Remember the Number Line?

Absolute Value (or modulus) arises when we are concerned with how far a value is from zero. For instance, 8 is simply a distance of 8 units from zero and -8 is also 8 units from zero. In other words, absolute value is about magnitude of a number regardless of sign.

Absolute value is shown using “bars” e.g. │8│ or │-8│

Both are equal to 8. An absolute value cannot be negative.

The graph of y =│x│from x = -10 to 10 is

y must always be positive.

Equations:

Example A

│x - 3│= 7

x - 3 can be either positive or negative. In either case, the absolute value is the same. There are therefore two solutions.

One is when x - 3 = 7 and one is when x - 3 = -7. In the first case, x = 10 and, in the second case, x = -4

Check those against the given equation. │10 - 3│= 7 is plainly correct and │-4 - 3│= │- 7│= 7 is also correct.

Example B

3│x +8│= 39

Divide both sides by 3 to get the absolute value expression by itself

│x + 8│= 13 and then note the two possible cases

Case (1) is x + 8 = 13 from which x = 5

Case (2) is x + 8 = -13 from which x = -21

Always check the solutions for accuracy. They are correct). Remember that │-13│= 13

Types of absolute value equation:

There are 4 main categories.

1. An absolute value expression is equal to a constant

e.g. │x + 3│= 9 produces x + 3 = 9 and x + 3 = -9 and so the solutions are x = 6 and x = -12

2. One absolute value expression is equal to a variable expression

e.g. │x│= │2x + 5│ produces x = (2x + 5) and x = -(2x + 5) and the solutions are x = -5 and x = -5/3

3. Absolute values on both sides of the equation

e.g. │x - 1│ = │3 - x│ produces x - 1 = (3 - x) and x - 1 = -(3 - x)

x - 1 = 3 - x produces the solution x = 2

x - 1 = -(3 - x) produces -1 = -3 which is contradictory and is therefore rejected

The only solution is x = 2

4. Multiple absolute values and a constant

e.g. │x + 2│ + │x - 3│= 7

This type of equation is covered in Part 2 of this topic.

General Note - There is no solution if an absolute value is stated to be equal to a negative number. Further, if an absolute value is equal to zero (0) there is only one solution because zero has no negative counterpart.

Inequalities:

Under what circumstances is this statement correct?

The left hand side can be either zero (0) or positive - (an absolute value cannot be negative).

If it is zero then the greater than or equal to requirement is met. Also, all positive numbers are greater than zero. Hence, the statement is true for ALL real values of x

Now consider

Again, the left side can either be zero or positive. If it is zero then the statement is false. If is it positive, the statement is again false because no positive number is less than zero. Hence, the entire statement is false. In other words, there is no solution (or the solution SET is empty.

These two examples are taken from the excellent Prime Newtons youtube channel - see ABSOLUTE VALUE INEQUALITIES with ZERO/NEGATIVES on one side

The following Math and Science video also offers an excellent explanation

Solving & Graphing Absolute Value Inequalities in Algebra, Part 1

Other Video:

Professor Dave Explains - Absolute Values: Defining, Calculating, and Graphing

The GoTutormath - Solving Absolute Value Equations: Everything You Need to Know!

Quadratic equation - Sum and Product of Roots:

The Carlyle circle of a quadratic equation is closely related to the sum and product of the roots - see François Vieta and Polynomials - by Under Northern Skies.

Example:

Consider the quadratic equation x² – 6x + 8 = 0

The sum of roots is given by -(-6)/1 = 6 and the product of roots is given by 8/1 = 8

The graph of the function is shown above and the roots are shown as x = 2 and x = 4.

To construct the Carlyle circle, find the diameter endpoints using point A(0,1) as the start and the point B(Sum of roots, product of roots) as the end. o

For this equation a is 0, 1 and B is 6. 8

Find the midpoint of the diameter and then draw the circle. The intersections of this circle with the x-axis provide the geometric roots of the equation, which are (x = 2) and (x = 4).

The name Carlyle Circle is after Thomas Carlyle (1795 - 1881).

Below is the visual representation of the Carlyle circle:

Find the midpoint and radius:

The midpoint of diameter AB is found from the coordinates of A (0, 1) and B (6, 8)

The radius of the circle is half the diameter. The length of AB is found from

from which the diameter (d) is

and the radius is approximately 4.61 units.

Another example:

2x² - 15x + 18 = 0

The sum of roots is -(-15)/2 = 7.5 and the product of roots is 18/2 = 9

Draw the diameter from point 0, 1 to point 7.5, 9

Find the midpoint and radius.

The Caryle circle crosses the x-axis at 3/2 and 6 which are the roots of the equation.

Links:

Carlyle Circle – GeoGebra - explore various Carlyle Circles.

Carlyle circle Facts for Kids

Carlyle circle - Wikipedia

François Viète (1540 - 1603) - Biography - MacTutor History of Mathematics. The surname is, perhaps more usually, written Vieta.

Vieta’s formulas (or Viète’s formulas) provide a beautiful relationship between the roots of a polynomial and its coefficients.

Standard form of polynomial f(x) - is given by

f(x) = a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + ... + a₁x + a₀

where x is the variable and a_n a_n-1 etc. are coefficients.

n is the degree of the polynomial.

Example:

f(x) = x³ – 4x² - 4x² + 16 is a polynomial of degree 3 (or cubic). The equation

x³ – 4x² - 4x² + 16 = 0 has 3 roots. Its graph is

In fact, this particular polynomial factorises to (x+2)(x-2)(x-4) and so the roots of the equation (x+2)(x-2)(x-4) = 0 are x = -2, 2, 4. That is shown by the red dots on the graph (above).

Quadratic:

The general quadratic equation is ax² + bx + c = 0. It has two roots r₁ r₂

Vieta’s formula applied to a quadratic is that

Sum of roots r₁ + r₂ = -b/a and the product of the roots r₁ r₂ is c/a

e.g. for the quadratic x² - 6x + 8 = 0 the sum of roots -b/a is -(-6/1) = 6 and the product of roots c/a is 8/1 = 8

e.g. for the quadratic 2x² + 3x - 14 = 0 the sum of roots -b/a is -3/2 and the product of roots c/a is -14/2 = -7

Cubic:

The general cubic equation is ax³ + bx² +cx + d = 0 has 3 roots x = r₁ r₂ r₃

Vieta’s formula applied to a cubic is

Sum of roots r₁ + r_{2 +} r₃ = -b/a

Sum of products of roots in pairs r₁ r₂ + r₁ r₃ + r₂ r₃ is c/a

Product of all roots r₁ r₂ r₃ is -d/a

e.g. the cubic equation x³ - 10x² + 29x - 20 = 0

Apply Vieta’s formula

Sum of roots = - (-10)/1 = 10

Sum taking products of roots in pairs = 29/1 = 29

Product of all three roots is -(-20)/1 = 20

The three roots are actually x = 1, 4 and 5 and so

Sum of roots = 1 + 4 + 5 = 10

Sum taking products of roots in pairs 29/1 = 29

Product of all roots = 1 x 4 x 5 = 20

Signs are Crucial:

When applying Vieta’s formula be careful with signs. They alternate between - and +

Sum of roots begins with -b/a

The sum of products of roots in pairs is +c/a

The product of all roots is -d/a

Video:

Prime Newtons - Vieta’s Formula

Art of Problem Solving: Vieta for Quadratics Part 1

The next diagram shows a problem from Geometria as shown on Youtube. The radius of the yellow circle is R, the radius of the green circle is r₁, the radius of the red circle is r₂ and r₁ > r₂

The aim is to prove that

The configuration:

Study of the diagram shows that this is a configuration of circles with a particular set of features.

The Green circle (centre A) and the Yellow Circle (centre O) are tangential at point (T). The straight line PN is tangential to the Green circle at (N) and passes through the centre of the yellow circle. The red circle has its centre (B) on the straight line from P to A. The red circle is also tangential to the Yellow circle at point (K) and OBK is a straight line. There is also a straight line tangent (PZX) to the red and green circles.

Theory:

Similar Triangles

The right triangle, the Pythagorean Theorem, and other results

Circle Theorems - Tangents which meet at the same point are equal in length

Equivalent Fractions - i.e. either multiplying or dividing BOTH the numerator and denominator of a fraction by the same amount.

Proof of the formula:

A) Similar triangles - △ PAN is similar to △ PBL - (three equal corresponding angles). Hence the ratio AN/BL = r₁/ r₂ = PN/PL

B) Find length ON - apply Pythagorean theorem to △ OAN so

OA² = AN² + ON²

OA = R + r₁and AN = r₁

(R + r₁ )² = r₁² + ON²

From which ON = √(R² +2Rr₁) NB: Principal square root

C) Find length OL - note that OB = (R - r₂) and BL = r₂

Then apply Pythagoras to △ OBL to get OL = √(R² - 2Rr₂)

D) Lengths PN and PL

PN = R + ON = R + √(R² +2Rr₁) and PL = R + OL = R + √(R² - 2Rr₂)

E) Return to the ratio PN/PL

r₁/ r₂ = PN/PL

Now substitute the expressions shown in (D) above for PN and PL and then use equivalent fractions to change the right hand side

(The equivalent fraction was was achieved by dividing BOTH numerator and denominator by √R)

F) Rearrange the formula to obtain an expression for R

This is actually a considerable algebraic task. I took the “shortcut” of using online formula rearrangement.

The result was

This proves the required formula.

(The key was to note the very particular configuration of the circles which made it possible to apply Pythagoras).

Example:

If r₁ = 3 and r₂ = 1.5 then R = 3.375 units

The diagram (below) shows a configuration of circles. One circle, centre A, has radius a units. Tangential to that circle is a smaller circle centre B and radius b. Hence, a > b

Tangents are drawn from an external point (P) to both of those circles.

A third circle, centre O and radius R, passes through P and the point (T) of tangency of the first two circles. Point O is on one of the tangents to the first two circles.

The task is to find a formula for R (the radius of the third circle) in terms of a and b.

Theory:

Similar Triangles

The right triangle, the Pythagorean Theorem, and other results

Circle Theorems - Tangents which meet at the same point are equal in length

Transitive property of equality

Construction:

BL and AN are drawn perpendicular to PN

BQ is parallel to line LN

It follows that BQNL is rectangle and △ BAQ is right-angled

TM is tangential (at T) to the first two circles.

Similar Triangles:

△ PAN is similar to △ PBL - (3 angles are the same). It follows that

Tangents to circles:

LM = MT (both are tangents to circle from external point)

MT = MN (both are tangents to circle from external point)

Hence LM = MT = MN and so, by the transitive property of equality, LM = MN.

Point M bisects LN.

Trapezoid BANL:

BQ = LN

In △ BAQ apply Pythagoras

(a + b)² = (a - b)² + BQ²

rearrange and simplify to get BQ = LN = 2√(ab)

Because M bisects LN it follows that LM = MN = √(ab)

Find PL:

Since LM = √(ab)

OL =OM - LM = R - √(ab)

PL is then R + OL = R + R - √(ab) = 2R - √(ab)

Find PN:

PN = PL + LN = 2R - √(ab) + 2√(ab) = 2R + √(ab)

The ratio PN/PL:

and so

Cross-multiply and rearrange to find R …

Note that a > b

That is the required formula.

Example:

If radius a = 4 units and radius b = 2 units then

R = √(4 x 2)(4 + 2) / 2(4 - 2) = 6√8 / 4

R = 4.24 (to 2 decimal places).

Please let me know via comments in the event that you find any problem with the mathematical reasoning above. The reasoning is entirely my own and I am keen to ensure that it is accurate. Thank you.

Many mathematical problems may be found on Youtube (e.g. Mind Your Decisions) and Facebook (e.g. Geometria) etc.

Georg Alexander Pick (1859 - 1942) was an Austrian mathematician who, in 1899, published an interesting theorem about “lattice polygons.”

A lattice polygon is a polygon whose vertices all sit on the integer coordinates of a grid - see next diagram.

Consider a rectangle of length 4 units and width 3 units. Area = 12 sq units. Now place the rectangle on a lattice made up of squares with 1 unit sides - (see Diagram). Our aim is to express the area of the rectangle (12) in terms of numbers of lattice points.

There are 6 lattice points inside the rectangle and 14 points on the perimeter. If we take the 6 interior points and add HALF the perimeter points we get 13. Deduct 1 and we have the area of the rectangle. See John D. Cook - Intuition for Pick’s Theorem.

Similar reasoning may be applied to ANY lattice polygon. Pick’s Theorem tells us that the Area (A) of a lattice polygon is given by the number of interior lattice points (letter I) + half the number of lattice points on the border (B) minus 1. That is

A = I + B/2 - 1

The theorem is a beautiful and simple mathematical result with numerous surprising applications in various fields - see “Connecting the dots with Pick’s Theorem” (Kristian Kiradjiev - University of Oxford).

A more complex example:

The diagram shows a form of star made up of a central square with identical right triangles placed on the sides of the square. The total area is Area of Square + 4(Area of a Triangle).

The sides of the square are 3 units. Each triangle has base 2 units and height 2 units. The area of the square is therefore 9 sq units. The area of each triangle is 2 sq units. The star has area 9 + (4 x 2) = 17 sq units.

Here is the star on a lattice.

There are 8 interior points (shown as X) and 20 boundary points (large dots). The volume is then

8 + 20/2 - 1 = 8 + 10 - 1 = 17 sq units.

The Equilateral Triangle:

It is a fact that one cannot draw an Equilateral Triangle on a lattice. A proof of this is shown in “Connecting the dots with Pick’s Theorem”

Euler Characteristic:

The Euler Characteristic (χ) describes the structure or shape of a space, regardless of how it is bent or stretched. For a POLYHEDRON (or a surface), the characteristic is calculated by the formula Number of Vertices (corners) - Number of Edges (lines) + number of faces (flat shapes). That is χ = V - E + F

For example, a cube has 8 corners, 12 edges and 6 surfaces. It has Euler Characteristic 8 - 12 + 6 = 2

Euler Characteristic and Pick’s Theorem are closely related as shown in this Youtube Video prepared by PBS Infinite Series.

Links:

Pick's Theorem (Theorem of the Day)

Pick's Theorem: Proof (Cut the Knot)

You are told that

a² – b = 133 (Equation 1)

b² – a = 133 (Equation 2)

and the task is to solve for a and b

You are also told that a ≠ b (i.e. a not equal to b)

Try the problem before reading further.

ooooo

Step 1 Subtract Equations 2 from 1and simplify

a² – b - (b² – a) = 0

a² – b² – b + a = 0

(a + b)(a - b) + (a - b) = 0

(a - b)(a + b + 1) = 0 and so either a - b = 0 or (a + b + 1) = 0

Since we are told that a is not equal to b we reject the first solution and so

(a + b + 1) = 0 from which a + b = -1 (Equation 3)

Step 2 Add Equations 1 and 2

a² – b + (b² – a) = 266

a² –b² – (a + b) = 266

a² + b² – (-1) = 266

a² + b² = 265

Step 3

From the general formula

(a + b)² = a² + b² + 2ab

(a + b)² = 265 + 2ab from which ab = -132

Step 4

From the general formula

(a - b)² = a² + b² - 2ab

(a - b)² = 265 + 264 = 529

and so, taking square roots,

(a - b) = ±23

Step 5 Two cases

a) If a - b = +23

a - b = +23

a + b = -1

adding gives

2a = 22 and a = 11 from which b = -12

b) If a - b = -23

a - b = -23

a + b = -1

adding gives

2a = -24 and a = -12 from which b = 11

The two solutions are therefore

a = 11 and b = -12

a = -12 and b = 11

This was an interesting and not entirely easy problem.

As usual the answers should be checked for accuracy.

I came across this problem on Facebook ….. You are told that

The task is to find the value of

Try the problem before reading further …..

ooooo

It is tempting to try to solve the equation for x and then evaluate the sum. BUT this is not actually necessary.

The following utilises the fact that A² - B² = (A + B)(A - B)

Let A = √(49 – x²) and B = √(25 – x²)

then A - B = 3

A² - B² = (A + B)(A - B) = 3(A + B)

also

A² - B² = 49 - x² - (25 - x²) = 24

and so 2(A + B) = 24 and A + B = 8

Solving the original equation:

A - B = 3

A + B = 8

add those to get 2A = 11 and so A = 11/2

Hence

√(49 – x²) = 11/2 and by squaring both sides

49 – x² = 121/4

x² = 49 = 121/4 = 18.75

As always verify the accuracy. Substitute 18.75 for x² in the original equation. The answer is correct.

The actual value of x is the square root of 18.75 and that is approximately ± 4.330 (to 3 decimal places).

This article highlights some interesting theorems related to circles.

Basic Circle Theorems:

Circle Theorems - GCSE Maths - Steps, Examples & Worksheet

Eight circle theorems (Tim Devereux)

The “circle theorems” shown in the diagram have been well-known for centuries.

Some other results:

Condensed List of All Formulas in Circles - MathBitsNotebook(Geo) - various formulas for circle angles

There are many other interesting items linked to circles such as The Arbelos etc.

Kissing circles:

Descartes' theorem - Wikipedia

Given three mutually tangent circles (black), there are, in general, two possible answers (red) as to what radius a fourth tangent circle can have.

Two possible answers? Algebraically, that points to some form of quadratic relationship. The problem was solved in 1643 by René Descartes.

The Descartes Circle Theorem - (Theorem of the Day)

Descartes' Theorem and Soddy Circle Radius Calculator | 4 Mutually Tangent Circles

Ptolemy’s theorem:

The Egyptian astronomer, mathematician and geographer Ptolemy (c. 100CE to c.170CE) left us an interesting theorem which is shown in the next diagram.

The Johnson circle theorem:

Roger Johnson (1890 - 1954) and George Tzitzeica (1874 - 1939) both discovered and proved the same theorem around 1916.

Johnson’s Theorem: You Probably Haven’t Heard of This Circle Theorem | by Tony Berard | Intro to Math | Medium

Johnson circle theorem

Draw three circles of radius r that intersect at a single point. Then draw a triangle connecting the remaining three points of intersection.

(Each pair of circles intersects in two points, one of which is the point where all three circles intersect, so there are three other intersection points.)

Then the circumcircle of the triangle, the circle through the three vertices, also has radius r.

The circumference (C) of a circle is 2πr units where π is the irrational number 3.14 …. and r is the radius of the circle.

The ratio of the circumference to the radius C/r = 2π

This led to the idea of angular measure. If the full circle is 2π then the 360 degrees of the circle came to be referred to as 2π radians or 2π rad.

Unlike degrees, which measure angles based on an arbitrary division of a circle (360 degrees), radians provide a natural, geometric way to measure angles directly related to a circle's radius.

Measuring angles in radians enables more exact calculations and commonly encountered angles are easy to express in radians. Thus, 360 degrees = 2π radians, 180 degrees = π rad, 90 degrees = π/2 rad, 60 degrees = π/3 rad, 45 degrees = π/4 rad and 30 degrees = π/6 rad.

It is also preferable to use radians to express angles greater than 360 degrees.

Suppose that a circular wheel of radius 3 metres turns at 30 revolutions per minute (rpm). A point on the wheel would turn through 10,800 degrees in a minute but that does not tell us the distance the point has travelled. However, during one revolution of the wheel the point would travel 2π radians = 2π x 3 metres = 6π metres. At 30 rpm that would mean 30 x 6π = 180π metres.

That is 565.49 metres (to 2 decimal places). That is equivalent to a linear speed of approx 9.42 metres per second.

The Byjus website shows

Conversion:

Links:

Radians and Degrees | Mathematics | Research Starters | EBSCO Research

Pi (π) - by Under Northern Skies - MyMathematics

Radian measure - Byjus

Video:

Master The Unit Circle & Radians - A Step-by-Step Guide

What is a Radian Angle? Convert Degrees to Radians & Radians to Degrees

My previous article showed how a formula for the volume of a pyramid (with a square base) could be derived from the previously observed fact that it takes 6 identical (congruent) pyramids to form a cube.

This article shows two methods of finding the formula without the need for previous knowledge about the required number of pyramids.

Both methods imagine that the pyramid is divided into slices (prisms) each of which has a specific volume. The sum of the volumes of the prisms approximates the pyramid volume and, as the number of prisms is increased to infinity, the total volume of the pyramid is found.

Method 1 - Volume of a Square Pyramid

Let the pyramid’s base length be L units and the vertical height be H. The pyramid volume is V. The pyramid is divided into N prisms each of identical thickness. The prisms are numbered from the top downwards 1, 2, 3. 4, …… n

Each prism has a thickness of H/N

The base is also divided by the same number (N). The length of the very top prism is L/N. The length of the second prism (N = 2) is 2L/N and the third prism (n = 3) is 3L/N and so on down to the lowest prism which has length NL/N

Each prism has a square base and so the volume of a prism is given by

(Prism number x L/n)² x H/N

By adding the volumes of the prisms we get

V ≈ (1 x L/N)² H/N + (2L/N)² H/N ……. + (NL/N)² H/N

Factor out (L/N)² H/N

(L/N)² H/N [1² + 2² + 3² + ….. + N²]

The square bracket [ ] contains the sum of the squares of N integers and that sum can be shown to be is given by

N(N + 1)(2N + 1) / 6 or N(2N² + 3N + 1) / 6 [See Sum of first n squares]

Hence, the volume of the pyramid is given by

V = (L/N)² x H/N [N(2N² + 3N + 1) / 6]

Which simplifies to L²H/6(2 + 3/N + 1/N²)

As N approaches infinity the terms containing N approach zero. The result is that, in the limit

V = L²H /3 or

1/3(Base Area x Height)

Method 2 - Volume of a Square Pyramid using integral calculus

Calculus (or “the calculus”) has an interesting history - Calculus history (MacTutor History of Mathematics) and, for a more detailed analysis, The History of Calculus by Arthur Rosenthal.

The following is one method of using calculus to find a formula for the volume of a pyramid with a square base.

A first point to note is that integral (with respect to x) of x squared is given by

where C is the constant of integration. Note: The constant of integration (usually written as C) may be ignored when evaluating definite integrals (integrals with a specified upper and lower range).

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Imagine our square pyramid is placed on an x. y cartesian coordinate system ( as shown).

The diagram shows pyramid ACD with a base length of A units

Line AB (length S) represents a small segment of the pyramid. The segment has thickness δx

The volume of the segment is therefore S²δx

Triangles OAB and OCD are similar and therefore S/X = A/H from which S = AX/H and so the volume of a segment can be written (AX/H)²δx

The volume of the pyramid is the summation of all of the segments from x = 0 to x = H

Applying integration we then have

and, with integration over the bounds 0 to H, that evaluates to

V = (A/H)² x [H³ / 3 - 0³ / 3] = (A/H)² x H³ / 3 = A²H / 3

but A²is the base area of the pyramid. Hence. the volume (V) of the square pyramid is (again) shown to be 1/3 x Base Area x Height

Links:

Volume of a Pyramid – Without Calculus – The Math Doctors

Cones_Pyramids_and_Spheres.pdf

General Proof: Why Volume of Pyramid is 1/3 of Volume of Prism (no Calculus)

How to Derive The Volume? Hard Geometry Problem

Sum of first n squares

Pyramids are found all across the world.

The Great Pyramid of Giza (Wikipedia) is the largest of the Egyptian pyramids and the most famous landmark of the Giza pyramid complex in Giza, Egypt. It is the oldest of the Seven Wonders of the Ancient World, and the only wonder that has remained largely intact. The pyramid has a square base and the apex is vertically above the centre of the base.

The (above ground) dimensions of the pyramid are 230.3 metres x 230.3 metres with a vertical height of 146.6 metres - (all dimensions approximate). An interesting fact emerges. Th perimeter of the base is 4 x 230.3 = 921.2 metres and that divided by the height (146.6) is 6.284 (to 3 decimal places) That is very close to 2 x π if we take π to be approx 3.142. Whether that was by accident or design is not known. [See Great Pyramid of Egypt at Giza].

A very early mathematical study of pyramids was undertaken by Liu Hui who lived in the state of Cao Wei during the Three Kingdoms period (220–280 CE) of China. His major contributions to mathematics were recorded in his commentary on The Nine Chapters on the Mathematical Art and include a proof of the Pythagorean theorem, theorems in solid geometry, an improvement on Archimedes's approximation of π, and a systematic method of solving linear equations in several unknowns.

What is the volume of a pyramid?

From early times it was known that six identical (congruent) pyramids could be placed together to form a cube.

The 6 pyramid arrangement requires one pyramid to be placed above another - apex to apex. The other pyramids are then placed in the pyramid-shaped spaces remaining.

Let H be the height of a pyramid and V the pyramid volume.

The height of the cube is twice the height of the pyramid and so the cube height is 2H.

The volume of the cube is base area x height. If B is the cube base area then its volume is B x 2H = 2BH

Since the six pyramids are identical it follows that each one has a volume of 1/6th the cube’s volume. Therefore

6V = 2BH and so

V = 1/3 x B x H

The volume of the pyramid is therefore one-third of its base area x its vertical height.

Volume of a Pyramid - GCSE Maths - Steps, Examples & Worksheet

Calculation:

Using the formula, the above ground level volume of the Great Pyramid is 1/3 x 230.3 x 230.3 x 146.6 and that is approximately 2,259,795 cubic metres.

A more general result:

Does the same formula apply to find the volume of any pyramid and not just one with a square base and apex vertically above the base centre?

YES - The formula applies to all pyramids regardless of their base shape or whether the apex is perfectly centred. Hence, it works for base shapes such as triangles, pentagons, hexagons or any polygon. Furthermore, the volume of a cone is also one-third base area x height which is

Furthermore, the apex does not have to be centred because the formula applies to “skew” pyramids where the apex is not directly over the centre of the base. (This can be seen as an application of what is known as Cavalieri’s Principle - named after Bonaventura Cavalieri 1598–1647).

Various proofs have been found for the formula applying to pyramids in general. For now, I will leave those to one side. The links below are worth exploring.

Links:

Volume of cubes and cuboids - KS3 Maths - BBC Bitesize

Volume of Rectangular Pyramid - Formula, Examples, Definition

Volume of a pyramid - Calculating the volume of a standard solid - National 5 Maths Revision - BBC Bitesize

Why is the volume of a pyramid what it is? | Flying Colours Maths

Papercrafts and other fun things: Six Square Pyramids Can Fit Perfectly Into a Cube

Volume Patterns for Pyramids

The Egyptian Triumph: The Volume of an Incomplete Pyramid

Pi and the Great Pyramid

Cavalieri's Principle: Visualizing Volume & Cross-Sections (Geometry Explained)

Cavalieri’s Principle and how it could transform area and volume | Cambridge Mathematics

The British Flag Theorem:

A proof of this theorem (by application of the Pythagorean Theorem) is set out at British Flag Theorem | Brilliant Math & Science Wiki.

The proof places P inside the rectaingle but the theorem also applies if P is on one of the sides of the rectangle AND it also applies if P is outside the rectangle.

Further still, the theorem can be applied in 3 dimensions - that is when P is elevated above the plane in which the rectangle lies.

BUT the theorem does NOT apply to a rhomboid (parallelogram).

A formula for a rhomboid (usually referred to as a parallelogram):

Rhomboid- Definition, Properties, Formulas, Solved Examples (Cuemath.com). In the following I use the usual word - parallelogram.

As already noted, the British Flag Theorem does not apply to a parallelogram but it is possible to show that (a² + b²) – (c² + d²) = 2st

Step 1 - apply the Pythagorean Theorem to give

a² = w² + z² b² =v² +y²

c² = u² + y² d² = x² + z²

Step 2 - apply those results to get an expression for (a² + b²) – (c² + d²)

(w² + z² + v² +y²) - (u² + y² + x² + z²) and that simplifies to

w² - x² + v² - u² and that may be written

(w + x)(w - x) + (v + u)(v - u)

Step 3 - note that w + x = s and that u + v = s and so we have

sw - sx + sv - su) = s[w - x + v + u) = s(w - u + v - x)

Now w - u = t and also v - x = t and so s(w - u + v - x) = 2st

The result is

(a² + b²) – (c² + d²) = 2st

Links:

Follow-up: British Flag Theorem

Tricky Geometry Problem: British Flag Theorem & Extension

Pi in the Sky : The British Flag Theorem

ABCD is a rectangle with side lengths a and b (as shown). E is a point on CD and F is a point on DA. Straight lines join B to E, E to F, F to B. The rectangle is therefore divided into four triangles with areas (in square units) of X, Y, Z and W (as shown). The problem is to find an expression for the area of W in terms of X, Y and Z.

Step 1: △ ABF

The area of △ ABF = Y and is ½AF.a

It follows that AF = 2Y/a and it then follows that FD = b - 2Y/a

Step 2: △ BCE

The area of △ BCE = X and is ½CE.b and so CE = 2X/b. It then follows that

ED = a - 2X/b

Step 3: △ DEF

The area of △ DEF = Z and is ½FD.ED and, from Steps 1 and 2, that is

Z = ½(b - 2Y/a)(a - 2X/b)

Step 4: Multiply both sides by 2 and then multiply the brackets and simplify

2Z = (b - 2Y/a)(a - 2X/b)

2Z = ab - 2X - 2Y + 4XY/ab

ab - 2X - 2Y - 2Z + 4XY/ab = 0

ab - 2(X + Y + Z) + 4XY/ab = 0

then multiply throughout by ab so as to remove ab from the denominator

(ab)²- 2ab(X + Y + Z) + 4XY = 0

This is a quadratic equation in variable (ab).

Step 5: Solve the quadratic equation

Use the quadratic formula to solve for (ab)

At this point we accept the + solution and reject the - solution. This is because the whole rectangle has to be larger than the sum of X + Y + Z. Hence

but since ab is also equal to (X + Y + Z) + W it follows that

and that is the required formula for the area W in terms of areas X, Y and Z

Note that X and Y are the areas of the two largest triangles.

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In the next diagram the triangle areas are defined as T, T + 1, T+ 2, T + 3 and we are required to find the value of T to three significant figures.

The formula derived above can be used to solve this problem. It is necessary to insert values for W, X, Y, Z into the formula. For Z write T, for X write T + 1, for Y write T + 2 and for W write T + 3

The result is

T + 3 = √ [ (T + 1 + T + 2 + T + 3)² – 4(T + 1)(T + 2)]

T + 3 = √ [ (3T + 3)² – 4(T + 1)(T + 2)]

T + 3 = √ [ (9T ²+ 18T + 9) – 4(T² + 3T + 2)]

Square both sides

(T + 3)² = (9T ²+ 18T + 9) – 4(T² + 3T + 2)

Multiply out the brackets and collect like terms

T² + 6T + 9 = 9T ²+ 18T + 9 – 4T² - 12T – 8

Leads to

4T² – 8 = 0

And

T = √2 square units which is 1.41 sq. units (to 3 significant figures)

( NB: The negative square root is rejected because T cannot be a negative area! )

Links:

If we apply the formula (above) to this problem we get

Orange Area = √ [ (9 + 5 + 4)² - 4(5 x 9) ]

That is √ [ 18² - 180 ] = √ [ 324 - 180 ] = √ 144 which is

Orange area = 12 square units

Lost Math Lessons Blog - Triangles inscribed in rectangles

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Parts of a Circle:

The various terms used in connection with circles are explained at Parts of a Circle - Definition, Formulas, Examples (Cuemath.com). That website offers this diagram

Chord of a Circle- GCSE Maths - Steps, Examples & Worksheet (Thirdspacelearning.com)

Area of a sector:

There are 360 degrees ( = 2 radians) in a circle - see Degrees (Angles)

The area of a circle is π R² see previous article Area of a Circle.

If a sector has a central angle θ degrees then the area of the sector will be a proportion of the full circle

Area of sector with θ as its central angle = θ/360 x π R²

For example: If a circle has radius 6 cm and the central angle of a sector is 60 degrees then the area of the sector is 60/360 x π R² = 1/6 x π 6² = 6π and so the area of the sector is 18.850 cm² (to 3 decimal places)

A solution to a problem:

I discovered the following interesting problem on Youtube at Presh Talwalker’s “Mind Your Decisions”

A square has sides 4 units. A quarter circle and a semicircle are drawn as shown. The area of overlap is coloured red. What is the area of the red overlap?

Please attempt the problem before reading further …..

My approach to solving the problem was as follows.

Step 1: Construct a tangent from A to the semicircle at point T. Also construct a straight line from A to the midpoint (M) of side CD. Then construct a tangent (MT) from M to the quarter circle.

Step 2: There are two congruent right triangles - △ ADM and △ AFT. (They are congruent because 3 corresponding sides are equal).

Step 3: RED area = Area of semicircle -YELLOW AREA - GREEN AREA

Step 4: Area of semicircle is half the area of a circle of radius 2. This is 2π sq units

Step 5: Yellow Area.

This is Area of of quadrilateral ATMD - area of the sector (ATD) of the quarter circle.

Quadrilateral ATMD is made up of two congruent triangles each of which has area 4 sq units and so Area ATMD = 8 sq units.

Area of the sector is ∠DAT / 360 x π R²

The radius of the quarter circle is 4 units and so Area Sector = ∠DAT/360 x π x 4²

The Yellow Area therefore has area

8 - [∠DAT/360 x π x 16]

Step 6: Angle DAT

∠DAM = ∠MAT

Using trigonometry, ∠DAM is the angle with tangent 2/4 = 0.5 and that is approx. 26.5651 degrees. ∠DAT = 2 x ∠DAM = 53.1302 degrees

Step 7: Green Sector

From the geometry of the figure, ∠CMT = ∠DAT = 53.1302 degrees

The semicircle is half of a circle with area π x 2 ² = 4π

The green sector has area

53.1302/360 x 4π

Step 9: RED Area

Red = Area of semicircle -YELLOW AREA - GREEN AREA

RED = 2π - {8 - [53.1302/360 x π x 16]} - [53.1302/360 x 4π]

Using an online calculator that worked out to approximately 3.847 sq units.

A not entirely straightforward problem. The Mind Your Decisions approach is somewhat different to mine and is at Viral question from China

The diagram shows a square ABCD of side length 4 units.

A circle (centre O) is within the square such that sides BC, CD and also diagonal BD are tangential to the circle.

Find the radius (r) of the circle.

This problem was published on X (Twitter) and it was interesting to see the various solutions offered. Some ended up with the need to solve a quadratic equation.

The following much simpler solution did not appear on X.

Solution:

Point 1). We should not assume that diagrams are to scale or otherwise drawn accurately unless, of course, we are told that they are.

Let the diagonals of the square cross at X. The diagonals of a square cross at right angles and so angle CXB is a right angle. Diagonal BD is tangential to the circle and so makes a right angle with the circle’s radius (XO). It follows that diagonal AC passes through both X and the centre of the circle (O).

Point 2). Construct a line from the circle centre (O) perpendicular to side BC. Let the line meet BC at point T

Point 3). BX = BT because both are tangents to the circle from point B. (For proof of this see Tangents - Higher - Circle theorems - BBC Bitesize).

OT = OX = r

△s BTO and BXO are congruent - (3 sides).

Point 4). AX = BX = √8 = 2√2 (Pythagoras’ theorem applied to △ AXE

Hence, from point 3, BX = BT = 2√2

Point 4). Since CT = r it follows that BT = 4 -r units

Point 5). From points 3 and 4

BT = 2√2 = 4 - r

Hence r = 4 - 2√2 = 2(2 - √2)

r = 2(2 - √2)

This is 1.175 units (to 3 decimal places).

*****

Circled symbols:

Circled symbols are often used to show that an arithmetical operation is in modular arithmetic. For instance, the plus sign becomes ⊕

Addition: ⊕

a ⊕ b (mod n) can be found by adding a and b and then finding the remainder when the sum is divided by n

e.g. 9 ⊕ 7 (mod 5)

9 + 7 = 16 and then 16/5 = 3 remainder 1

Hence 9 ⊕ 7 ≡ 1 (mod 5)

e.g. 15 ⊕ 20 (mod 7)

15 + 20 = 35 and then 35/7 = 5 remainder 0

Hence 15 ⊕ 20 ≡ 0 (mod 7)

e.g. 32 ⊕ 47 (mod 6)

32 + 47 = 79 and 79/6 = 13 remainder 1

15 ⊕ 20 ≡ 1 (mod 6)

The general rules for addition are …

• If A + B = C, then A (mod n) + B (mod n) ≡ C (mod n)

• If A ≡ B (mod n), then A + k ≡ B + k (mod n) for any integer ‘k’

• If A ≡ B (mod n) and C ≡ D (mod n), then A + C ≡ B + D (mod n)

• If A ≡ B (mod n), then -A ≡ -B (mod n)

Subtraction: ⊖

a ⊖ b (mod n) can be found by subtracting b from a and then finding the remainder when divided by n. If the result is negative then add n until the result is positive.

e.g. 8 ⊖ 11 (mod 5)

8 - 11 = -3 and so we add 5 to get 2

Hence 8 ⊖ 11 ≡ 2 (mod 5)

e.g. 4 ⊖ 9 (mod 6)

4 - 9 = -5 and so add 6 to get 1

4 ⊖ 9 ≡ 1 (mod 6)

The general rules for subtraction are similar to those for addition.

Multiplication: ⊗︀

a ⊗︀ b (mod n) can be found by multiplying a and b and then finding the remainder when the product is divided by n

e.g. 17 ⊗︀ 14 (mod 3)

Method (1) 17 x 14 = 238 and 238 / 3 = 79 remainder 1

Hence 17 ⊗︀ 14 ≡ 1 (mod 3)

Method (2) 17 mod 3 = 2 and 14 mod 3 = 2 and so 17 ⊗︀ 14 (mod 3 ) becomes 2 ⊗︀ 2 (mod 3)

2 ⊗︀ 2 (mod 3) ≡ 4 (mod 3) ≡ 1 (mod 3)

e.g. 12 ⊗︀ 23 (mod 8)

12 mod 8 = 4 and 23 mod 8 = 7

4 ⊗︀ 7 (mod 8) ≡ 28 (mod 8) ≡ 4 (mod 8)

e.g. 15 ⊗︀ 81 (mod 12)

15 mod 12 = 3 and 81 mod 12 = 9

15 ⊗︀ 81 (mod 12) ≡ 3 ⊗︀ 9 (mod 12) ≡ 27 (mod 12) = 3

The general rules for multiplication are …

• If A ⋅ B = C, then A (mod n) ⋅ B (mod n) ≡ C (mod n)

• If A ≡ B (mod n), then kA ≡ kB (mod n) for any integer ‘k’

• If A ≡ B (mod n) and C ≡ D (mod n), then AC ≡ BD (mod n)

Division: ⨸

Modular system do NOT support DIRECT division and division is not always possible.

In some cases, division is possible by multiplication by the modular multiplication inverse of the divisor (i.e. the given the modulus). Such an inverse exists if a and b are COPRIME - that is to say that the greatest common divisor (gcd) of a and b is 1

In other words, two or more integers are coprime when their only shared factor is 1

Coprime numbers need not be PRIME numbers themselves.

An example of coprime numbers are 9 and 14.

Two consecutive numbers (e.g. 21 and 22) are always coprime.

A further point relating to division is that, just as in ordinary arithmetic, division by zero is not defined.

e.g. 4 ⨸ 12 (mod 6)

Here, 12 mod 6 = 0 and so 4 ÷ 12 (mod 6) is not possible - division by zero.

Division may require further consideration later.

Links:

Modular Arithmetic: Addition and Subtraction

How to Multiply in Modular Arithmetic - Cryptography - Lesson 5 - YouTube

Highest Common Factor (HCF) - by Under Northern Skies

Of deeper interest:

This completely changed the way I see numbers | Modular Arithmetic Visually Explained

Tashko - Modular arithmetic - the math of remainders and cycles

n^2 +1 is never divisible by 7 - Use modular arithmetic to prove this statement

Modular arithmetic is a system of arithmetic for INTEGERS where numbers "wrap around" after reaching a certain value, called the modulus. This form of arithmetic is mainly concerned with remainders - i.e. the amount left after wrapping around. It is often referred to as "clock arithmetic.

Take an integer A and divide it by integer B. There will be a quotient Q and a remainder R which may be equal to zero. Examples are (a) 13/10 = 1 remainder 3 and (b) 14/7 = 2 remainder zero.

The 12 hour clock:

A clock marked with 12 hours (1 to 12) offers a basic example of modular arithmetic.

Eg. 1]

5 hours after 10:00 is 15:00 but the clock only shows numbers up to 12. So, in this case, the clock displays the time as 3 o’clock - that is, (10 + 5) -12 = 3

Another way of looking at that is to say 15/12 = 1 remainder 3

Eg. 2]

If it is now 11:00 then the time in 83 hours can be found by adding 83 to 11 = 94 and dividing by 12. That is 94/12 = 7 reminder 10. Thus the time after 83 hours will be 10:00

In the example of the 12 hour clock, the number 12 is known as the modulus. Examples 1 and 2 are written

15 mod 12 = 3

94 mod 12 = 10

General point:

A modulus can be any positive number.

Problem:

Today is Friday. What day of the week will it be in 923 days time?

This is easily found by using modulus 7 since there are 7 days in a week.

923 mod 7 = 6 and so the answer would be Thursday.

Positive numbers:

What is 7 mod 5

It is easy to see that 5 divides once into 7 and leaves remainder 2. The answer is therefore 7 mod 5 = 2

Negative numbers:

What is -6 mod 5

Find the number nearest to -6 which is both smaller than -6 and divisible by 5

That number is -10

Subtract that number from -6

-6 - (-10) = 4

Hence -6 mod 5 = 4

Congruence:

The concept of congruence in modular arithmetic may be illustrated by an example.

17 mod 3 = 2

32 mod 3 = 2

56 mod 3 = 2

It can be said that 17 and 32 are congruent (symbol ≡) under modulus 3 because they both yield the same remainder when divided by 3.

This is written 17 ≡ 32 (mod 3)

Similarly, 32 and 56 are congruent under mod 3 and that is written

32 ≡ 56 (mod 3)

17 and 56 are also congruent and so

17 ≡ 56 (mod 3)

Note the brackets around mod 3. This shows that mod 3 applies to both sides -

e.g. 17 ≡ 56 (mod 3) therefore refers to both 17 mod 3 and 56 mod 3

Having illustrated the meaning of congruency it can be stated that - Two integers a and b are congruent modulo n if their difference (a - b) is an integer multiple of n

i.e. (a - b) = kn where k is an integer

a and b will leave the same remainder when divided by n

Links:

An Introduction to Modular Arithmetic | NRICH

Fun With Modular Arithmetic – BetterExplained

Modular Arithmetic - GeeksforGeeks

Modular Arithmetic (w/ 17 Step-by-Step Examples!)

Modular Arithmetic - Properties and Solved Examples

Article - Modular arithmetic explained: basics, applications and examples - Casio Calculators

Modulo Calculator

Video:

■ What is Modular Arithmetic? | An introduction to the strange world of mathematical time-telling

Modular Arithmetic: Addition and Subtraction

An interesting geometric problem was first posed in 1922 by Edward Mann Langley (1851 - 1933), the founder of the Mathematical Gazette. The problem is referred to as Langley’s Adventitious Angles. Various solutions have been put forward to this classic isosceles triangle geometry puzzle. For present purposes, we seek a solution using only basic geometry. The following diagram sets out the problem.

ABC is an isosceles triangle with angles of the values shown.

At first sight, it appears that finding angle BEF would be simply a matter of using the fact that the angles of a triangle add to 180 degrees. In fact, this seemingly innocent problem is more difficult than its appearance. Straightforward use of angle chasing is not sufficient to solve the problem.

Adventitious angle problems relating to isosceles triangles require us to find a particular angle (known as the derived angle) when we are given certain other angles. Langley proposed the particular case of (20⁢°,60⁢°,50⁢°;30⁢°). The derived angle (BEF) is 30⁢°

Interestingly, only a limited number of cases produce a derived angle with an integer value.

The properties of isosceles triangles are set out at Math.net Isosceles Triangle. The adjective “adventitious” means something not expected or not planned.

One solution (Wikipedia):

A solution is presented at Wikipedia

Mind Your Decisions:

I recommend viewing the solution that is well-presented visually on the estimable Mind Your Decisions - Youtube

Image: Edward Mann Langley

The diagram shows three identical red circles within a right triangle. Two side lengths of the triangle are given as 5 and 12 units. The problem is to find the total area of the three red circles.

Before reading further, try to solve the problem. My solution is below.

Some initial points:

1. The area of a circle is π R² see previous article Area of a Circle. For the purposes of this problem we take π = 3.142

2. The Pythagorean Theorem enables us to find the third side of right triangle if we know the lengths of the other two sides - previous article Pythagorean Theorem

3. Tangents to a circle form a right angle with the circle radius at the point of tangency

4. The lengths of two tangents drawn to a circle from the same external point are of equal length. In the next diagram CA = CB See Cue Math Tangents

   

My solution:

I hope you tried the problem. At first it seems somewhat tricky but the solution is quite straightforward if the initial points (above) are used.

Step 1 The hypotenuse of the large right triangle is 13 units. This follows from the Pythagorean theorem.

Step 2 Find the various lengths as shown in the diagram.

Step 3 Construct a line from A to the centre of the right hand circle - this is line AC

Two right triangles are formed ACD and ACB

In △ ACD we have (applying Pythagoras) AC² - (5r)² + (5 – r)² = 26r²- 10r + 25

In △ ACB we have (applying Pythagoras) AC² = (5r + 1)² + r² = 26r²+ 10r + 1

Hence 26r²- 10r + 25 = 26r²+ 10r + 1 from which r = 1.2

Step 4 The combined area

The three red circles therefore have area

3 x π x 1.2²

and that is approximately 13.57 sq units