## Wednesday, September 30, 2015

### Unizor - Geometry3D - Pyramids - Problem 3

Unizor - Creative Minds through Art of Mathematics - Math4Teens

To effectively learn from problem solving, try to solve these problems just by yourself, then listen to a lecture and then try to solve them again by yourself.

Pyramids - Problem 3

Given a triangular pyramid SABC. A point on each side edge is chosen as follows.
Point A' divides edge SA so that AA'/A'S = ka
Point B' divides edge SB so that BB'/B'S = kb
Point C' divides edge SC so that CC'/C'S = kc
In what ratio does a plane that contains these three division points divide the volume of a pyramid?

The ratio of the volume of the bottom part (unevenly truncated pyramid) to the volume of the top part (pyramid with side edges shorter than original) is
(1+ka)·(1+kb)·(1+kc) − 1

### Unizor - Geometry3D - Pyramids - Problem 2

Unizor - Creative Minds through Art of Mathematics - Math4Teens

To effectively learn from problem solving, try to solve these problems just by yourself, then listen to a lecture and then try to solve them again by yourself.

Pyramids - Problem 2

Given a triangular pyramid SABC with all side edges equal in length:
SA = SB = SC
All angles of the side faces at its apex S are right:
∠ASB = ∠BSC = ∠CSA = 90o.
The altitude from apex S onto base triangle ΔABC is SH=h.
Express in terms of altitude h all the edges, area of each face and volume of the pyramid.

Side edge: h√3
Base edge: h√6
Area of side face: 3h²/2
Area of base face: 3h²√3/2
Volume: h³√3/2

### Unizor - Geometry3D - Pyramids - Problem 1

Unizor - Creative Minds through Art of Mathematics - Math4Teens

To effectively learn from problem solving, try to solve these problems just by yourself, then listen to a lecture and then try to solve them again by yourself.

Pyramids - Problem 1

Given a triangular pyramid SABC with side edges that form an arithmetic progression with known first member a and unknown difference x:
SA = a, SB = a+x, SC = a+2x
All angles of the side faces at the apex are right:
∠ASB = ∠BSC = ∠CSA = 90o.
The volume of a pyramid is v.
What are the lengths of side edges?

The difference of the arithmetic progression is
x = ( −3a2+√a4+48av ) ⁄4a
So, the side edges are:
SA=a, SB=a+x, SC=a+2x,
where the value of x in terms of a and v is determined above.

## Thursday, September 24, 2015

### Unizor - Geometry3D - Symmetry - Problems 1

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Symmetry in 3D Space - Problems 1

Problem

Given are plane γ, point A outside it and point P lying on it: A∉γ; P∈γ.

Point B is symmetrical to point A relative to plane γ.
Point C is symmetrical to point A relative to point P.

Prove that points B and C are symmetrical relative to a line d perpendicular to plane γ and going through point P.

## Monday, September 21, 2015

### Unizor - Geometry3D - Axis Symmetry

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Symmetry in 3-D - Symmetry around an Axis

In
this lecture we will discuss the symmetry about an axis (that is,
relative to a straight line) in three-dimensional space that assumes the
existence of an axis of symmetry.

Recall from the symmetry on a
plane that point A', symmetrical to point A relative to an axis of
symmetry (straight line) s, can be constructed by dropping a
perpendicular AP from point A onto axis s (P∈s, AP⊥s) and extending this
perpendicular beyond point P by the same length, thus obtaining point
A'.

In three-dimensional space symmetry about an axis requires analogous construction.
If
we are given point A and an axis of symmetry (straight line) s, then
point A', symmetrical to point A relatively to axis s, is located on a
continuation of perpendicular AP to axis s (P∈s, AP⊥s) beyond point P by
the same distance as the length of segment AP.

The construction
of a symmetrical point is, obviously, reversible. If we start from point
A', drop a perpendicular to an axis of symmetry A'P and extend by the
same length, we will be at point A. It follows from the uniqueness of a
perpendicular from a point onto a straight line. So, if point A' is
symmetrical to point A relatively to axis s, point A is symmetrical to
point A' relatively to the same axis of symmetry s.

The other
viewpoint on symmetry around an axis in three-dimensional space is
considering the symmetrical reflection relative to an axis as a result
of rotation in space around this axis by 180o.

Rotation of point A in space around an axis s by an angle φ can be constructed as follows:
(a) draw a plane β perpendicular to axis s through point A (s⊥β, A∈β);
(b)
draw a line on plane β connecting point O of intersection of plane β
and axis s (O = β∩s) to point A; since OA∈β and s⊥β, OA⊥s;
(c) within plane β rotate segment OA by angle φ around point O, so point A would take position A'.
In
particular, if φ=180o, point A' would be on continuation of segment OA
beyond point O by the same distance as the length of segment OA, which
exactly corresponds to the rules of construction of a point symmetrical
relative to an axis.

We can say now that two points A and A',
symmetrical relatively to axis s, are centrally symmetrical within plane
β drawn perpendicularly to axis s trough point A. The center of
symmetry within plane β is a point of its intersection with axis s.

The
symmetry can be viewed also as an operator on points of
three-dimensional space. Given an axis of symmetry, this operator
transforms each point into its image constructed by the rules above. We
will sometimes relate to symmetry as a transformation assuming exactly
this type of operation.

There are a few simple theorems we'd like to present about symmetry about an axis in three-dimensional space.

Theorem 1
Symmetry about an axis preserves the distance between points.

Proof
Consider two points A and B and axis of symmetry s, so each point has its symmetrical counterpart - A' and B' correspondingly.
We have to prove that AB=A'B'.
Draw
plane γ trough point A perpendicular to axis s, intersecting it at
point P, and plane δ through point B also perpendicular to our axis s,
intersecting it at point Q:
A∈γ, γ⊥s (⇒ A'∈γ)
B∈δ, δ⊥s (⇒ B'∈δ)
Let C and C' be projections onto plane γ of points B and B' correspondingly.
Obviously, BB'C'C is a rectangle.
First,
let's prove that projections AC and A'C' of segments AB and A'B' onto
plane γ are congruent. Then it will easily follow the congruence of
segment AB and A'B'.
But congruence between AC and A'C' is obvious
since points C and C' are centrally symmetrical relatively to the same
center of symmetry P as points A and A' (a trivial statement of the
plane geometry).
Since AC=A'C', right triangles ΔABC and ΔA'B'C' are
congruent - their catheti AC and A'C' are congruent and their catheti BC
and B'C' are both equal to a distance between two parallel planes γ and
δ.
Therefore, hypotenuses AB and A'B' are congruent. The length of a
segment AB is preserved by symmetry relative to an axis. Since the
lengths of segments are preserved, angles between intersecting lines are
preserved as well, as it's sufficient to include any angle into a
triangle and symmetry around an axis will preserve the length of its
sides and, therefore, all its angles.
End of proof.

Consequently,
symmetry around an axis is an invariant transformation, that is the
transformation that preserves lengths of segments and values of angles.

Theorem 2
An object symmetrical to a straight line relatively to some axis is a straight line.

Theorem 3
An object symmetrical to a plane relatively to some axis is a plane.

## Friday, September 18, 2015

### Unizor - Geometry3D - Cuboctahedron

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Cuboctahedron

Given a cube ABCDA'B'C'D'. We cut-off all its vertices as follows.
Let points P, Q and R be midpoints of, correspondingly, edges AB, AD and AA'. The plane going through these three points cuts-off a triangular pyramid APQR from our cube.
Similar procedure we perform for each remaining vertex. The remaining solid after all these cuts is cuboctahedron.
What is the ratio of a volume of remaining part of a cube (cuboctahedron) to its original volume?

## Thursday, September 17, 2015

### Unizor - Geometry3D - Truncated Pyramid

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Truncated Pyramids

Imagine a pyramid. Pick any point on any of its side edges and draw a plane through this point parallel to a base plane. Cut-off a piece of a pyramid between this new plane and an apex. Whatever is left is called a truncated pyramid.

Let's do a concrete construction of a truncated pyramid.

Assume that the original pyramid was SABCD with apex S and quadrilateral ABCD that belongs to a base plane β.
Pick some point A' on edge SA between apex S and vertex A on the base plane β.
Now draw a plane γ through point A' parallel to base β.
Plane γ intersects edges SB, SC and SD at points B', C' and D' correspondingly, so we have quadrilateral A'B'C'D' within plane γ that bounds our truncated pyramid from above.
We can say now that object ABCDA'B'C'D' is a truncated pyramid.

Since β ∥ γ,
the corresponding edges lying on parallel planes are parallel:
AB ∥ A'B',
BC ∥ B'C',
CD ∥ C'D',

Each side face of a truncated pyramid is, therefore, a trapezois.
Also, from the above property follows similarity of corresponding triangles:
ΔSAB ~ ΔSA'B',
ΔSBC ~ ΔSB'C',
ΔSCD ~ ΔSC'D',
The center of scaling for all these similarities is apex S and, since pairs of these triangles share sides, they all share the same scaling factor.
That same scaling factor is between the bases of two pyramids - triangles ΔABC and ΔA'B'C'.
Therefore, two pyramids, small pyramid SA'B'C'D' that we cut-off and big original pyramid SABCD are similar with the same scaling center and factor.

Drop a perpendicular from apex S onto base β.
It falls into point H, so SH⊥β.
This same perpendicular is also perpendicular to plane γ since planes β and γ are parallel.
Let the intersection of this perpendicular with γ be point H', so H'∈AH and SH'⊥γ.

It's easy to prove that the same scaling with a center at apex S and the same factor as above for pyramids transforms point H' into H (good exercise for self-study).

We see that similarity between pyramids includes not only all edges but altitudes SH' and SH as well.
Now we can proceed with calculating the volume of a truncated pyramid as a difference between volumes of two pyramids - the original one SABCD and the one we cut-off from it, SA'B'C'D'.

Assume that the factor of scaling we discussed above is f.
Then all linear elements of two pyramids are in this ratio: LenSA/LenSA'=f (edge factor)
(the same for all other edges)
Let the lengths of altitudes SH and SH' be h and h' correspondingly.
Then h/h'=f (altitude factor).
Let the areas of bases of two pyramids, big SABCD and small SA'B'C'D' be s and s' correspondingly.
As was explained in the topic 3-D Similarity, the areas of similar objects are proportional to a square of a scaling factor:
Then s/s'=f².

The volumes of small and big pyramids are:
VolumeSA'B'C'D' = s'·h'/3
VolumeSABCD = s·h/3
Therefore, the volume v of a truncated pyramid ABCDA'B'C'D' equals to
v = s·h/3 − s'·h'/3

This is a good formula, but not good enough. It would be much cleaner to express the volume of an object in terms of elements of this object. In the formula above s and s' are such elements - areas of top and bottom bases of a truncated pyramid, but h and h' are not elements of a truncated pyramid since they involve a distance from apex S, which is outside of our truncated pyramid) to its bases.
Our goal now is to replace dependency on these parameters with a dependency on the height of a truncated pyramid itself - the distance between its bases a.

Let's summarize what we know and what has to be done in terms of elements of a truncated pyramid.
We know that
h/h' = f
h − h' = a
s/s' = f²
We have to express the volume v=s·h/3−s'·h'/3 in terms of s, s' and a.

Consider the top three equations above as a system of three equations with three unknown h, h' and f.
Solving it and substituting the solutions for h and h' into a formula for volume will produce the desired result.
v = (a/3)·(s+√s·s'+s')

## Wednesday, September 16, 2015

### Unizor - Geometry3D - Pyramids - Volume of Any Pyramid

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Volume of Any Pyramids

In the previous lecture we have proved, based on the Cavalieri's principle, that the volume of a triangular pyramid equals to a product of the area of its base and one third of an altitude. Let's expand this to all other types of pyramids.

Let's consider a pyramid with any polygon as a base. For instance, hexagonal pyramid SABCDEF with apex S and hexagon ABCDEF as a base. To evaluate its volume, draw diagonals AC, AD and AE. Draw planes through each of these diagonals and apex S. These three planes dissect our pyramid into four triangular pyramids SABC, SACD, SADE and SAEF.

All these four triangular pyramids share the same altitude since their bases belong to the same plane and they have a common apex. Let the length of this altitude be h.
The volumes of these pyramids are:
VolumeSABC = AreaΔABC·h/3;
VolumeSACD = AreaΔACD·h/3;
VolumeSAEF = AreaΔAEF·h/3;

Therefore, the volume of our original hexagonal pyramid equals to
VolumeSABCDEF =
(AreaΔABC+AreaΔACD+
= AreaABCDEF·h/3

Again, we see that
THE VOLUME OF A PYRAMID EQUALS TO A PRODUCT OF THE AREA OF A BASE AND ONE THIRD OF THE ALTITUDE.

Example 1
Given a tetrahedron (right regular triangular pyramid with all edges of the same length) with the length of each edge d.
What is its volume?

d³√2/12

Example 2

The base of the Egyptian (square) pyramid is a square with the length of a side d. The side edge of a pyramid is s.
Calculate the volume of this pyramid.

[d²·√(s²−d²/2)]/3

Example 3

The base of the Egyptian (square) pyramid is a square with the length of a side d. The volume of a pyramid is V.
What is the length of its side edge?

√[(3V/d)²+d²/2]

Example 4

A regular right hexagonal pyramid has an altitude equal to a radius of a circle that circumscribes a regular hexagon at its base and equals to d.
Calculate the volume of this pyramid.

d³√3/2

Example 5

Given a triangular pyramid SABC (vertex S is an apex, triangle ΔABC is a base).
Let P be a midpoint of segment AB and Q be a midpoint of segment AC.
Draw a plane through points S, P and Q. It cuts our pyramid into two parts.
Find the ratio of their volumes.

Ratio of a bigger part to a smaller is 3.

Example 6

Given a tetrahedron (right regular triangular pyramid with all edges of the same length).
What part of its volume is within an octahedron obtained by connecting all midpoints of its edges?

One half.

## Tuesday, September 15, 2015

### Unizor - Geometry3D - Pyramids - Volume from Cavalieri Principle

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Volume of Pyramids and Cavalieri's Principle

We will construct a prism from our pyramid that has the same base, altitude and an edge. Then we will show that our prism takes only one third of the volume of this prism.

Assume that SABC is a triangular pyramid with a base plane β that contains triangle ΔABC and apex (top vertex) S.
Let's add a few elements to this pyramid to make a prism.
Construct a plane γ parallel to plane β going through apex S of a pyramid. This plane will contain the top base of a prism we construct.
From points B and C we draw lines parallel to edge SA. These lines intersect with plane γ at points B' and C' correspondingly.
Consider an object bounded by two base planes β and γ and side faces SABB', SACC' and BB'C'C. Obviously, it's a triangular prism, we recommend to prove it as a self-study exercise.

All side faces of this prism are parallelograms.
Draw diagonal B'C in parallelogram BB'C'C.
Consider now two new triangular pyramids inside our prism:
CSB'C' with apex C and base SB'C' and
CSB'B with apex C and base SB'B.

These two pyramids, combined with our original pyramid SABC, dissect the prism into three pyramids. We will prove that the volumes of all three are the same and, therefore, the volume of our original pyramid equals to one third of the volume of prism with the same base and altitude.

First of all, consider pyramids SABC (original one with base ABC) and CSB'C'. They have the same altitudes (the distance between planes β and γ) and congruent bases ABC and SB'C'. Turning pyramid CSB'C' upside down, putting its base SB'C' on plane β and placing its apex into point S, we see two pyramids with congruent bases lying on the same plane and coinciding top vertices. This is exactly the situation discussed in Theorem B of the lecture Mini Theorems 1 of a previous topic 3-D Similarity. Therefore, the volumes of these two pyramids are equal.

Next, consider pyramids SABC (the original one) and CSBB'. Since any vertex in a triangular pyramid can be considered its apex, while the other three vertices form a base, consider our original pyramid SABC with an apex S and base ABC as a pyramid CSAB with apex C and base SAB. Now two pyramids CSAB and CSBB' have common apex C and their bases are two halves of a parallelogram ASB'B'. This is exactly the situation discussed in Theorem C of the lecture Mini Theorems 1 of a previous topic 3-D Similarity. Therefore, the volumes of these two pyramids are equal.

We see that, on one hand, the volume of the original pyramid SABC equals to the volume of pyramid CSB'C' and, on the other hand, the volume of that same pyramid (considered as CSAB with apex C) equals to the volume of pyramid CSBB'. Since these three pyramids combined form a prism, the volume of each pyramid is one third of the volume of a prism, which, as we know, equals to a product of the area of its base and the altitude.

Again, we see that
THE VOLUME OF A PYRAMID EQUALS TO A PRODUCT OF THE AREA OF A BASE AND ONE THIRD OF THE ALTITUDE.

## Wednesday, September 9, 2015

### Unizor - Geometry3D - Pyramids - Volume as Limit

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Volume of Pyramids as Limit

We strongly recommend to review a lecture that introduces a concept of a pyramid under a topic "Elements of Solid Geometry".
It introduces a pyramid-related terminology we will use in this and other lectures.

We also suggest to review a lecture Area as Limit in the previous topic 3-D Similarity, since we are going to use an analogous method to evaluate the volume of a pyramid, just extend it to a three-dimensional case.

The most practical aspect of a theory of pyramids is their volume. This lecture will study this issue and we will derive a formula for a volume of the pyramid.
We will approach this problem from two different angles, both not absolutely rigorous, but intuitively acceptable. The rigorous proof is based on more advanced topics studied in Calculus.

Approach 1

We will approximate a volume of a pyramid with a volume of an object that consists of little steps around this pyramid and getting close to its slanted side faces.

Assume that SABC is a triangular pyramid with a base plane β that contains triangle ΔABC and apex (top vertex) S.

Drop an altitude SH from apex S onto a base plane β (point H is the base of this perpendicular).
Let's divide segment SH into N equal parts and label the division points (sequentially, from S to H) as H1, H2, ...,HN-1. We will identify point H as HN for convenience.

Draw N-1 planes parallel to base plane β through each division point on altitude SH. The plane #n, that we will call βn, going through point Hn, intersects our pyramid at triangle ΔAnBnCn, where n is any integer number from 1 to N-1. We will identify point A, B and C as AN, BN and CN correspondingly for convenience.

Construct a short right prism using triangle ΔAnBnCn as a bottom base and the plane βn−1 just above it as a plane where the top base is located. Let this top base be triangle ΔA'nB'nC'n.

It's intuitively acceptable (and we are not going to prove it rigorously now because of complexity) that the composition of all these short prisms resembles the shape of a pyramid, that the resemblance is better when the number of prisms increasing, while the height of each decreasing and that the combined volume of these prisms approximates the volume of a pyramid with the approximation becoming more and more precise as the number of prisms N grows to infinity.

So, let's evaluate the combined volume of these prisms and determine the limit it tends to as N→∞.

As we know, the volume of a prism equals to a product of the area of the base by its height.
The height of each prism is the same and equals to 1/N of the height of a pyramid AH.
The area of the base of the prism #n is the area of the triangle ΔAnBnCn. To evaluate it, consider similarity between this triangle and the base of a pyramid - triangle ΔABC. The similarity is very easy to prove based on a scaling with a center at the apex S and the factor n/N. As we know (see 3D Similarity topic), similar flat objects have their areas proportional to a square of the scaling factor. Therefore, the area of triangle ΔAnBnCn equals to the area of triangle ΔABC multiplied by a factor n²/N².

Let the height of our pyramid AH be equal to h and the area of its base triangle ΔABC be equal to s. Then the prism #n has volume equal to
vn = (s·n²/N²)·(h/N)
Simplifying this, we get
vn = (s·h/N³)·n²

The next step is to summarize this expression for all n from 1 to N.
Since s, h and N are constants, Σn∈[1,N](vn) =
= (s·h/N³)·Σn∈[1,N](n²)

It's easy to derive that Σ[n²] = N(N+1)(2N+1)/6

Using this in the formula for a volume of an object that contains short prisms stacked on the top of each other, we obtain the following:
Σn∈[1,N](vn) =
= (s·h/N³)·N(N+1)(2N+1)/6 =
= s·h/3+s·h·/2N+s·h·/6N²

As N→∞, the above expression tends to s·h/3, which is the formula for a volume of a pyramid:
A PRODUCT OF THE AREA OF A BASE AND ONE THIRD OF THE ALTITUDE.

## Tuesday, September 8, 2015

### Unizor - Geometry3D - Similarity - Area as Limit

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Area as Limit

Let's prove the well known formula for an area of a triangle using the similarity and the limit theory.
While it's pretty easy to prove it geometrically by doubling the area of a triangle to an area of a parallelogram and transform a parallelogram into a rectangle with a known formula for an area, it's important to go through this other method as an exercise before we use it for calculating a volume of a pyramid.

Assume we have some triangle ΔABC. To evaluate its area, we will construct a stack of rectangles around it and, as we increase the number of these rectangles to infinity, we will evaluate the limit of their combined area, assuming that it gets closer and closer to a true area of a triangle.

Let AH be an altitude of this triangle with base H lying on line BC. Let its length be h and let the length of a base BC be a.

Let's divide segment AH into N equal parts and label the division points (sequentially, from A to H) as H1, H2, ...,HN-1. We will identify point H as HN for convenience.
Draw N-1 lines parallel to base BC through each division point on altitude AH. The line #n, going through point Hn, intersects side AB at point Bn, where n is any integer number from 1 to N-1. We will identify point B as BN for convenience. That same line #n intersects side AC at point Cn. We will identify point C as CN for convenience.

Draw a short perpendicular BnB'n from each point Bn to line Bn−1Cn−1.
Draw a short perpendicular CnC'n from each point Cn to line Bn−1Cn−1.

Consider a rectangle BnCnC'nB'n. Let's calculate its area for each n and summarize all these areas to approximate the area of a triangle ΔABC.

All these rectangles have the same height, that is equal to h/N. The width of these rectangles are all different. However, we can use the similarity between triangles ΔABnCn and ΔABC. Since their altitudes AHn and AH relate as n/N, we conclude that the ratio between their bases BnCn and BC is the same.
Therefore, the length of BnCn equals to a·n/N.

Now we can calculate the area of the nth rectangle:
Sn = (a·n/N)·(h/N) = a·h·n/N²

Summarizing this by all n from 1 to N, we get the approximation for the area of our triangle ΔABC:
SΔABC ≈ Σ(a·h·n/N²) =
= (a·h/N²)·Σ(n),
where the summation is performed for all n from 1 to N. The latter represents a sum of arithmetic progression that is equal to N·(N+1)/2 (see Algebra - Sequence and Series - Arithmetic Progression in this course).
So, the resulting approximation is:
SΔABC ≈ (a·h/N²)·N·(N+1)/2 =
= [(N+1)/N]·(a·h/2) =
= (1+1/N)·(a·h/2) =
= a·h/2 + a·h/(2·N)

Recall that we assumed that, as N increases, the total area of all rectangles gets closer and closer to the area of the original triangle. If N tends to infinity, the latter formula for approximate area of the triangle tends to a·h/2 since the second term tends to 0.
Therefore, we can conclude that the area of a triangle is:
SΔABC = a·h/2,
which is exactly the one we all know from plane geometry and purely geometric considerations.
The end.