Unizor - Geometry2D - Apollonius Problems - Lines and Circles

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

The problems below are easily solvable using the transformation of inversion (symmetry relative to a circle). We recommend to review a previous lecture dedicated to this topic.
The main technique we will use is transformation of a circle passing through a center of an inversion circle into a straight line.
Here is a quick reminder of how inversion works.
An inversion circle with radius R and center O is given. Then any point P (other than center O) is transformed into point P' such that P' lies on a ray from center O to given point P and OP·OP'=R²
We have proven that any circle passing through a center of an inversion circle is transformed by an inversion into a straight line and a circle not passing through a center of inversion is transformed into a circle.

Apollonius Problems -
Lines and Circles

Problem LLC
Construct a circle tangential to two given lines, a and b, and tangential to a given circle c, presuming that both lines are located outside of circle c.

Solution
Analysis:
Assume, our circle is constructed. It is tangential to lines a and b and to circle c.
We can easily reduce this problem to the one we solved before - construction of a circle tangential to two lines and passing through a point.
To accomplish this, increase the radius of a presumably constructed circle by the radius of a given circle and shift both given lines, a and b, away from a given circle in between them by the same value to new positions - a' and b'.
Now a new circle, concentric with the one we have to construct and having a radius greater than that by a radius of a given circle c, will be tangential to lines a' and b' and passing through a center of a given circle c.
Construction:
From a center of circle c drop a perpendicular onto line a and extend it by the radius of circle c beyond its base on line a to point A. Draw line a' parallel to line a through point A.
From a center of circle c drop a perpendicular onto line b and extend it by the radius of circle c beyond its base on line b to point B. Draw line b' parallel to line b through point B.
Construct a circle tangential to lines a' and b' and passing through a center of c.
Decrease the radius of this new circle by a radius of circle c leaving a center in place. This is a required circle.

Problem LCC
Construct a circle tangential to a given line a and tangential to two given circles, c and d, presuming that none of the given elements lies inside the other.

Solution
Analysis:
Assume, our circle is constructed. It is tangential to line a and to circles c and d.
Increase the radius of a presumably constructed circle by the smaller radius among two given circles (let's say, it's circle c), shift line a away from circle c parallel to itself by this same radius to a new positions a', and concentrically shrink circle d by the same radius to a new circle d'.
Now a new circle, concentric with the one we have to construct and having a radius greater than that by a radius of a given circle c, will be tangential to line a', to circle d' and passing through a center of a given circle c.
We can construct it by introducing an inversion with a center at the center of circle c that transforms line a' into some circle aa, circle d' - also into some circle dd, and the circle concentric to the one we need to construct - into a line tangential to these two circles. aa and dd.
Construction:
From a center of circle c drop a perpendicular onto line a and extend it by the radius of circle c beyond its base on line a to point A. Draw line a' parallel to line a through point A.
Draw a circle d' concentric to circle d with a radius smaller than original by a radius of circle c.
Draw an inversion circle with a center at the center of circle c and invert relative to it line a' into circle aa and circle d' into circle dd.
Construct a line tangential to circles aa and dd.
Invert this line into a circle.
Decrease the radius of this new circle by a radius of circle c leaving a center in place. This is a required circle.

Unizor - Geometry2D - Apollonius Problems - Points and Circles

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

The problems below are easily solvable using the transformation of inversion (symmetry relative to a circle). We recommend to review a previous lecture dedicated to this topic.
The main technique we will use is transformation of a circle passing through a center of an inversion circle into a straight line.
Here is a quick reminder of how inversion works.
An inversion circle with radius R and center O is given. Then any point P (other than center O) is transformed into point P' such that P' lies on a ray from center O to given point P and OP·OP'=R²
We have proven that any circle passing through a center of an inversion circle is transformed by an inversion into a straight line and a circle not passing through a center of inversion is transformed into a circle.

Apollonius Problems -
Points and Circles

Problem PPC
Construct a circle passing through two given points, A and B, and tangential to a given circle c, presuming that both points are located outside of circle c.

Solution
Analysis:
Assume, our circle is constructed. It passes through points A and B and is tangential to circle c.
Using point A as a center of inversion and any radius draw a circle that will serve as an inversion circle.
Transform all elements relative to this inversion circle.
Point B will transform into point B', circle c will be transformed into circle c' and a circle we assumed as constructed (passing through a center of inversion A) will be transformed into a line that will be a tangent from point B' to circle c'.
Construction:
Draw an inversion circle q with center at point A and some radius R.
Construct images of point B and circle c relative to inversion circle q.
Draw a tangent from point B' to circle c'.
Transform a tangent through inversion relative to our inversion circle q. The resulting circle would be tangential to circle c and pass through points A and B.

Problem PCC
Construct a circle passing through a given points, A and tangential to two given circles, c and d, presuming that none of the given elements lies inside the other.

Solution
Analysis:
Assume, our circle is constructed. It passes through point A and is tangential to circles c and d.
Using point A as a center of inversion and any radius draw a circle that will serve as an inversion circle.
Transform all elements relative to this inversion circle.
Circle c will be transformed into circle c', circle d will be transformed into circle d', and a circle we assumed as constructed (passing through a center of inversion A) will be transformed into a line that will be a tangent to both circles c' and d'.
Construction:
Draw an inversion circle q with center at point A and some radius R.
Construct images of circles c and d relative to inversion circle q.
Draw a tangent to circles c' and d'.
Transform a tangent through inversion relative to our inversion circle q. The resulting circle would be tangential to circles c and d.

Unizor - Geometry2D - Inversion

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Inversion

Before addressing construction problems of Apollonius, where a circle is one of the given elements, to which a constructed circle must be tangential, we will define a special transformation on a plane that allows to convert these types of problems into problems that involve only lines and points.
In other words, we will try to simplify our task using this transformation and to reduce its solution to a known solution with only points and lines given.
This is a rather clever method which is not easy to come up with by yourself. Yet, it is relatively simple and allows easily to solve Apollonius problems that involve given circles.

The transformation we would like to address is called inversion or symmetry relatively to a circle.
Any transformation of the points on a plane prescribes how each point is transformed. For instance, symmetry relatively to a straight line transforms all points lying on one side of a plane (let's call it here "left of the symmetry line") to points on another side ("right of the symmetry line") and all points from the "right" side to corresponding points on the "left". This transformation has a rule that the source and the image points lie on a perpendicular to a symmetry line and on the same distance from it. Points on the symmetry line are transformed into themselves.

The definition of inversion is somewhat similar, but slightly more complex. We divide the plane by a circle of certain radius R with a center O (inversion circle) into two parts - inside and outside of this circle.
All points of the inside area (except a center of an inversion circle) are transformed into points outside and all outside points are transformed to inside. The rule of transformation is as follows.
For any point P inside of an inversion circle we connect a center of an inversion circle with this point by a ray and find on this ray on the outside of a circle such a point P' that
OP·OP' = R²

Similarly, for any point P' outside of an inversion circle we find point P inside that satisfies the same condition.
Obviously, if point P inside an inversion circle is transformed into point P', then this point P' is transformed by an inversion to point P. Also obvious is that all points on the inversion circle are transformed into themselves. These properties qualify the transformation of inversion to be called "symmetry relative to a circle".

The closer point P lies to a center - the farther from a center will be P' to preserve the main rule of inversion.
The center of an inversion circle does not participate in the transformation, though sometimes one might say that a center is transformed into "infinity", which only means that, as we move point P infinitely closer to a center, its image P' moves infinitely far from a center.

Inversion of a Straight Line

Theorem
Any straight line lying outside of an inversion circle is transformed into a circle inside the inversion circle that passes through a center of inversion.

Inversion of a Circle

Lets consider a circle of radius r concentric with an inversion circle and lying inside it. For any point P on it the segment OP will have a length r. Therefore, the distance from a center of an inversion circle to an image of this point P' must be R²/r for the main rule of inversion to hold. As point P goes around a circle of radius r, its image P' should always be on a distance R²/r from point O - that is, all image points must lie on a circle of radius R²/r concentric with an inversion circle.
We see now that a circle, concentric with an inversion circle is transformed into another concentric circle.

The interesting theorem we are going to prove is that the fact of transformation of a circle into a circle is much broader and expands to circles that are not concentric with an inversion circle. This is the main subject of this lecture.

Theorem
Any circle lying inside an inversion circle and not passing through its center is transformed into a circle outside the inversion circle.

Unizor - Geometry3D - Spheres - Problems 3

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spheres - Problems 3

Just as a reminder, here are a few formulas related to spheres. They are all derived in the previous lectures and are helpful in solving these problems.
Volume of a sphere: 4πR³/3
Surface area of a sphere: 4πR²
Volume of a cap: πH²(3R−H)/3
Surface area of a dome: 2πRH
Volume of a sector: 2πR²H/3

Problem A
Diameter of Jupiter is 11 times greater than diameter of Earth.
Diameter of Mars is half of diameter of Earth.
What is the ratio of surface and volume of Jupiter to those of Mars, assuming all the planets are spherical?

Answer:
The ratio of the surface areas of Jupiter to that of Mars is 484:
SJupiter = 484·SMars
The ratio of their volumes is 10648:
VJupiter = 10648·VMars

Problem B
A cylindrical glass of radius R is filled with water to the middle.
A ball (heavier than water) of a radius r (smaller than R) is dropped into this glass.
By how much the height of the water in a glass will rise?

Answer:
The water will rise by
h = 4r³/(3R²)

Problem C
An air-filled ball made of some heavier than water material is floating in the water such that exactly half of it is above the water.
According to Archimedes' principle of flotation, the weight of a ball equals to the weight of the water it displaced.
The ball's material unit weight is B, the water's unit weight is W.
Assume that the air inside the ball is weightless.
The radius of a ball is R
Construct the equation to solve to determine the width of the ball's wall.

Solution:
Assume that the width of the ball's wall is X. Then the internal radius of the ball is R−X.
The ball's weight equals to its unit weight B multiplied by the volume occupied by its wall, which is a difference between the volumes of two spheres, one has an external diameter of a ball, another is inside it and has its internal diameter:
B·[4πR³/3−4π(R−X)³/3]=
= 4πB[R³−(R−X)³]/3
This weight should be equal to the weight of displaced water, which has a unit weight W and the volume equal to half of the volume of a ball:
W·2πR³/3
The above two weights, that of the ball and that of the displaced water, must be equal, which produces the following equation for X:
4πB·[R³−(R−X)³]/3 =
= 2πW·R³/3
Simplifying this equation, we get
2B·[R³−(R−X)³] = W·R³
Generally speaking, this is an equation of the third degree and can be solved numerically.

Problem D
A sphere is inscribed into a regular tetrahedron, it's tangential to all its four faces.
Prove that the radius of this sphere is 1/4th of the altitude of a tetrahedron.

Unizor - Geometry2D - Apollonius Problems - Points and Lines

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Generally speaking, all problems related to construction of a circle tangential to three other geometrical objects, points, straight lines or circles, is referred to as one of the Apollonius problems. Obviously, there is a history behind it. We will not talk about this history and, for those interested, the Web is your best source for details.
Here we will spend some time to address these problems using some old, previously addressed, and some new techniques, wherever they are applicable.

Apollonius Problems 1

Problem PPP
Construct a circle passing through three given points, A, B and C, not lying on the same line.

Solution
This problem had been addressed before. Here is a recap.
A locus of points equidistant from two given points is a line passing through a midpoint of a segment connecting these two points and perpendicular to it (prove it!).
If three points A, B and C are not lying on the same line, the midpoint perpendicular to AB intersects the midpoint perpendicular to BC (prove it!), and their intersection is equidistant from all three points - A, B and C.
Incidentally, the consequence of this is that a perpendicular to a midpoint of AC also passes through the same intersection point, so all three midpoint perpendiculars intersecting at one point - a center of a circle passing through three given points (prove it!).

Problem PPL
Construct a circle passing through two given points, A and B, and tangential to a straight line d, presuming that both points are located on the same side from the line.

Solution
Solution is based on the Tangent-Secant Power theorem that states that, if from a point P outside a circle drawn a tangent with a point of tangency T and a secant intersecting a circle at points A and B, then
PA·PB = PT² (prove it!).
Given PA and PB, we can find PT as an altitude towards a hypotenuse in a right triangle, where the base of this altitude divides a hypotenuse into segments equal to PA and PB.
Knowing that, in case AB is not parallel to line d, we find their intersection P and find PT, thereby determining a point of tangency of line d and a circle.
Once this point T is found, two segment bisectors for segments AB and AT would give a center of a circle.
If AB ∥ d, point T lies on intersection of d and segment AB's bisector.

Problem PLL
Construct a circle passing through a given point A and tangential to two straight lines p and q.

Solution
A solution can be based on a transformation of scaling.
Consider a general case of an angle formed by lines p and q with a vertex M and point A inside this angle.
Assume a circle tangential to lines p and q is constructed and it passes through point A.
Connect vertex M with point A and consider a scaling of this picture relative to center of scaling positioned at vertex M with any factor. Obviously, lines p and q will be transformed into themselves, an old circle tangential to these lines will be transformed into a new circle also tangential to the same lines, and point A on an old circle will be transformed into another point A' lying on line MA an on a new circle.
Based on this, a construction can be started with any circle tangential to lines p and q (its center is, obviously, on a bisector of an angle formed by these lines). Consider a line MA and its intersection with this circle at point A' (two points will lead to two solutions).
Now all we have to do is apply a scaling that transforms point A' into point A. This scaling will transform a circle we started with into the one needed.
A case of parallel lines p and q is trivial since we can easily determine the radius of a circle we want to construct - it is half the distance between these lines.

Problem LLL
Construct a circle tangential to three straight lines p, q and r.

Solution
Lines should not be all parallel. If two are, the case is simple since we know the radius of a circle. So, consider a general case of three lines forming a triangle.
In this general case we deal with a problem of inscribing a circle into a triangle. As we know, it's center lies on intersection of its angles' bisectors.

Unizor - Geometry3D - Spheres - Problems 2

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spheres - Problems 2

Just as a reminder, here are a few formulas related to spheres. They are all derived in the previous lectures and are helpful in solving these problems.
Volume of a sphere: 4πR³/3
Surface area of a sphere: 4πR²
Volume of a cap: πH²(3R−H)/3
Surface area of a dome: 2πRH
Volume of a sector: 2πR²H/3

Problem
Given a sphere of some unknown radius.
A cylindrical hole is drilled through it such that the axis of a cylinder goes through a center of a sphere. The height of a hole from edge to edge is h.
What is the volume of remaining part of a sphere?

Answer
V = πh³/6

Solution

The drilled out part of a sphere is a cylinder with two caps on both ends.
To calculate their volumes we need the radius of a sphere R, the radius of a cylinder r, the height of each cap H and the heights of a cylinder h with only the height of a cylinder h given.
Challenging, is not it?

From Pythagorean Theorem:
R² = r² + (h/2)²
The heights of a cylinder and caps is related to the radius of a circle:
R = H+h/2
So, we have only two equations with three unknown - R, r and H.

Let's express r² in terms of known h and unknown H:
r² = (H+h/2)²−(h/2)² = H²+Hh

Let's proceed with calculations of a volume of the remaining after drilling part of a sphere, hoping that the resulting formula would be a function of only cylinder's height h.

The volume of a sphere is
Vsphere = 4πR³/3
The volume of a cylinder is
Vcyl = πr²h
The volume of each cap is
Vcap = πH²(3R−H)/3
The volume of a remaining after drilling part of a sphere is
V = Vsphere − Vcyl − 2Vcap =
=4πR³/3−πr²h−2πH²(3R−H)/3
Let's substitute in this expression
R = H+h/2 and
r² = H²+Hh
The result will be:
V = 4π(H+h/2)³/3 −
− π(H²+Hh)h −
− 2πH²[3(H+h/2)−H]/3 =
= π{4H³+6H²h+3Hh²+h³/2 −
− 3H²h−3Hh² −
− 4H³−3H²h}/3 =
= πh³/6

Unizor - Geometry3D - Spherical Sectors

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spherical Sector

Consider a spherical cap that is a part of a sphere of radius R with a center O.
Connect each point of a base circle of a cap with center O by a radius of a sphere. All these radiuses form a conical surface with a base circle as a directrix.
An original spherical cap and a cone formed as described above constitute a spherical sector.


A spherical sector is defined by two parameters - radius of a main sphere R, from which it is a part, and the height of a cap on its top H. For "true" spherical sectors, the ones we will be considering here, height H is smaller or equal to radius R of a main sphere.

Since perpendicular from a center of a sphere onto a base circle of a cap falls into a center of this circle, our cone is a right circular one - the only type we considered in this course.

Volume of a Sector

The radius of the base circle (in terms of the radius of a main sphere R and the height of a spherical cap H) equals, by Pythagorean Theorem, to
L = √R²−(R−H)² = √2RH−H²
This is a cone's radius.
The cone's height is, obviously, R−H.
So, the cone's volume is
Vcone = πL²(R−H)/3 =
= π(2RH−H²)(R−H)/3 =
= π(2R²H−3RH²+H³)/3

As we know from the previous lecture about spherical caps, the volume of a cap is
Vcap = = πH²(3R−H)/3

From the two formulas for volumes of a cap and a cone we can determine the volume of a spherical sector:
Vsector = = πH²(3R−H)/3 +
+ π(2R²H−3RH²+H³)/3 =
= 2πR²H/3
A very short formula indeed!

Just for checking, if H=R, our spherical sector occupies exactly half a sphere. The volume of a sphere, as we have determined before, is 4πR³/3.
So, half a sphere has a volume 2πR³/3.
This is exactly a formula we get from the volume of a spherical sector if H=R.

Area of a Dome

Recall the technique we used to calculate the area of a sphere Ssphere, knowing its volume Vsphere.
We considered a sphere's surface as approximated by an infinitely large number of inscribed polyhedrons, whose faces are infinitely small. Then the volume of a sphere would be approximated by a sum of volumes of all the pyramids obtained by connecting each vertex of each face of each polyhedron with a center of a sphere.
Then the volume of a sphere Vsphere would be equal to an area of its surface Ssphere multiplied by one third of its radius R:
Vsphere = Ssphere·R/3.
This consideration allowed us to derive from a formula for the volume of a sphere -
Vsphere=4πR³/3
- its surface area:
Ssphere=4πR².

Exactly the same considerations connect the area of a dome with a volume of a spherical sector: Vsector = Sdome·R/3.
Therefore,
2πR²H/3 = Sdome·R/3
Now we can calculate the area of a dome:
Sdome = 2πRH

Unizor - Geometry3D - Spherical Caps

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spherical Cap

A spherical cap is formed when a plane cuts through a sphere. This cut divides a sphere into two parts, and each can be called, technically, a spherical cap. Usually, however, it's the smaller one that is considered a "true" cap. The part of a cutting plane that is inside a sphere is a base of a spherical cap, and we know from the previous problems that it is a circle. The part of a sphere that belongs to a cap forms its dome.

A spherical cap is defined by two parameters - radius of a main sphere R, from which it is cut, and the height H defined as the length of the longest segment from a dome onto a base along a perpendicular to it. For "true" spherical caps, the ones we will be considering here, height H is smaller or equal to radius R of a main sphere.

Volume of a Cap
Let's start with a volume of a spherical cap of radius R and height H using already familiar process of approximation of this volume as a sum of volumes of cylinders of the same but small altitude and different radiuses stacked on each other.

The radius of the base circle (in terms of the radius of a main sphere R and the height of a spherical cap H) equals, by Pythagorean Theorem, to
L = √R²−(R−H)² = √2RH−H²

Let AB be a segment that lies along the height of our cap. Divide segment AB into N equal parts by points A1, A2,,, AN-1 from the "top" of the dome down (for uniformity, we can designate point A, the center of a base circle, as AN and point B at the "top" of the dome as A0) and draw planes parallel to a base through each such division point.
The intersection of each plane and a sphere is a circle with a center at a corresponding division point Ak because, if we take any two points X and Y on this intersection, lengths of AkX and AkY are equal since right triangles ΔOAkX and ΔOAkY are congruent by a common cathetus OAk and congruent hypotenuses OX=OY=R.

These planes slice our spherical cap into layers. Each layer resembles a cylinder in a way that it is bounded from top and bottom by two parallel planes and is somewhat rounded in shape, but it's not a true cylinder because its side surface is not formed by straight lines parallel to the same generatrix.

The next step is to make a cylinder within each layer preserving it's circular top base and replacing the side surface with a cylindrical surface by dropping perpendiculars from each point on the upper base towards its bottom base.

Now it's time to make a leap of faith and consider a reasonable, intuitively obvious statement that, as N→∞, the total volume of cylinders tends to some limit that we can call the volume of our spherical cap.

Our task is to calculate a sum of volumes of the N cylinders as a formula, that depends on radius of a sphere R, height of the cap H and the number of division points N, and to find its limit as N→∞, which will depend only on radius R and height H.
Let's calculate a volume of a cylinder #k and then summarize it by k∈[1,N].

By Pythagorean theorem, for any point X on a circle with center Ak
AkX² = OX² − OAk²
Point X is on a sphere, therefore OX=R
Point Ak is kth point of division of segment AB into N equal parts, therefore
OAk=R−H·k/N.

Hence, the square of a radius of a base of the kth cylinder equals to AkX² = R²−(R−H·k/N)²
Its altitude is, obviously, H/N.
Therefore, the volume Vk of the kth cylinder is
π[2RHk/N−(Hk/N)²]·H/N =2πRH²k/N² − πH³k²/N³

Now we have to summarize the volumes of all N cylinders and find the limit of this sum as N→∞.
We will use a symbol Σ for summation by k from 1 to N

V = Σ(Vk) = πRH²(N+1)/N − πH³(N+1)(2N+1)/6N²

Note that we have used a known and previously derived expressions for a sum of numbers from 1 to N:
Σk = N(N+1)/2
and for a sum of squares from 1 to N:
Σ(k²)=N(N+1)(2N+1)/6

The limit of (N+1)/N is 1. The limit of (N+1)(2N+1)/6N² is 2/6=1/3.

Therefore, the volume of the spherical cap equals to
Vcap = πRH² − πH³/3 = πH²(3R−H)/3

Unizor - Geometry3D - Spheres - Problems 1

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spheres - Problems 1

Problem A
Explain that a sphere of radius R divides an entire three-dimensional space into two parts - points outside of a sphere that are at a distance greater than R from its center and points inside a sphere that are at a distance smaller than R from its center.

Explanation
(not really a rigorous proof)
Choose any point M outside a sphere and connect it with its center O. The fact that point M is outside a sphere means that segment OM intersects a sphere at some point P lying in between points O and M.
That is, OP⊂OM.
Since the length of segment OP equals exactly to R, the length of segment OM must be greater than R.
Choose any point M inside a sphere and connect it with center O. The fact that point M is inside a sphere means that a continuation of segment OM beyond point M intersects a sphere at some point P, so point M is lying in between points O and P.
That is, OM⊂OP.
Since the length of segment OP equals exactly to R, the length of segment OM must be smaller than R.

Problem B
A plane intersects a sphere, but does not pass through its center.
Prove that a base of a perpendicular from a center of this sphere to this plane is inside the sphere.

Proof
Let the base of a perpendicular from the center of our sphere O to a plane be point M.
Choose any point A on an intersection of a sphere and a plane.
Since OM is perpendicular to an entire plane, it's perpendicular to line MA on it.
Therefore, triangle ΔOMA is a right triangle with catheti OM and MA and hypotenuse OA.
Since hypotenuse is longer than cathetus, the length of OA, equaled to R, is greater than the length of OM.
According to a previous problem, point M must lie inside the sphere.

Problem C
Given a sphere of radius R with center O.
Plane δ intersects this sphere.
Prove that the intersection of this plane with a sphere is a circle of a radius not greater than R and a center being a base of a perpendicular from a sphere's center O onto plane δ.
Determine the radius of this circle, if plane δ is positioned at a distance d (smaller than R) from a center of a sphere O.

Answer:
Drop a perpendicular from center O onto a plane δ. Its base - point M - is inside a sphere, according to a previous problem. The length of segment OM is d.
Choose any two points A and B on intersection of our sphere with plane δ.
We know that OM⊥δ
⇒ OM⊥AM; OM⊥BM
Consider two right triangles ΔAOM and ΔBOM.
They are congruent by common cathetus OM and equal hypotenuses OA and OB, each being a radius of a sphere. Therefore, two catheti AM and BM are equal.
Since A and B are any two points on the intersection of a sphere and plane δ, all points on this intersection are equidistant from point M - a center of a circle and a base of a perpendicular from a sphere's center O onto plane δ.
The radius AM=r can be calculated by Pythagorean theorem as
r² = R² − d²
r = √R² − d²
Technically, this proof is not applicable, if the plane δ passes through a center of a sphere O because in this case we cannot form triangles used above in the proof. However, this situation does not need any special proof since what we have to prove is contained in the definitions of a sphere and a circle. All points lying on the intersection of our sphere and plane δ are lying in one plane and equidistant from one point on this plane - point O - since they belong to a sphere. Therefore, these points form a circle.
The radius of this centrally located circle is exactly R, which is greater than the radius of any other circle formed by intersection of a plane and a sphere represented by a formula above.
By the way, that formula for d=0 (which technically is not applicable) gives the radius of a circle R, which corresponds our intuition that, as a plane moves closer and closer to a center, the radius of a circle it cuts from a sphere becomes greater and greater with a limit R.

Problem D
Consider a sphere of radius R and center O and a plane δ tangential to this sphere, that is the one that has only one point of intersection with this sphere - point P.
Prove that a radius OP connecting a center of a sphere O with a point of tangency of a plane with this sphere P is perpendicular to a plane δ.

Proof
All points of plane δ, except point of tangency to a sphere P, are located outside a sphere. Therefore, their distance from a center of a sphere O is greater than the radius of a sphere R (see Problem A above). Only a point of tangency P is at a distance equal to a radius of a sphere R.
We see now that segment OP is the shortest among all segments connecting center O to points on plane δ. Since we know that a perpendicular to a plane from a point is the shortest connection between them, and any other connection between center O and points on plane δ is greater than R, OP⊥δ.

Unizor - Geometry3D - Cones - Problems 2

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Right Circular Cones - Problems 2

In this course we will be dealing only with right circular cones and will call them simply cones.

Problem A
A doctor has recommended for a sick girl to take some liquid medicine a couple of times a day by 15-20 grams each time.
Mother was going to use a small conical glass that could hold exactly 20 gram of liquid for this. She filled it up, but a girl said that she remembered the taste of this medicine and did not like it.
Then the mother suggested a compromise: "How about you drink only half a glass?"
The girl agreed, drank some medicine and left some in a glass. What she left took exactly half the height of a glass.
How much medicine did a girl drink?

Answer:
17.5 grams - as doctor ordered.

Problem B
A semicircle of a radius R is rolled into a cone.
What is a volume of this cone?

Answer:
V = πR³√3/24

Problem C
A right circular cylinder of a radius r is inscribed into a cone of a radius R (greater than r) and an altitude H such that the lower base of a cylinder lies in the same plane as the base of a cone with the centers of these bases coinciding, the cylinder height ends at an intersection of a side of a cylinder with a side of a cone.
What is a volume of the cylinder?

Answer:
V = πr²(R−r)H/R

Problem D
Given a cone with a radius of a base R and height H.
Determine a central angle of a circular sector obtained if we cut a side of a cone along a generatrix and roll it out on a plane (in radians).

Answer:
φ = 2πR/√R²+H²

Problem E
Consider a right circular cylinder. Two congruent cones are inscribed into it. One cone has its base coinciding with the bottom base of a cylinder and has its apex at the center of an upper base of a cylinder. Another cone has its base coinciding with the upper base of a cylinder and has its apex at the center of a bottom base of a cylinder.
Find the ratio of a volume of the common part of these two cones (part of their intersection) to a volume of one of them.

Answer:
1/4

Unizor - Geometry3D - Cones - Problems 1

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Right Circular Cones - Problems 1

In this course we will be dealing only with right circular cones and will call them simply cones.

Problem A
Find a volume of a truncated cone obtained by cutting-off its top by a plane parallel to its base in terms of two radiuses of two bases R1 and R2 and its height H (the distance between bases).

Answer:
πH(R1²+R1R2+R2²)/3

Problem B
The side surface of a cone, rolled out on a plane, is a circular sector with a central angle 120o and area S.
Determine the volume of this cone.

Answer:
2S√6πS /27π

Problem C
Given an area S of a side surface of a cone and a distance d from a center of its base to any generatrix, find a volume of a cone.

Answer:
S·d /3

Problem D
Side surface area of a cone is twice as big as an area of its base.
A section of a cone obtained by cutting it by a plane passing through its main axis (that is, a plane that goes through an apex and a center of a base) has an area S.
Find a volume of a cone.

Answer:
(27^1/4)πS√S/9

Problem E
An altitude of a cone of a volume V is divided into N equal parts by points Ai, where i∈[1,N−1]. For convenience, let A0 be an apex of a cone and AN - a center of its base.
Then we draw a plane through each division point parallel to a base of a cone. These planes divide the cone into N parts, all of them except the top are truncated cones.
Find the volume of a truncated cone #k between planes going through points Ak−1 and Ak.

Answer
V[k³−(k−1)³]/N³

Unizor - Geometry3D - Sphere - Volume and Surface Area

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Volume and Surface Area of a Sphere

Volume

Let's start with a volume of a sphere of a radius R and a center at point O using already familiar process of approximation of this volume as a sum of volumes of other geometric objects with known formulas for volume. We will use cylinders of the same but small altitude and different radiuses stacked on each other as an approximation of a sphere.

Imagine a plane that cuts our sphere of a radius R in two equal halves by going through its center. We will calculate the volume of a half a sphere and then double it.

From a center of a base O draw a perpendicular to this plane "upward" to intersection with our half a sphere at point P on its "top". The length of this perpendicular OP is, of course, R, since this is a radius of an original sphere.

Divide this perpendicular into N equal parts by points A1, A2,,, AN-1 (for uniformity, we can designate point P at the end of this perpendicular as AN) and draw planes parallel to a base through each such division point. The intersection of each plane and a sphere is a circle with a center at a corresponding division point Ak.

The next step is to make a cylinder within each layer preserving it's circular top base and replacing the side surface with a cylindrical surface by dropping perpendiculars from each point on the upper base towards its bottom base.

By Pythagorean theorem, for any point X on a circle with center Ak
AkX² = OX² − OAk²
Point X is on a sphere, therefore OX=R
Point Ak is kth point of division of a radius OP into N equal parts, therefore OAk=R·k/N.
Hence, the square of a radius of a base of the kth cylinder equals to
AkX² = R²−(R·k/N)²
The above is a radius of the kth cylinder.
Its altitude is, obviously, R/N.
Therefore, the volume of the kth cylinder is
Vk = π[R²−(R·k/N)²]·R/N =
= πR³/N − πR³·k²/N³
Now we have to summarize the volumes of all N cylinders and find the limit of this sum as N→∞. Doubling this (since we were dealing with half a sphere) will give us the volume of a sphere.
The result, the volume of the whole sphere of radius R will then be
Vsphere = 4πR³/3

Surface Area

Let's inscribe a convex polyhedron into a sphere by choosing a sufficiently large number of points on its surface (they will be vertices of a polyhedron) and connecting each point with its closest neighbors (these connections will be edges of the polyhedron and each triangle formed by these edges will be its face).
Connecting all the vertices with a center of a sphere, we divide a polyhedron into pyramids with different faces and, generally speaking, different altitudes.

As we add new points on a sphere, evenly distributing them in the empty spots on a sphere's surface, we will increase the number of faces of these polyhedrons, increasing the number of pyramids these polyhedrons are divided. What's interesting is that altitudes of all the pyramids will be closer and closer to a radius of a sphere and the surface area of a polyhedron becomes closer and closer to a surface area of sphere.

In the limit, as all faces are getting smaller and smaller, and the distance between a center of a sphere and all these faces becomes closer and closer to a radius of a sphere, we can say that the volume of a polyhedron (that approximates the volume of a sphere) approximately equals to one third of its surface area (that approximates the surface area of a sphere) multiplied by the radius of a sphere (that approximates the altitudes of all pyramids). The approximation is better as the largest face area of a polyhedron tends to zero.

The above logic is an intuitive base for the following statement about the relationship between the volume of a sphere Vsphere of a radius R and its surface area Ssphere:
Vsphere = Ssphere·R/3

Knowing the formula for the volume of a sphere, we can derive from this the formula for the surface area:
4πR³/3 = Ssphere·R/3
form which immediately follows:
Ssphere = 4πR²

Unizor - Geometry3D - Cones - Area and Volume

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Area and Volume of a Cone

Let's define two important parameters that fully characterize a cone.
They are:
(a) radius of a base circle, which we will refer to as radius of a cone and denote as R,
(b) altitude or height of a cone (the distance from an apex to a bottom circular base), which we denote as H.
Note that the distance from an apex to any point on a circular base it is connected with by a straight line on a side conical surface is constant and, according to Pythagorean Theorem, is equal to
L = √R²+H²

Surface Area

There are different approaches to defining an area of a cone. More rigorous approach involves full force of the theory of limits, but we would suggest here a different approach.

First of all, consider the side surface of a cone.
Since this side surface is formed by a straight lines connecting points of a circular directrix with an apex, it is intuitively obvious that, if we cut the side surface of a cone along one of these straight lines, we will be able to "flatten" it on a plane without stretching or squeezing, that is without any change to its area.

As a result of this transformation, we will obtain a circular sector with radius equal to a distance between an apex and each point on a circular base of a cone that we have calculated above as
L = √R²+H².

Another characteristic of this sector, that we will use to determine its area, is the length of an arc of this sector. Obviously, it is equal to a circumference of a base of a cone, that is
C = 2πR.

Therefore, the area of a side surface of a cone is equal to the area of a circular sector with a radius
L = √R²+H²
and an arc length
C = 2πR.

The sector's area is a part of an area of a circle of the same radius L. The ratio between the sector's area and the area of a circle, which this sector is a part of, equals to the ratio between their arcs.

We know the arc of a sector, it is equal to 2πR.
We also know the arc (that is, circumference) of a circle, which our sector is a part of, it is equal to 2πL.
And we know this circle's area, it is equal to πL².
We can determine the side area Sside of a cone from the following ratio:
Sside /πL² = 2πR /2πL

The solution to this is
Sside = πR·L = πR√R²+H²

To determine a full area of a cone, we have to add an area of its base Sbase=πR².
The final result for a full area of a cone Sfull:
Sfull = Sside + Sbase =
= πR√R²+H² + πR² =
= πR(R+√R²+H²)

Volume

The situation with volume of a cone is similar to that of a volume of a cylinder, and we will not be able to escape considerations based on the limit theory.

Let's inscribe into a circular base of a cone a regular N-sided polygon. Then construct a pyramid with this polygon being a base and the same apex as that of a cone. We obtain a pyramid inscribed into a cone.

Without rigorous proof, it is intuitively obvious that, as we increase the number of vertices N, the regular polygon inscribed into a circular base of a cone becomes closer and closer to a circle itself, and the pyramid, based on this polygon inscribed into a circular base of a cone, becomes closer and closer to a cone. So, the volume of a cone is a limit of the volumes of inscribed in this manner pyramids as N→∞.

Since a volume of a pyramid is one third of a product of an area of its base by height and, as N→∞, the area of the N-sided polygon inscribed into a circle of a radius R tends to the area of a circle itself, that is πR², while the height H remains constant, we conclude that the volume of a pyramid tends to
V = πR²·H /3

Unizor - Geometry2D - Area of a Circle

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Area of a Circle

1. Definition

No matter how small our area measurement unit (a square with a side of 1) is, we will always have problems to measure exactly the area of a circle of a radius 1.
So, we have no choice, but to resort to a process that leads to an area of a circle as its limit.
This process might involve filling a circle with as many as possible squares of a side 1, then choose a smaller unit square of a side size, say, 1/10 and fill the empty area and continue this process getting more and more precise value with smaller and smaller measurement units.

In this lecture I'd like to introduce a different method that leads to a formula for the area of a circle more directly through another process.

Let's inscribe a regular polygon into a circle. We can start with a hexagon, similarly to a process we used to evaluate a circumference of a circle, or any other regular polygon. Next, we will double the number of edges of this polygon on every step by constructing perpendiculars to all edges and taking new vertices at points of intersection of these perpendiculars with a circle.

Now we state without rigorous proof that inasmuch as the perimeter of these polygons tends to a circumference of a circle, their area tends to some limit that can be used as a definition of the area of a circle.

Moreover, it can be proven that the limit of the area of polygons exists and is always the same, regardless of which polygon we start with and how we increase the number of its vertices, as long as the longest edge of polygons decreases to zero.

With this theorem we can state that the definition of an area of a circle is correct. It exists as a limit of areas of polygons transforming during a process described above and is unique.

2. Formula for Area of a Circle

Our next task is to evaluate the limit of the area of these polygons. That limit will be the area of a circle.

Let's examine a polygon with N sides obtained at some point during our process of increasing the number of vertices. Connect all vertices of our N-sided polygon with a center of a circle and consider the area of a polygon as a sum of areas of all N triangles with one vertex at the center of a circle and two other vertices being adjacent vertices of a polygon.
All these triangles are congruent and the area of each of them equals to
a·h/2
where a is the length of the edge of a polygon serving as an edge of a triangle and h - the altitude of each triangle dropped from the center of a circle onto an opposite edge.

Since we have N such triangles in a polygon, the total area of a polygon equals to
N·a·h/2
Now notice that N·a is a perimeter of our polygon p. So, we can say that the area of a polygon equals to
p·h/2

Now it's time to observe the trend of this area. As the number of vertices increases to infinity, the perimeter of inscribed polygons tends to a circumference of a circle that is equal to 2πR, where R is the radius of a circle.
At the same time the altitude h tends to be closer and closer to a radius of a circle.

So, as we approach the limit, the area of a circle will be equal to
2πR·R/2 = πR²

Unizor - Geometry2D - Circumference of a Circle

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Circumference of a Circle

1. Definition

The first question to discuss related to a circumference of a circle is "What is a circumference of a circle?"
Unit of measurement used for segments is not good for measuring length of curves. So, we have to come up with a different definition.

Typical approach to this lies in approximation. We will rely a lot on intuitively obvious statements with full understanding that proving them rigorously might present some problem.
It's tempting to define the circumference of a circle as follows:
a) start a process by inscribing a regular hexagon into a circle and setting the first approximation to a circumference of a circle as a perimeter of this hexagon;
b) on each step we draw perpendiculars to all sides of a polygon obtained on a previous step and take intersections of these perpendiculars with a circle as new vertices of a polygon with twice as many edges as on a previous step;
c) continue this process of doubling the number of edges to infinity, and the limit of the perimeters of our polygons is, by definition, a circumference of a circle.

There are a few problems with this definition. For example, it is intuitively obvious that, if we start with an inscribed square instead of a hexagon, we should also gradually approach some limit by doubling the number of edges of a polygon. But is this limit the same as if we start with a hexagon?
Issues of existence of a limit and its uniqueness can be rigorously addressed, but are rather complex, and we just point the result that, no matter how our process is arranged, as long as the maximum distance between neighboring points goes to zero, the limit of the perimeters of polygons exists and is the same. That justifies the definition of a circumference as such a limit.

2. Similarity of All Circles

Therefore, if a circumference of one circle is greater than its radius by some factor, the same factor is applied to any other circle.

3. About π

Traditionally, the factor between a circumference and a diameter of a circle is designated a Greek letter π. It is a constant for all circles. It's an irrational real number and can be approximated to any precision. To four decimal places it's equal to 3.1416.

4. Iterations

If d[N] is the length of a polygon, inscribed into a circle, on Nth iteration of replacing each of its sides with a pair of equal but smaller ones, this recursive equality is held:
d[N+1] = √{2 − √[4 − (d[N])²]}

5. Conclusion

As you see, our first process started with hexagon and we got one set of formulas for a perimeter of polygons obtained by doubling the number of edges.
The second process started with a square and formulas were different.
But in both cases a few first steps of iterative process led to very close results. This is exactly how it should be, because, no matter how we organize the process of approximation of the circumference of a circle with perimeters of polygons, as long as the maximum edge length tends to zero, perimeter tends to the circumference of a circle.

Unizor - Geometry3D - Similarity of Cylinders, Cones, Spheres

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Similarity of Cylinders, Cones and Spheres

We have discussed the fact that scaling of a straight line transforms it into a straight line and scaling of a plane transforms it into a plane.
We have also discussed that angles between lines and dihedral angles between planes are preserved by scaling.

Based on this, we can say that any polyhedron is transformed by scaling into a polyhedron with the same number of vertices, edges and faces, with the same angles between edges and with the same dihedral angles between faces. So, the general shape of a polyhedron is preserved.

In this topic we will attempt to prove that the scaling of other geometrical objects - cylinders, cones and spheres - preserves their type.

Cylindrical Surface

Cylindrical surface is characterized by its directrix and generatrix. In particular, we draw a line parallel to generatrix through each point on a directrix.
Now let's scale each such line. A transformed image of this line by a scaling will be, as we know, a straight line parallel to an original. Therefore, no matter how a directrix is transformed, the new object will still be a cylindrical surface because it will consists of straight lines parallel to the same generatrix.

Right Circular Cylinder

Recall that all angles are preserved by scaling. In a right circular cylinder any line forming its side surface is parallel to a generatrix, which, in turn, is perpendicular to a plane that contains a circular directrix and both base planes. In the image of this right circular cylinder all such side lines will be still parallel to an original generatrix and an image of a plane that contains a circular directrix and images of both bases would be corresponding parallel planes. Therefore, perpendicularity between a generatrix of a cylinder and a plane where its directrix lies is preserved. Therefore, scaling transforms a right circular cylinder into a right circular cylinder.

Right Circular Cone
A right circular cone is the one whose altitude (a perpendicular form an apex onto a circular base) is falling into a base's center.
Let point S be an apex of an source right circular cone, plane β - its base plane, point O - a center of its circular base and points A and B - two points on a base circle.
Since this is a right circular cone, SO⊥β, SO⊥OA and SO⊥OB.
Let S', β', O', A' and B' be images of corresponding points after scaling.
Since all angles are preserved, S'O'⊥O'A' and S'O'⊥O'B'. Therefore, S'O'⊥β', that is, S'O' is an altitude of a cone's image. Hence, an image of a right circular cone is a right circular cone.

Sphere

Since equality of the lengths of two segments is preserved by scaling, images of two points on a sphere, equidistant from its center by definition of a sphere, will be equidistant from an image of a center. Therefore, an image of a sphere after scaling is a sphere.

Unizor - Geometry3D - Cylinders - Not So Easy Problems

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Cylinders - Not So Easy Problems

The problems presented here are steps towards the last problem and their combined solutions are supposed to constitute a complete solution to that last problem.

The Last Problem

There is a cylinder inscribed into a cube ABCDA'B'C'D' as follows.
The main axis of a cylinder, connecting centers of its circular bases, lies along a main diagonal of the cube AC'.
One base of a cylinder, a circle, is tangential to three faces of a cube, ABCD, AA'B'B and AA'D'D, that share vertex A, one end of a cube's main diagonal.
An opposite base of a cylinder, a circle, is tangential to three other faces of a cube, C'D'A'B', C'CBB' and C'CDD', that share the opposite end of the main diagonal, vertex C'.
A section of a cylinder along its main axis is a square (that is, a diameter of its base equals to its height).
Find the ratio of a volume of a cylinder to a volume of a cube.

Analysis:

First of all, let's think about the problem in general. There is no information about the size of a cube, which implies that the ratio of a volume of a cylinder to a volume of a cube is independent of a size. Is it true?
All the cubes are similar to each other. All the cylinders with a diameter of a base equal to a height also are similar to each other. So, it looks like changing the size of a cube might proportionally change the size of an inscribed into it cylinder, so the ratio of volumes is constant and the problem does make sense. These are logical considerations rather than proof.
The real solution would be to assign to a cube some dimensions and calculate the dimensions of an inscribed into it cylinder. Then we can calculate the ratio of the volumes and the result should be independent of the initial size of a cube.

Let's assume that the cube has all edges of a length d. Hence, its volume is d³. This is the simple part of the calculations.

With a cylinder it's not as simple. The bases of a cylinder are perpendicular to the main diagonal of a cube. From considerations of symmetry, the plane perpendicular to the main diagonal of a cube should intersect its edges on equal distance from the vertex that is the end of a diagonal. So, if the plane of the base of a cylinder that is closer to vertex A, which we will call δ, intersects edge AB at point P, edge AD at point Q and edge AA' at point R, segments AP, AQ and AR should be of the same length. To prove it is the subject of Problem 1 below.

Assume, this is proven, and continue our analysis.

Next step is to understand that a circular base tangential to three faces of a cube that share vertex A (and lying within a plane δ defined by triangle ΔPQR) is a circle inscribed into this triangle PQR. This follows from the fact that our circular base completely belongs to plane δ, so its common points with any one of three faces of a cube should belong to intersection of that face and plane δ. But this intersection is a line - a side of a triangle ΔPQR, so a circle has only one common point with a side of a triangle, that is tangential to it.

Let's assign a size x to three segments AP, AQ and AR. They completely define plane δ where a base of a cylinder lies. Using this length, we can calculate the radius of a base circle tangential to three faces that share vertex A as a radius of a circle inscribed into triangle ΔPQR - see Problem 2 below.

Another plane, the one where the other base of this cylinder (closer to vertex C') is located, intersects three other edges of a cube, C'B', C'D' and C'C at, correspondingly, points P', Q' and R'.
It should be obvious from the formula connecting the radius of an inscribed into triangle ΔPQR circle to x that C'P'=C'Q'=C'R'=x. This simple conclusion is the consequence of the fact that both circles at the bases of a cylinder are congruent.

The final component is the height of a cylinder. This might be calculated based on the length of a main diagonal of a cube and the height of a pyramid with vertex A and base triangle ΔPQR - see Problem 3 below. This height should be subtracted twice from the length of a diagonal (once for a pyramid with apex A, another - for a pyramid with apex C') to get the height of a cylinder.
Equating this height of a cylinder to a diameter of its base should give us the value of x in terms of the length of a cube's side d (see Problem 4).

With all this calculated, we can determine the volume of a cylinder in terms of d and compare it with a volume of a cube to get the ratio of a volume of a cylinder to a volume of a cube.
The variable d should be reducible from a formula.

Unizor - Geometry3D - Cylinders - Easy Problems

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Cylinders - Easy Problems

Problem 1

Prove that an intersection of a cylinder with any plane, that is parallel to its bases and is located in between them, is a circle.

Problem 2

Prove that an intersection of a cylinder with any plane that contains a generatrix at its two positions around the base is a rectangle.

Problem 3

A cube with an edge length of d is inscribed into a cylinder such that one face of a cube (a square) is inscribed into one base of a cylinder (a circle) and an opposite face of a cube is inscribed into an opposite base of a cylinder.
What is the total area and a volume of a cylinder?

Answer:
Area = 2πd²(1+√2)/2
Volume = πd³/2

Problem 4

Prove that the line connecting two centers of bases of a cylinder (center line) is its axis of symmetry.

Problem 5
Given a cylinder of a radius R and height H. A plane is drawn parallel to its center line on a distance d from it (less than a radius).
What is the area of a section of a cylinder cut by this plane?

Answer:
2H√R²−d²

Problem 6

A regular tetrahedron with all edges equal to d is inscribed into a cylinder such that one of its faces (an equilateral triangle) is inscribed into one base of a cylinder (a circle) and an opposite vertex of a tetrahedron coincides with a center of an opposite base of a cylinder.
What is the total area and a volume of a cylinder?

Answer:
Area = 2πd²(1+√2)/3
Volume = πd³√6/9

Unizor - Geometry3D - Cylinders - Area and Volume

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Area and Volume of a Cylinder

For the definition of a cylinder and corresponding terminology please refer to topic "Elements of Solid Geometry" in this part of a course.
In short, a cylinder is formed by
(1) a generatrix (straight line) moving along a circular directrix perpendicularly to a plane where this circular directrix is located, thus making a side surface and
(2) two planes parallel to a plane that holds a directrix - top and bottom bases of a cylinder, they bound the cylinder from both ends..

Let's define two important parameters that fully characterize a cylinder.
They are:
(a) radius of a base circle, which we will refer to as radius of a cylinder,
(b) altitude or height of a cylinder (the distance between the top and bottom circular bases).

There are different approaches to defining an area of a cylinder. More rigorous approach involves full force of the theory of limits, but we would suggest here a different approach.

First of all, consider the side surface of a cylinder.
Since this side surface is formed by a straight line (generatrix) moving along a circular directrix always perpendicularly to the same plane where this directrix is located and, therefore, parallel to itself at different positions, it is intuitively obvious that, if we cut the side surface of a cylinder along one of these straight lines, we will be able to "flatten" it on a plane without stretching or squeezing, that is without any change to its area.

As a result of this transformation, we will obtain a rectangle with the width equal in length to a circumference of a circular base of a cylinder and the height equal to a height of a cylinder - both are known variables for any given cylinder.
Therefore, the area of a side surface of a cylinder is equal to the area of a rectangle and can be easily calculated.

The circumference of a circular base of a cylinder of a radius R equals to 2πR. If the height of a cylinder equals to H, the area of the side surface would be equal to 2πR·H.
The full area of a cylinder should include two areas of circular bases of radius R, each equal to πR².
That makes the full area of a cylinder of a radius R and height H equal to
2πR·H+2πR² = 2πR(R+H)

The situation with volume of a cylinder is less obvious and we will not be able to escape considerations based on the limit theory.

Let's inscribe into a circular base of a cylinder a regular N-sided polygon. Then construct a right prism with this polygon being a base and the height equal to a height of a cylinder. We obtain a prism inscribed into a cylinder.

Without rigorous proof, it is intuitively obvious that, as we increase the number of vertices N, the regular polygon inscribed into a circular base of a cylinder becomes closer and closer to a circle itself, and the prism, based on this polygon inscribed into a circular base of a cylinder, becomes closer and closer to a cylinder. So, the volume of a cylinder is a limit of the volumes of inscribed in this manner prisms as N→∞.

Since a volume of a prism is a product of an area of its base by height and, as N→∞, the area of the N-sided polygon inscribed into a circle of a radius R tends to the area of a circle itself, that is πR², while the height H remains constant, we conclude that the volume of a prism tends to πR²·H

Unizor - Geometry3D - Pyramids - Problem 6

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Pyramids - Problems 6

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.

Problem A

Given a triangular pyramid SABC with apex S and side edges having sizes:
SA = a,
SB = b,
SC = c.
Inside its base ΔABC we choose any point P.
From this point P we draw three lines parallel to three side edges of a pyramid until lines intersect the opposite faces:
PA' ∥ SA where A'∈ΔSBC,
PB' ∥ SB where B'∈ΔSAC,
PC' ∥ SC where C'∈ΔSAB.
Let the length of these new segments are
PA' = a',
PB' = b',
PC' = c'.
Prove that
a'/a+b'/b+c'/c = 1

Unizor - Geometry3D - Pyramids - Problem 5

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Pyramids - Problems 5

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.

Problem A

Given a regular tetrahedron SABC.
Draw a median AM within its triangular base ABC from vertex A to opposite side BC (M∈BC).
From its apex S draw a plane γ perpendicular to median of the base AM.
Assume that the area of a section of a pyramid formed by this plane equals to s.
Pick a midpoint P of median AM and construct another plane δ that is parallel to plane γ. It also cuts some section inside a pyramid.
What is the area of this section?

Answer:
9·s /16

Unizor - Geometry3D - Pyramids - Problems 4

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Pyramids - Problems 4

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.

Problem A

Given a right rectangular prism ABCDA'B'C'D' with three edges with common vertex A having length x, y and z.
Let's connect with straight lines vertices
A and C,
A and B',
A and D',
C and B',
C and D',
B' and D'.
What is the volume of tetrahedron ACB'D' formed by these vertices?

Answer:
x·y·z /3

Problem B

Given a triangular pyramid SABC with pairs of opposite edges equal to each other as follows:
SA = BC = a,
SB = AC = b,
SC = AB = c.
What is the volume of this pyramid?

Answer:
√[(m−a²)·(m−b²)·(m−c²)] /3
where m = (a²+b²+c²) /2

Hint:
Use the Problem A above.

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?

Answer:

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.

Answer:

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?

Answer:
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.

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.

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.

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?

Answer: 5/6

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',
AD ∥ A'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',
ΔSAD ~ ΔSA'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')

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;
VolumeSADE = AreaΔADE·h/3;
VolumeSAEF = AreaΔAEF·h/3;

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

Answer:
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.

Answer:
[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?

Answer:
√[(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.

Answer:
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.

Answer:
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?


Answer:
One half.

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.

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.

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.

Unizor - Geometry3D - Similarity - Cavalieri - Theorems 1

Notes to a video lecture on http://www.unizor.com



Mini Theorems 1



Let's prove the condition of Cavalieri's principle (see the previous
lecture explaining this principle) in a couple of very simple cases
using the similarity. This would justify our decision to present the
Cavalieri's principle among topics related to similarity.



Mini Theorem A

(two-dimensional case to introduce the approach to a proof)



Given a straight line d on a plane and two segments on it, AB and CD, that have equal lengths.

Let S be a point on this plane outside the line d.

Prove that the condition of Cavalieri's principle is held for triangles ΔSAB and ΔSCD.

If proven, we can state that the areas of these triangles, according to Cavalieri's principle, are the same.



Mini Theorem B

(three-dimensional case that will be used in calculation of the volume of a pyramid)



Given a plane δ and two triangles on it, ΔABC and ΔDEF, that have equal area.

Let S be a point in space outside this plane δ.

Prove that the condition of Cavalieri's principle is held for triangular pyramids SABC and SDEF.

If proven, we can state that the volumes of these pyramids, according to Cavalieri's principle, are the same.





Mini Theorem C

 

Given a plane δ and parallelogram ABCD on it with diagonal AC.

Let S be a point in space outside this plane δ.

Using the Cavalieri's principle, prove that the volumes of two pyramids, SABC and SACD, are equal.

Unizor - Geometry3D - Cavalieri's Principle

Unizor - Creative Minds through Art of Mathematics - Math4Teens

The
purpose of this lecture is to introduce the Cavalieri's principle that
will be used to derive the formula for a volume of a pyramid, the next
topic of this course.

The problem with a volume of a pyramid, as
with many other measures of geometric objects is that absolutely
rigorous derivation of the formulas for their area and volume can be
obtained only within a framework of higher levels of mathematics,
usually addressed in colleges. However, using the theory of limits on an
intuitive level, as presented in this course in a topic Algebra -
Limits is sufficient to derive these formulas for high school students.

The
Cavalieri's principle, actually, hides the complexity of this issue by
postulating certain property of geometric objects. This principle is not
an axiom mathematicians accept, it is a theorem that can be proven with
above mentioned higher levels of mathematics using integration.
However, accepting this principle as an axiom at this stage allows to
shortcut the derivation of formulas for area and volume of geometric
objects. The Cavalieri's principle sounds very natural and is
intuitively obvious. So, we will use it without any hesitation.

There
are two related to each other parts of the Cavalieri's principle - one
for area of a geometric object on a plane and another for volume of
solid objects in three-dimensional space.

Two-dimensional case

Assume we have two geometric objects X and Y on a plane.
Assume
also that there is a base line d on this plane such that any line h
parallel to this base line has the following property:
its
intersection with object X (a segment or a set of segments for
irregularly shaped object X) has the same linear measure as its
intersection with object Y.
Then the areas of objects X and Y are equal.

This
principle is a sufficient condition for equality of the areas of
objects X and Y, but by no means a necessary one. There are objects of
equal areas for which this principle is not true. For instance, take a
2x2 square and 1x4 rectangle. Their areas are equal, but you cannot find
a base line d such that any line parallel to it produces sections of X
and Y of the same linear measure.

Three-dimensional case

Assume we have two solid geometric objects X and Y in a three-dimensional space.
Assume
also that there is a base plane δ on this plane such that any plane γ
parallel to this base plane has the following property:
its intersection with object X (a flat object) has the same area as its intersection with object Y.
Then the volumes of objects X and Y are equal.

This
principle is a sufficient condition for equality of the volumes of
objects X and Y, but by no means a necessary one. There are objects of
equal volumes for which this principle is not true. For instance, take a
2x2x2 cube and 1x4x2 right rectangular prism. Their volumes are equal,
but you cannot find a base plane δ such that any plane parallel to it
produces sections of X and Y of the same area.