equal to the length of the segment PO. If P is on the same side of P₀P₁ as P₂, the second coordinate will be positive.

You may wonder what is the point of introducing various types of coordinates. As we shall see later, different types are useful in different situations.

Division of segment into parts

Before closing this chapter, we need to close a loose end. In the chapter about rational numbers it was stated that it is possible to divide a segment into arbitrary number of parts. The questions is how? Imagine we want to divide the segment AB into 5 equal parts.

Let’s draw any straight line going through the point A and place on it five uniformly spaced points: I₁, I₂, I₃, I₄, I₅. The distance between them doesn’t matter, all that matters is: |AI₁|=|I₁I₂|=|I₂I₃|=|I₃I₄|=|I₄I₅|_.

Note, that when we connect B with I₅ and through every other point I, we pass a line parallel to BI₅, according to the Thales theorem these lines divide AB into five equal parts.

Summary

In this chapter we have found a way of specifying positions of objects on a flat surface. There is more than one way, but it always boils down to providing just two numbers.

In the process we have found what properties the space must have to be considered flat. Everything that was stated follows logically from just seven postulates. If we could demonstrate that any of them was violated, we would have to question our conclusions.

The space we live in is three, not two-dimensional, so in the next chapter we will tackle the problem of pinpointing locations in three-dimensional space.