## Friday, May 22, 2015

### Lines and Planes - Problems 4: UNIZOR.COM - Geometry3D

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Problem

Construct a common perpendicular h to two given skew lines a and b.

Analysis

Assume a common perpendicular h to skew lines a and b is constructed. Let points of its intersection with these lines be A and B correspondingly. Then h must be an intersection of two planes: plane σ that is perpendicular to line a at point A and plane τ that is perpendicular to line b at point B:

σ⊥a; A∈σ

τ⊥b; B∈τ

h = σ∩τ

We don't know positions of points A and B. Consider, instead of point B, any other point B' on line b and construct plane τ' perpendicular to line b at this point B':

B'∈b; τ'⊥b; B'∈τ'

Since planes τ and τ' are parallel their intersections with plane σ are parallel. So, if we know plane σ, using any plane τ' perpendicular to line b allows us to construct a line h'=σ∩τ' that is parallel to a common perpendicular h we need.

Let's make another step analogous to the previous one. We don't know the position of point A. Consider, instead of it, we choose any point A' on line a and draw a plane σ' perpendicular to line a at this point:

A'∈a; σ'⊥a; A'∈σ'

It is parallel to plane σ and, therefore, lines of intersection of plane τ' with these two planes, σ and σ', are parallel. Let new line of intersection between τ' and σ' be h":

h" = σ'∩τ'

Since h ∥ h' and h' ∥ h", lines h and h" are parallel.

We came to a situation, when the construction of any two planes σ' and τ' correspondingly perpendicular to lines a and b allows us to construct a line h" which is parallel to the one we need to construct.

Next step is to shift in space our line h" parallel to itself to a place where it intersects both lines a and b.

We can do it in two steps:

Step 1 involves drawing a line parallel to a given line h" through a given point A' within plane σ'. We know how to do it from plane geometry. Let this new line that is perpendicular to a, parallel to h" and going through A' be line p.

Step 2 can be accomplished by drawing a plane ρ through lines a and p and finding a point of intersection of this plane with line b. This point must be point B on line b - the base of a common perpendicular h we are constructing. Dropping a perpendicular from point B to line a within plane ρ produces a common perpendicular h.

Construction

1.[A',σ': A'∈a; σ'⊥a; A'∈σ']

2.[B',τ': B'∈b; τ'⊥b; B'∈τ']

3.[h": h"=σ'∩τ']

4.[p: p∈σ'; A'∈p; p ∥ h"]

5.[ρ: a∈ρ; p∈ρ]

6.[B: B=ρ∩b]

7.[h: B∈h; h∈ρ; h⊥a]

More details are presented in the notes to this lecture at UNIZOR.COM

## Tuesday, May 19, 2015

### Unizor - Geometry3D - Lines and Planes - Problems 3

Unizor - Creative Minds through Art of Mathematics - Math4Teens

We would like to address this construction problem in a very detailed way to exemplify the rigorousness required from mathematics, if we want to approach this subject on an advanced level.

This rigorousness requires to

(1) analyze the construction problem to properly plan the construction on a step by step level;

(2) present the construction steps and

(3) prove that this construction leads to required results.

Problem

Drop a perpendicular a to a given plane σ from a given point M outside of a given plane.

Analysis

We don't know yet how to drop a perpendicular to a plane from a point outside of it. But, introducing the concept of a line that is perpendicular to a plane (see "Line ⊥ Plane" lecture), we have constructed a plane perpendicular to a line at a given point on this line and also constructed a line perpendicular to plane at a given point on this plane.

Using the latter construction, we can choose any point N on a given plane σ (N∈σ), construct a line b perpendicular to a plane at this point (b⊥σ; N∈b) and, finally, construct line a parallel to line b, but passing through a given point M (a ∥ b; M∈a).

By the Theorem 1 from the lecture "2 Lines ⊥ Plane", this line a should satisfy the requirements of a problem - passing through point M and be perpendicular to plane σ.

Construction

Step 1. Choose any point N∈σ.

Step 2. Draw a perpendicular b to plane σ through point N on this plane as described in lecture "2 Lines ⊥ Plane".

Step 3. Draw a line a through point M parallel to line b as described in Exercise 1 of lecture "Line ∥ Line".

Proof of construction

Since line b is perpendicular to plane σ by construction and line a is parallel to line b, line a is perpendicular to plane σ.

Proof of uniqueness

Assume that we have two lines a and a' such that

a⊥σ, a∩σ=P

M∈a;

a'⊥σ, a'∩σ=P';

M∈a'.

' Draw a plane through intersecting lines a and a'. It will intersect plane σ on a line connecting P and P'.

Now lines MP and MP' are both perpendicular to line PP', which is impossible.

Therefore, points P and P' are one and the same, and the perpendicular from point M to plane σ is unique.

### Unizor - Geometry3D - Lines and Planes - Problems 2

Unizor - Creative Minds through Art of Mathematics - Math4Teens

We would like to address this construction problem in a very detailed way to exemplify the rigorousness required from mathematics, if we want to approach this subject on an advanced level.

This rigorousness requires to

(1) analyze the construction problem to properly plan the construction on a step by step level;

(2) present the construction steps and

(3) prove that this construction leads to required results.

Problem

Given a straight line a in the three-dimensional space and a point M outside of it.

Construct a plane σ that passes through a given point M and is perpendicular to a given line a.

Analysis

Assume that such plane σ has been constructed (a⊥σ, M∈σ) and its intersection with line a, a base of a perpendicular, is point P (P=a∩σ).

Construct a new plane τ that is defined by line a and point M (a∈τ, M∈τ). Plane τ contains both points P (since P∈a and a∈τ) and M (by construction). Hence, plane τ contains line a and segment MP.

Segment MP is also contained in plane σ because point P is a base of the perpendicular to plane σ.

So, segment MP belongs to the intersection of planes σ and τ (MP∈σ∩τ).

In plane τ segment MP is perpendicular to line a (MP⊥a) since line a is perpendicular to an entire plane σ and, therefore, to any line on this plane that passes through a base point P.

Therefore, we can construct segment MP even without plane σ but using only plane τ (defined by given line a and point M) and drawing a perpendicular from point M to line a within that plane.

To determine the position of plane σ, it is sufficient to have two intersecting lines going through a base point P. The point P and one such line, segment MP, we have just constructed by dropping a perpendicular from point M to line a within plane τ.

To construct another line also perpendicular to a at the base point P, construct another plane ρ through line a that is different from plane τ. In that plane draw a perpendicular NP to line a at point P.

Now we have two lines defined by segments MP and NP perpendicular to line a. These two lines define a plane perpendicular to line a and containing both of these lines, in particular segment MP and point M in it.

Therefore, this plane is the one we need - plane σ.

Construction

Step 1. Construct a plane τ by given line a and point M.

Step 2. Drop a perpendicular MP from point M to line a, point P is the base of this perpendicular.

Step 3. Construct any other plane ρ through a, different from plane τ.

Step 4. Draw a perpendicular NP in plane ρ to line a at point P.

Step 5. Construct a plane σ by lines defined by segments MP and NP.

Proof of construction

(a) σ∋MP

⇒ σ∋M

(b) a⊥MP; a⊥NP;

σ∋MP; σ∋NP

⇒ a⊥σ

Proof of uniqueness

Assume there exist two planes σ and σ', both containing point M and both perpendicular to line a. Let points P and P' be intersections of line a with these planes, correspondingly.

Assume, P≠P' (case P=P' is considered below).

Draw a plane τ through line a and point M. It intersects plane σ along segment MP, which is perpendicular to line a. It also intersects plane σ' along segment MP', which is also perpendicular to line a.

So, within plane τ we have two perpendiculars MP and MP' from the same point M to line a, which is impossible in plane Euclidean geometry.

Therefore, points P and P' of intersection of line a with planes σ and σ' are one and the same.

Now we drop the initial assimption that P≠P' as invalid and consider only the case when P=P'.

Our planes σ and σ' have now two common points - given point M and the intersection with line a - point P=P'.

To say that planes σ and σ' are identical, we need the third common point.

Draw a plane ρ through line a that is different from plane τ. It intersects plane σ along a line that goes through point P and it intersects plane σ' also along a line that goes through point P. If we assume that these lines of intersection are different, we would have two different perpendiculars to line a at its point P within one plane ρ, which is impossible in plane geometry.

Therefore, these two lines perpendicular to line a at point P within plane ρ are one and the same. Now we have plenty of new points along this line that are common between planes σ and σ', which completes the proof that these planes are one and the same.

The uniqueness of a plane perpendicular to a line and passing through a point outside of this line is proven.

## Monday, May 18, 2015

### Unizor - Geometry3D - Lines and Planes - 2 Planes Perpendicular to a Line

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Theorem 1

If one of two parallel planes is perpendicular to some line, the other one is perpendicular as well.

Proof

Assume that planes γ and δ are parallel and line a is perpendicular to plane γ:

γ ∥ δ;

a⊥γ.

We will prove that a⊥δ.

Let points P=a∩γ and Q=a∩δ be intersection points of line a with two given planes.

Construct a new plane σ that contains line a. It also contains points P and Q of intersection of line a with planes γ and δ and, therefore, intersects γ along some line PP' and intersects δ along some line QQ'. Notice that a⊥PP' since a⊥γ.

If a plane σ intersects parallel planes γ and δ along lines PP' and QQ' correspondingly, these lines are parallel (see Theorem 2 of "Plane ∥ Plane" lecture), that is PP' ∥ QQ'.

Similarly, construct a different new plane τ that contains line a. It also contains points P and Q of intersection of line a with planes γ and δ and, therefore, intersects γ along some line PP" and intersects δ along some line QQ". Notice that a⊥PP" since a⊥γ.

If a plane τ intersects parallel planes γ and δ along lines PP" and QQ" correspondingly, these lines are parallel (the same Theorem 2 of "Plane ∥ Plane" lecture as above), that is PP" ∥ QQ".

We have now the following relationships between the lines.

In plane σ:

a⊥PP' and PP' ∥ QQ', from which follows that a⊥QQ'.

In plane τ:

a⊥PP" and PP" ∥ QQ", from which follows that a⊥QQ".

Since a⊥QQ' and a⊥QQ", we conclude that a⊥δ.

End of proof.

Theorem 2

If two planes, γ and δ are perpendicular to the same line a, they are parallel to each other.

Proof

Given: a⊥γ; a⊥δ.

To prove: γ∥δ.

Let point P=a∩γ and point Q=a∩δ.

Construct a new plane α that contains line a. It also contains points P and Q of intersection of line a with planes γ and δ and, therefore, intersects γ along some line PP' and intersects δ along some line QQ'.

Within plane α both lines, PP' and QQ', are perpendicular to line a and, consequently are parallel to each other.

Similarly, construct a different new plane β that contains line a. It also contains points P and Q of intersection of line a with planes γ and δ and, therefore, intersects γ along some line PP" and intersects δ along some line QQ".

Within plane β both lines, PP" and QQ", are perpendicular to line a and, consequently are parallel to each other.

Now we have a pair of lines PP' and PP" on plane γ correspondingly parallel to a pair of lines QQ' and QQ" on plane δ. According to one of the theorems proven earlier (see Theorem 1 in "Plane ∥ Plane" lecture), this is a sufficient condition for these planes to be parallel to each other.

End of proof.

Theorem 3

If from a point M lying outside line a two different planes, γ and δ, are constructed in such a way that a⊥γ and a∩δ≠∅, then plane δ is NOT parallel to plane γ and is NOT perpendicular to line a.

Proof

Obviously, γ and δ are NOT parallel because they intersect at point M.

Let point A=a∩γ and point B=a∩δ. Assume, these are two different points on line a.

If a⊥δ, we have two perpendiculars from point M to line a, segments MA and MB, which is impossible in plane Euclidean geometry. So, a is NOT perpendicular to δ. Actually, it is slant or oblique line relative to this plane.

End of proof.

### Unizor - Geometry3D - Lines and Planes - 2 Lines Perpendicular to a Plane

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Theorem 1

If one of two parallel lines is perpendicular to some plane, the other one is perpendicular as well.

Proof

Assume that a and b are parallel lines and plane γ is perpendicular to a:

a ∥ b;

a⊥γ.

We will prove that b⊥γ.

Let the base of line a on plane γ (that is, their intersection a∩γ) be point A and the intersection b∩γ be point B. The latter intersection must exist because, otherwise, line b would be parallel to plane γ, and line a, as parallel to b, would also be parallel to γ, which contradicts their perpendicularity.

Lines a and b lie in the same plane because they are parallel, let's call this plane δ. In plane δ lines a and b are parallel and line AB (lying in both planes γ and δ) is perpendicular to a (since a⊥γ). Therefore, according to theorems of plane geometry, AB⊥b.

Since line AB belongs to plane γ and passes through intersection point B of b and γ, we have one line on γ that b is perpendicular to.

To complete the proof, we need a second line on γ that is perpendicular to b.

Let's draw two parallel lines c and d in plane γ through, correspondingly, bases A and B. We know that a⊥c (since a⊥γ). Compare an angle formed by a pair of lines a and c with an angle formed by a pair of lines b and d. The former is the right angle and both angles have correspondingly parallel sides. Therefore, these angles are congruent and the latter is the right angle as well.

So, the line d is the second line on plane γ that b is perpendicular to. Together with already proven AB⊥b, this is a sufficient condition for line b to be perpendicular to plane γ.

End of proof.

Theorem 2

If two lines are perpendicular to the same plane, they are parallel to each other.

Proof

Assume that a and b are two lines and plane γ is a plane perpendicular to both of them:

a⊥γ;

b⊥γ.

We have to prove that a ∥ b.

Let points A and B are, correspondingly, bases of perpendiculars a and b:

A=a∩γ,

B=b∩γ.

Construct a line b' that passes through point B and is parallel to line a. According to Theorem 1 above, b'⊥γ. If b and b' are not one and the same line, we have two perpendiculars to plane γ at the same point B, which we have proven before is impossible, the perpendicular to a plane at a point on it is unique.

Therefore, b=b' and, therefore, a∥b.

End of proof.

Theorem 3

If from a point M lying outside plane γ two different lines, a and b, are constructed in such a way that a⊥γ and b∩γ≠∅, then line b is NOT parallel to line a and is NOT perpendicular to plane γ.

Proof

Obviously, a and b are NOT parallel because they intersect at point M.

If b⊥γ, lines a and b would have been parallel (see Theorem 2 above), which we have just rejected.

So, b is NOT perpendicular to γ. Actually, it is slant or oblique line relative to plane γ.

End of proof.

## Thursday, May 14, 2015

### Unizor - Geometry3D - Lines and Planes - Perpendicular Lines and Planes

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Definition 1

A line a is called perpendicular to a plane μ if it intersects this plane at some point M and any line within this plane μ that passes through a point of intersection M is perpendicular to line a.

Definition 2

A non-perpendicular line intersecting a plane is called slant or oblique.

Definition 3

A point of intersection of a line perpendicular to a plane with this plane is called a base of a perpendicular.

Construction 1

Given a line a and a point A on it.

Construct a plane passing through point A perpendicular to line a.

Mini-Theorem 1

There is one and only one plane perpendicular to a given line at a given point.

Construction 2

Given a plane β and a point A on it.

Construct a line a perpendicular to plane β intersecting it at point A.

Mini Theorem 2

There is one and only one line perpendicular to a given plane at a given point.

Theorem

If line a intersects plane μ at some point M (μ∩a=M) and is perpendicular to two different lines, b and c, lying within this plane μ and passing through a point of intersection M (b∈μ; c∈μ; b∩a=M; c∩a=M; b⊥a; c⊥a), then it is perpendicular to any other line n on the plane μ that passes through point M (n∈μ; n∩a=M ⇒ a⊥n) and, consequently, is perpendicular to an entire plane μ (a⊥μ).

## Tuesday, May 12, 2015

### Unizor - Geometry3D - Lines and Planes - Problems 1

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Problem 1

Given a straight line a in three-dimensional space and a point M outside of it.

Construct a line b that passes through a given point M parallel to a given line a.

Solution

There is one and only one plane μ that can be constructed by a given line a and a point M outside it. After constructing this plane μ=μ(M,a), we should construct a line b within this plane μ that passes through a given point M and parallel to a given line a. It's a plane geometry construction that we, supposedly, are familiar with.

That is the line we need.

Problem 2

Given a plane μ and a point M outside of it.

Construct a plane ν that passes through a given point M parallel to a given plane μ.

Solution

On a given plane μ construct two intersecting straight lines a and b.

Construct two corresponding lines a' and b' that pass through a given point M parallel to two lines a and bon the plane μ.

These two new lines a' and b' determine one and only one plane ν that is parallel to a given plane μ, according to a theorem proven in a lecture about parallel planes.

Problem 3

Given two non-intersecting non-parallel lines (skew lines) a and b in three-dimensional space.

Construct a plane μ that passes through line a parallel to line b.

Solution

Choose a point M on line a and construct a new line b' that passes through this point M parallel to line b.

The two intersecting lines a and b' determine one and only one plane μ that is parallel to line b', according to a theorem proven in a lecture about parallel lines and planes.

Problem 4

Given two non-intersecting non-parallel lines (skew lines) a and b in three-dimensional space and a point M not lying on any of them.

Construct a line c that passes through a given point M and intersects both line a and line b.

Solution

The line that connects point M and both skew lines a and b must lie on the plane that is defined by point M and line a as well as on the plane defined by the same point M and another line b.

Therefore, it's the line of intersection of these two planes, each of which can easily be constructed by a point and a line it should contain.

## Monday, May 11, 2015

### Unizor - Geometry3D - Lines and Planes - Parallel Planes

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Prior to listening this lecture, we recommend to recall the main axioms and already proven theorems from previous lectures.

Definition

Two planes in three-dimensional space are called parallel if they do not intersect (that is, have no common points).

There are a few theorems related to parallel planes that we will prove in this lecture.

Theorem 1

(sufficient condition for two planes to be parallel)

Assume that we have two planes γ and δ.

Plane γ contains two intersecting straight lines aγ and bγ.

Plane δ contains two intersecting straight lines aδ and bδ.

These lines are correspondingly parallel, that is

aγ ∥ aδ and bγ ∥ bδ.

Prove that planes are parallel, that is γ ∥ δ.

Proof

Since aγ ∥ aδ, line aγ and plane δ are parallel. Similarly, since bγ ∥ bδ, line bγ and plane δ are parallel.

So, both intersecting lines aγ and bγ are parallel to plane δ.

Assume that planes γ and δ intersect. Their intersection must be some straight line c, according to Axiom 2 introduced earlier in the course. This line c belongs to both planes γ and δ.

Now we have a situation when line aγ is parallel to plane δ and lies in plane γ that intersects plane δ along a line c. According to Theorem 2 of the previous lecture about lines parallel to planes, the line of intersection c is parallel to line aγ.

Similarly, we have a situation when line bγ is parallel to plane δ and lies in plane γ that intersects plane δ along a line c. Therefore, c is parallel to line bγ.

As a result, line c in plane γ must be parallel to two lines, aγ and bγ, intersecting each other. This is not possible in the plane Euclidian geometry. Therefore, our assumption that planes γ and δ intersect is invalid, these planes must have no common points, that is they are parallel.

End of proof.

Theorem 2

Consider two parallel planes γ and δ (symbolically, γ ∥ δ). There is the third plane α that intersects both of them along lines aγ and aδ correspondingly:

γ∩α = aγ

δ∩α = aδ

Prove that aγ ∥ aδ.

Proof

The definition of parallel lines includes two requirements:

(a) lines must lie in the same plane; in our case both lines lie in plane α.

(b) they must not intersect; in our case they do not intersect since they belong to parallel planes.

End of proof.

Theorem 3

Consider two parallel planes γ and δ (symbolically, γ ∥ δ). There are two parallel lines a and b intersecting both these planes.

Let the points of intersection be as follows:

a∩γ = Aγ

a∩δ = Aδ

b∩γ = Bγ

b∩δ = Bδ

Prove that AγBγBδAδ is a parallelogram.

Proof

Line AγBγ and line AδBδ are parallel, according to Theorem 2 above.

Line AγAδ and line BγBδ are parallel, according a preposition of this theorem.

All four points lie in one plane since this is a condition of lines AγAδ and BγBδ being parallel.

Therefore, AγBγBδAδ is a quadrilateral with parallel opposite sides. According to one of the theorems of geometry of quadrilaterals on the plane, this is a parallelogram.

Theorem 4

(extension of Theorem 1 above)

Assume that we have two planes γ and δ.

Plane γ contains two rays originating from the same point Aγ:

AγBγ and AγCγ.

Plane δ contains two rays originating from the same point Aδ:

AδBδ and AδCδ.

These rays are correspondingly parallel and similarly directed, that is

AγBγ ∥ AδBδ and AγCγ ∥ AδCδ.

Prove that angles ∠BγAγCγ and ∠BδAδCδ lie in parallel planes (that is, γ ∥ δ) and are congruent.

Proof

The fact that γ ∥ δ is a subject of Theorem 1 in this lecture. So, it's already proven.

To prove congruence of angles, we will construct triangles where they belong to.

Choose points Bγ, Cγ, Bδ and Cδ such that AγBγ=AδBδ and AγCγ=AδCδ.

Now we have:

AγBγBδAδ is a parallelogram since AγBγ ∥ AδBδ and are equal in length (by construction).

Therefore,

AγAδ ∥ BγBδ and equal in length.

AγCγCδAδ is a parallelogram since AγCγ ∥ AδCδ and are equal in length (by construction).

Therefore,

AγAδ ∥ CγCδ and equal in length.

Based on the two statements above,

BγCγCδBδ is a parallelogram since BγBδ ∥ CγCδ and are equal in length.

Therefore, BγCγ ∥ BδCδ and equal in length.

Hence, two triangles, ΔAγBγCγ and ΔAδBδCδ are congruent by three sides, from which follows the congruence of angles ∠BγAγCγ and ∠BδAδCδ.

End of proof.

## Friday, May 8, 2015

### Unizor - Geometry3D - Lines and Planes - Line Parallel Plane

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Theorem 1. If a straight line a lying outside of a plane γ is parallel to a straight line b that lies in a plane γ then line a and plane γ are parallel.

Theorem 2. If a straight line a lying outside of a plane γ is parallel to this plane and another plane δ contains line a and intersects plane γ, the straight line b of intersection of planes γ and δ is parallel to line a.

Theorem 3. Given a straight line a in three-dimensional space and a point P outside it.

Prove that there is one and only one line b that passes through point P and is parallel to line a.

Notice: Here we deal with three-dimensional space, not with the plane where this statement is, actually, a form of the 5th Euclid's postulate.

Theorem 4. Given a straight line a that is parallel to two intersecting planes γ and δ.

Prove that line a is parallel to line b of intersection between γ and δ.

Theorem 5. Given a straight line a in 3D space and two other lines, m and n, each parallel to line a.

Prove that lines m and n are parallel to each other.

### Unizor - Geometry3D - Lines and Planes - Parallel Lines

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Two straight lines in three-dimensional space are called parallel if

(1) they lie in the same plane and

(2) they do not intersect.

Exercise 1:

Construct a straight line l in three-dimensional space parallel to a given straight line a and passing through a given point P outside of a given line.

Solution:

According to one of the axioms that we discussed in previous lectures, we can construct a plane γ passing through a given line a and a given point P. It exists and is unique, according to this axiom. Then we construct a straight line l parallel to line a and passing through point P within this plane γ using the techniques known from two-dimensional geometry. Obtained line l is a line parallel to a given line a because (1) it lies in the same plane γ with line a and (2) does not intersect a (since it's parallel within plane γ).

Exercise 2:

Prove that there is one and only one line parallel to a given line a and passing through a given point P outside of a given line.

Proof:

The existence of one such line l follows from the construction exercise above.

To prove the uniqueness of such a constructed line, assume there is another line m also parallel to a, also passing through point P.

Since, by definition of parallel lines, lines a and m lie in the same plane (let's call it δ), we have a situation that there are two planes, γ (as constructed to obtain line l) and δ (containing lines a and m), both of which contain a given line a and a given point P. According to one of the axiom, there is one and only one plane that passes through a given line and a given point outside it. So, plains γ and δ are identical, that is they constitute one and the same plane. But in this plane, according to axioms of two-dimensional geometry, there is one and only one line parallel to a given line and passing through a point outside it. Therefore, lines l and m are identical.

That proves the uniqueness of a line parallel to a given line and passing through a given point outside it in the three-dimensional space.

Important part of a definition of parallel lines is the requirement that they must lie in the same plane. Without this requirement two lines might lie in space without intersecting each other and not be parallel in any sense. For instance, a road and a bridge over it might represent such two lines. They do not intersect but cannot possibly be considered as parallel. These non-parallel non-intersecting lines in three-dimensional space are called skew lines.

## Monday, May 4, 2015

### Unizor - Geometry3D - Elements - Construction in 3D

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Construction in three-dimensional space is a more complicated task than on a plane. Two-dimensional construction on a plane can be modeled using paper, straight ruler and compass. In three dimensions this is not that easy. Lots of things we just have to imagine.

Let's start with a simple principle. If we have to construct something on a plane that already exists in three-dimensional space, we will just use the same techniques of construction as we used before when studying the geometry on a plane.

Next is to be able to construct a plane in three-dimensional space. Here we will use axioms and simple theorems we introduced in the beginning and just say that a plane should be considered constructed if, at least, one of these conditions is met:

(a) there are three points that are known to belong to a plane;

(b) there is a line and a point outside of this line that are known to belong to a plane;

(c) there are two intersecting lines that are known to belong to a plane;

(d) there are two parallel lines that are known to belong to a plane.

Each of the above conditions is sufficient to define one and only one plane in three-dimensional space, and that's why we consider our job of constructing a plane completed if one of them is satisfied. So, when asked to construct a plane, we can always resort to one of the above cases and consider the plane is constructed if one of those conditions is met.

But there are other three-dimensional figures we might need to construct. Let's examine the most common ones.

(1) A right prism can be considered constructed if we can construct a polygon in its base and know its altitude (height).

(2) A right circular cylinder can be considered constructed if we can construct a circle in its base and know its altitude (height).

(3) A right regular pyramid can be considered constructed if we can construct a polygon in its base and know its altitude (height).

(4) A right circular cone can be considered constructed if we can construct a circle in its base and know its altitude (height).

(5) A sphere can be considered constructed if we know the position of its center and its radius.

Just as an example, if a problem states that we have to construct a sphere tangential to all faces of a given regular tetrahedron from inside, we have to find a center and a radius of this sphere - and the problem is considered solved.

Construction in three-dimensional space is a more complicated task than on a plane. Two-dimensional construction on a plane can be modeled using paper, straight ruler and compass. In three dimensions this is not that easy. Lots of things we just have to imagine.

Let's start with a simple principle. If we have to construct something on a plane that already exists in three-dimensional space, we will just use the same techniques of construction as we used before when studying the geometry on a plane.

Next is to be able to construct a plane in three-dimensional space. Here we will use axioms and simple theorems we introduced in the beginning and just say that a plane should be considered constructed if, at least, one of these conditions is met:

(a) there are three points that are known to belong to a plane;

(b) there is a line and a point outside of this line that are known to belong to a plane;

(c) there are two intersecting lines that are known to belong to a plane;

(d) there are two parallel lines that are known to belong to a plane.

Each of the above conditions is sufficient to define one and only one plane in three-dimensional space, and that's why we consider our job of constructing a plane completed if one of them is satisfied. So, when asked to construct a plane, we can always resort to one of the above cases and consider the plane is constructed if one of those conditions is met.

But there are other three-dimensional figures we might need to construct. Let's examine the most common ones.

(1) A right prism can be considered constructed if we can construct a polygon in its base and know its altitude (height).

(2) A right circular cylinder can be considered constructed if we can construct a circle in its base and know its altitude (height).

(3) A right regular pyramid can be considered constructed if we can construct a polygon in its base and know its altitude (height).

(4) A right circular cone can be considered constructed if we can construct a circle in its base and know its altitude (height).

(5) A sphere can be considered constructed if we know the position of its center and its radius.

Just as an example, if a problem states that we have to construct a sphere tangential to all faces of a given regular tetrahedron from inside, we have to find a center and a radius of this sphere - and the problem is considered solved.

### Unizor - Geometry3D - Elements - Polyhedrons

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Polyhedrons are objects in three dimensional space, all faces of which are polygons connected to each other at common edges and edges are connected at common vertices.

The simplest example of a polyhedron is a tetrahedron, all 4 faces of which are triangles. It has 4 vertices and 6 edges.

We can imagine pentahedrons with 5 faces or hexahedrons with 6 faces. So, the number of faces is a characteristic of a polyhedron that is important for its classification.

Obviously, all prisms and pyramids are particular types of polyhedrons.

We will mostly be dealing with convex polyhedrons - those whose surface divides the entire three dimensional space into "inside" and "outside" of a polyhedron, in such a way that all points of any segment connecting two points on its surface are located on its surface or inside of it.

A remarkable property of convex polyhedrons is the correspondence between the number of its vertices, edges and faces expressed in the famous Euler's formula:

V − E + F = 2

A particular kind of polyhedrons are regular ones. All their faces are regular polygons congruent to each other with all congruent angles between the faces. Examples of such regular polyhedrons are regular tetrahedron (all faces are equilateral triangles) and cube - regular hexahedron with square faces. There are others as well, known since antiquity.

### Unizor - Geometry3D - Elements - Spheres

Unizor - Creative Minds through Art of Mathematics - Math4Teens

A sphere is a three-dimensional surface that consists of all points in space located on the same distance R (called a radius of this sphere) from some fixed point O (called a center of this sphere).

This definition is a three-dimensional analogue of a two-dimensional circle.

Obviously, this definition is not perfect. Most importantly, we have not defined a concept of a distance in three-dimensional space. We also have to discuss the existence of such points for any real value of a radius R and a center O. These issues are very important and far from being clear. We will discuss them in further lectures. For now we would like to resort to intuitive understanding of these concept.

The surface of our planet is a rough approximation of a sphere and we will use some geography to exemplify the concepts we introduce in this lecture.

If a straight line passes through a center of a sphere, it intersects this sphere at two opposite points. A segment of this line between two opposite points of intersection with a sphere is a diameter.

If our planet is considered as an approximation of a sphere, its axis of rotation is such a line and two poles, North and South, are intersection of this line with a sphere.

Any segment that connects two points on a sphere along a straight line is called a chord (similarly to a two-dimensional case with a chord of a circle).

Diameter is the longest chord in a sphere.

If a plane cuts through a sphere, the intersection is a circle. If this plane passes through a center of a sphere, the intersection will be called a great circle, its radius will be equal to a radius of a sphere itself. All Earth's meridians and an equator are approximations of great circles.

A great circle has the largest radius among all circles formed by an intersection of a sphere and a plane.

Part of a great circle between any of its two points is a spherical arc (or simply arc if a context indicates that it's on a surface of a sphere).

A plane intersecting a sphere cuts it in two parts, each is called a spherical cap (or a spherical dome). This plane is called then a base plane for a spherical cap and a circle of its intersection with a sphere will be called a base circle of a spherical cap.

The radius of a base circle is considered a radius of a spherical cap. The altitude (or height) of a spherical cap is a perpendicular to a base plane from a center of a base circle to the surface of a sphere.

A spherical sector is an object bounded by a spherical cap and a conical surface with a circle of a spherical cap as a directrix and a center of a sphere as an apex.

## Saturday, May 2, 2015

### Unizor - Geometry3D - Elements - Cones

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Consider a plane α that we will call a base plane and a circle c with a center O and radius R on this plane. This circle will be used as a directrix of a conical surface we will construct.

Although non-circular directrix can be considered, we will unlikely deal with such. So, unless otherwise specified, a directrix is assumed to be a circle, conical surface in this case is classified as circular.

Also assume there is a point S outside this plane that we will use as an apex of our conical surface.

Let's construct a conical surface σ using circle c as a directrix and point S as an apex.

Cone is an object in solid geometry formed by a part of a conical surface σ between a base plane α and an apex S and a part of the base plane inside that conical surface - circle c on the base plane α.

The only vertex of a cone is its apex.

There are no edges of a cone.

Circle c is called a base of a cone.

A perpendicular from apex S to a base of a cone is its altitude or height.

If a perpendicular from apex S to a base of a cone falls into its center O, a cone is called a right cone.

In most cases we will be dealing with right circular cones calling them just cones unless otherwise specified.

Now let's imagine that we have another plane β parallel to base α and intersecting a cone's altitude somewhere between an apex and a base.

A geometric object consisting of a conical surface between planes α and β and parts of these planes lying inside the conical surface - a circle on the base plane α (we can call it "bottom") and a corresponding circle on the plane β (we can call it "top") - is called a truncated cone.

All the terminology of cones is the same for truncated cones. The altitude of a truncated cone is a distance between "top" and "bottom" bases along a common perpendicular.

### Unizor - Geometry3D - Elements - Pyramids

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Consider a plane α that we will call a base plane and a polygon ABCDEF on this plane (we specify 6-sided polygon, but it's not important, any polygon will do). This polygon will be used as a directrix of a conical surface we will construct.

Also assume there is a point S outside this plane that we will use as an apex of our conical surface.

Let's construct a conical surface σ using polygon ABCDEF as a directrix and point S as an apex.

Pyramid is an object in solid geometry formed by a part of a conical surface σ between a base plane α and an apex S and a part of the base plane inside that conical surface - polygon ABCDEF on the base plane α.

Points A, B, C, D and all others are called vertices of a pyramid. Apex S is also a vertex, though a special one.

Segments AB, BC, DS and all others are called edges of a pyramid.

Polygon ABCDEF is called a base of a pyramid.

Triangles ABS, BCS and others are called sides of a pyramid.

Base and sides of a pyramid are generically referred to as faces. Sides are also referred to as lateral faces.

A perpendicular from an apex to a base of a pyramid is its altitude or height.

Another classification of pyramids is by their bases.

If a base is a triangle, the pyramid is called triangular pyramid.

If a base is a rectangle, the pyramid is called rectangular pyramid.

If a base is a square, the pyramid is called square pyramid (Egyptian pyramids are of this kind).

If a base is a hexagon, the pyramid is called hexagonal pyramid.

If a base is a regular N-sided polygon and a perpendicular from the apex onto the base falls in a center of this regular N-sided polygon, the pyramid might be called N-pyramid.

Now let's imagine that we have another plane β parallel to base α and intersecting a pyramid's altitude somewhere between an apex and a base.

A geometric object consisting of a conical surface between planes α and β and parts of these planes lying inside the conical surface - polygon ABCDEF on the base plane α (we can call it "bottom") and a corresponding polygon A'B'C'D'E'F' on the plane β (we can call it "top") - is called a truncated pyramid (like the one on the 1 dollar bill of the US currency - one of the Masonic symbols).

All the terminology of pyramids is the same for truncated pyramids. The altitude of a truncated pyramid is a distance between "top" and "bottom" bases along a common perpendicular.

### Unizor - Geometry3D - Elements - Conical Surface

Unizor - Creative Minds through Art of Mathematics - Math4Teens

The subject of this lecture is to introduce a concept of a conical surface in three-dimensional space.

Let's assume that we have a curve c in the three-dimensional space and, separately, a fixed point S outside it.

A curve c does not necessarily is a "flat" one (that is, we do not require all its points to lie on the same plane).

Now imagine that through each point on this curve c we construct a straight line connecting it to a fixed point S outside it. The set of all points of all these lines forms a surface that we call conical.

Let's introduce some terminology related to conical surfaces.

A curve c determines the points through which we draw lines connecting these points to point S. This curve c is called directrix because it directs the position of each line we draw.

The lines that we draw from each point of curve c through a fixed point S form the conical surface and are sometimes called rulings.

The fixed point S connected with each point on a directrix c with rulings, is called an apex or a vertex (a more general term applicable also to other points of geometrical objects).

If a directrix is a closed curve lying on some plane, the area it encompasses on a plane may be referred to as a base of a conical surface.

Let's exemplify this construction without any rigorous proof of characteristics of constructed objects.

If directrix c is a straight line, the result of our construction will be a plane containing both directrix c and apex S.

If directrix c is a circle and apex S is a center of this circle, the result of our construction will be a plane this circle belongs to.

If directrix c is a circle and apex S is positioned outside of a plane containing this circle such that a perpendicular from a apex onto this plane falls in the center of a circle, the result of our construction will be an infinitely long cone propagating to infinity on both sides of a apex.

If, instead of a circle, we choose as a directrix a polygon lying on a plane and apex S is positioned outside of a plane containing this polygon such that a perpendicular from a apex onto this plane falls inside a polygon, the resulting surface would be an infinitely long conical surface, part of which between the polygon and the apex forms a pyramid.

Notice, we did not define yet concepts or perpendicularity and parallelism between planes and straight lines. In this introductory lecture we just assume that students have intuitive understanding of these concept.

Interesting property of any conical surface is that, if it's made of paper, it can be flattened on a flat plane without stretching (which is not the case with a spherical surface that we will introduce in subsequent lectures).

Another viewpoint to a conical surface is that it can be considered as a surface formed by a line passing through a fixed apex and a point moving along a directrix.

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