This document contains a work readiness pack for an A Level maths tutor, covering various math topics: 1. Series, the volume of a torus, and relating the arc of a circle to the chord of regular polygons are discussed. 2. Diagrams of a torus and polygons are included. Formulas for the volume of a torus and area of irregular ovoids are derived. 3. Methods for measuring angles using a polygon protractor and relating sides of triangles to angles are explored. Finding lengths in slices of a 24-sided polygon is addressed.

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20220705 to 20220926. Work Readiness Pack. From Sharon Xiao Liu (sl51926).
Main body of the document (original maths work)
Series
U_(1) = 1. U_(2) = (U_(1) + 1)^2.
(Subscript font, is denoted by an underscore and brackets.)
(Raising a number to a power, is denoted by ^.)
U_(n+1) = (U_(n) + 1)^2.
U_(4) = (((U_(1) + 1)^2 + 1)^2 + 1)^2.
The nth term, is calculated by doing +1)^2, n-1 times.
Volume of a torus
Let us describe the torus of interest. A torus, is a ring doughnut shape. The centre hollow, when viewed from above, has
the shape of a circle, which is expressed as x^2 + y^2 = r^2, where the axes are viewed from above. x is the x-coordinate,
of a point on the centre circle. y is the y-coordinate, of the same point, on the centre circle. r is the radius of the circle, in
the centre of the torus. (Later on, we’ll use these symbols again, but for different meanings, which will be defined.)
Checked, up to here. I’m not going to go through all of this, with a fine toothcomb. I’ll bear this need to polish the work,
in my mind, later. Going through all of this, and explaining it, is extremely boring.
The centre is a circle (x^2 + y^2 = r(_(centre))(^2)). The outer diameter is a circle (x^2 + y^2 = r(_(edge))(^2)). The cross-
section of the torus, has an outline which is represented by x^2 + y^2 = r(_(cross-section))(^2).
Description of the torus: The horizontal inner and outer circles, are in the xz plane. The horizontal circles, stack up along
the y plane. The vertical cross-sectional circles, are kind of in the xy plane, but rotate around the y axis.
The volume of a torus, is a constant, multiplied by the inner radius, and the cross-sectional radius. The constant, is found
by iteration.
The torus, is bounded by a box. This normally, involves pi, to sort out. Pi times a times b, is sufficient to describe an oval.
So then, what is the equation, to describe a flat ovoid, which has four axes, with different lengths? Is this, pi times the
square root, of ab, and the square root of cd? We are looking at the area of the flat ovoid. Or, is it some expression, of
brackets, which leads to a quadratic? What is, the area of a flat ovoid, which has the axes, 1, 2, 3 and 4? The highest
power, is still to the power of 2. When a = b, pi times a times b, is pi times r squared. Perhaps the area of a flat ovoid, is
pi times the square root of abcd. If a = b = c = d, you get pi times r squared. (a_(1) + b_(1))(c_(1) + d_(1)) = square root of
abcd. Expand the expression. There is addition, in this quadratic, because a_(1) and b_(1) are on the same line. Perhaps,
the equation is pi times (a + b)(c + d). However, a + b is too long. Is (a_(1) + b_(1))(c_(1) + d_(1)) = ab, when (a_(1) +
b_(1))/2 = a. The expression is wrong, as when a_(1) and b_(1) are swapped around, a different area might be formed.
Actually, it still works. The expansion of (a_(1) + b_(1))(c_(1) + d_(1)) = a_(1) times c_(1) plus a_(1) times d_(1) plus b_(1)
times c_(1) plus b_(1) times d(1). Axis a, for an ellipse = a_(1) + b_(1).
The area of a flat irregular ovoid, could be pi times the square root (abcd). This is because, when a = b, and c = d, for an
oval, then the area of an oval, is pi times ab. When a = b = c = d, the area of a circle, is pi times r squared. So from these
simplifications, the formula for the area of a flat irregular ovoid, appears to be working. Whichever way, that you rotate
or flip the four axes, the area stays the same. The axes, could be 1234, or 3214, or 4321, but with 3124, 1 is neighbouring
3 and 2. In the original 1234 shape, 1 is neighbouring 3 and 4. We could look at the differences, going clockwise or
anticlockwise. Do a score, for the left of the curve, and the right of the curve, around one point (meaning the limit of the
axis). 1234: 1: to the left of 1, there is 4 and 3; to the right of 1, there is 2 and 3. Because there are the same 4 numbers,

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the flat irregular ovoid, has the same area. We have now proven, that this equation is correct. These four axes (1234), do
exist. The formula, definitely is not pi times the square root ((a+b)/2 times (c+d)/2), because of the way the apogee are
related to each other, going around the shape. The area is proportional, to the axes.
There might be a possibility, that a multi-spoked splat, is made from pi times the cube root of the lengths of the radii.
If you investigate the bounding box, around a curve, you could apply the maths equation, to regular curves (where all
four axes, are the same). For the area of a bounding square, to be converted to the area of its inner circle, you multiply
by pi over 4. This constant is the same, to convert a rectangle to an inner oval. This is because, the area of a square is
(2r) squared. The area of the inner circle, is pi times r squared. The area of a rectangle is 2a times 2b. The area of the
inner oval, is pi times a times b.
We should now, look at the isosceles triangle, and its inner circle sector. You could look at a sector, which subtends 1
degree. The area of the isosceles triangle, is tan(x) times the radius squared. The area of the inner circle sector, is
(1/360) times pi times the radius squared. To convert from the isosceles triangle, to the sector, we multiply by ((x/360)
times pi)/(tan(0.5*x)), where x is the angle subtended. Check the area equation, for the isosceles triangle.
Work out tan(x), by drawing a curve, from the origin, to x=90 degrees and y=infinity. Tan(x) = opposite/adjacent. o^2 +
a^2 = 1. If you know either the opposite, or the adjacent, you would know tan(x). However, to find x, you would need to
know, how ratios link to the angle. When the angle is 0, the opposite is 0, and the adjacent is 1, so the ratio is 0. When
the angle is 90 degrees, the adjacent is 0, and the opposite is 1, so the ratio is infinity (a vertical line). When the angle is
45 degrees... When the angle is 30 degrees... When the angle is 60 degrees...
Table which shows the relationship between the angles and ratios of the sides, in a right-angled triangle.
Angle (theta, or x) Opposite side length Adjacent side length Hypotenuse length Ratio
We discover, that the specific lengths, are not important, as we are interested in just the ratio.
Angle (theta, or x) Opposite side, divided by the
adjacent side
0 0/1 = 0
30 1/(root 3)
45 1/1 = 1
60 (root 3)/1
90 1/0 = inf
All of these points, should be drawn out, with a smooth curve connecting them. Is it possible, to do this in Scalable
Vector Graphics? The work readiness pack, should contain the SVG curve torus-20220712.html . It is not recommended,
to put images in this document, as the document loads slowly on OneDrive, anyway.
Work out tan(x), and the angle, by using a polygon as a protractor. This polygon, which has 360 vertices, is constructed
using straight line equations. The polygon initially has four vertices. Then, this is bisected further, to create eight
vertices. The number of vertices, is 2 to the power of n, with n=2 initially. The coordinates of the initial 4 vertices, are
known. Then, an equation is produced, linking two corners together. The midpoint of this line, is found, and then a new
linear equation, linking the origin, to the midpoint is found. The length of this line, is extended, until it is the same as the
original 4 lines, coming out from the origin. So now, we have 8 vertices. A right-angled triangle, (an angle from it), can fit
against this polygon protractor, and the angle, or the sides, can be found from the grid. The right-angled triangle, can be
converted into an isosceles triangle, so it fits onto the polygon protractor. The right-angled triangle, can be standardised,
so the hypotenuse is a standard length, as it is on the polygon protractor. There can be 360 vertices, the polygon, so that
the angle in the right-angled triangle of interest, is measured, correct to the nearest degree.

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Go into numbers, to complete the theory of the polygon protractor. It definitely works, though!
The volume of a torus, if it is a single layer, can be calculated as follows. The inner radius, is 2. The outer radius, is 3. The
volume of a torus, can be related to the volume of a sphere. The volume of a torus, can be formed, by rolling a sphere,
around in a circle. The path of the sphere, is on a radius of 2.5 , and diameter of 5. The length of the path, is pi times 5.
When the sphere travels around a circumference of pi times 5, the width to the inner side, is formed of overlaps, and
the width to the outer side, is dispersed. When a sphere rolls, what volume, does it cover? When the sphere, is placed
sequentially, in a line, clearly the volume of the track, is more than the volume of the spheres. The torus, can be treated,
like a cylinder. Hence, the volume of the torus, is pi times 5, times pi times 0.5 squared.
Scanned diagrams

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Relating the arc to the chord
A circle can have a square diamond in it. The lengths of the sides of the diamond, are the same. So, the lengths of the
arcs are pi times 2r, divided by 4. The lengths of the sides of the diamond, are the square root of (r squared, plus r
squared). A circle can have a hexagon in it. The lengths of the sides of the hexagon, are the same. So, the lengths of the
arcs are pi times 2r, divided by 6. The lengths of the sides of the hexagon, are r.
By virtue of bisection, and the kite shape (applied to one side of the regular polygon), we do find the arcs and polygon
sides, for a 12-sided polygon.
So, what are the angles, for an irregular triangle? What are the arcs and polygon sides, for a 24-sided polygon, directly?
The side for an irregular triangle, can be extended, to make an isosceles triangle. Then, the angle surrounded by the two
equal sides, can be found.
A 24-sided polygon’s single slice, has sides of r. Modify pythagoras, for isosceles triangles. When an isosceles triangle, is
bisected, it becomes a right-angled triangle. What is the length, of the third side, of a 24-sided polygon’s single slice?
Measure the angles, in radians. Total angle is 2 times pi. So one angle, is 2 times pi, divided by 24. Which is pi divided by
12.
You can find out all of the angles, in a 24-sided polygon’s single slice. The number of slices is 24. So the total angles
within the slices, is 24 times 180 degrees. Can you build a single slice, using right-angled triangles? You could draw out
the triangle, and using an angle and length of the hypotenuse, measure the third side, of a 24-sided polygon’s single
slice. Can you work out the relative sides of a triangle, given three angles of the triangle? Yes. By drawing out the
triangle. You need two angles of the triangle, only. The bigger the subtending angle, the bigger the side. Angle 1: angle 2:
angle 3 = side 1: side 2: side 3. Angle 1 + angle 2 + angle 3 = 180 degrees. Try testing this, with a right-angled triangle. 30:
60: 90 = 3: 4: 5. 3: 6: 9 = 3: 4: 5. This doesn’t work. Try the angles in radians. 30/360 times 2 times pi: 60/360 times 2

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times pi: 90/360 times 2 times pi = 3/36 times 2 times pi: 6/36 times 2 times pi: 9/36 times 2 times pi = 3: 6: 9. This
doesn’t work.
Back to the isosceles triangle. We know the third angle, and the lengths of two sides (which is r). If we know the third
angle, is it possible, knowing also the coordinates of two points, to find out the coordinates of the third point? The angle
is 1/24 out of 360 degrees. Draw a text diagram. Angle 1: angle 2: angle 2 = side 1: side 2: side 2. Relate x squared plus y
squared equals r squared, to angles in a triangle. Draw a right-angled triangle, within a circle. Draw an isosceles triangle,
to the left of the right-angled triangle. You then know the third angle, of the isosceles triangle, and also the coordinates
of three points. So, for a 24-sided polygon’s slice, we know the origin angle, of the adjacent triangle, and its hypotenuse.
Relate sin angle, cosine angle, and tan angle, to each other. Sine angle, is opposite over hypotenuse. Sine angle, times
adjacent over opposite, is cosine angle.
In the kite, of a slice of the 24-sided polygon, we know the origin angle. The right-angled triangle’s origin angle, is half of
the origin angle, of the isosceles slice. Sine angle = half of the third side / r. Sine rule: sine 90 = 5/5 = 1. sine 30 = 3/5.
Sine 60 = 4/5. 5/1 = 3/(3/5) = 4/(4/5). Sine angle/half of the third side = sine (side isosceles angle)/bisecting side. 1/r =
sine (side isosceles angle)/bisecting side. We have found the length, of the bisecting side.
To find the other angles, apart from half of 60 degrees, and 360/12/2 degrees, you continue bisecting. 360/12=3, and
then if you divide by 2, you get 1.5 degrees. Then, you are accurate to the nearest 1.5 degrees.
Animations: A rotating cube, a bouncing cube.
Type on the computer, until 11am on the 26th of September 2022. Keep reading this. Don’t log off, until 11am.
A rotating cube, is easy. It is x^2 + y^2 = r^2, plus the transformations of one corner. This is, when it is rotated, around a
centre, which is located anywhere.
A bouncing cube. When a cube bounces, it flips. If it rotates as it bounces, it continues to rotate, after the bounce. The
cube falls in an arc. What is the equation, of the arc? It is an upside down parabola, y = x^2. Prove it is a parabola. A
physical object, falls in a straight line, when it is going towards the radius of the earth. That’s one-dimensional. When a
cube bounces, that’s two dimensional. Therefore, the x is raised to the power of 2. y = x^3, occurs in a spiral galaxy.
Dimensions come about, when multidirectional forces, are added. When you multiply a single dimension, by itself, this
does also qualify as another dimension added, when compared to adding an axis (e.g. the z-axis). This is how you get a
representation of the higher dimensions, on the xy plane.
When a cube rotates, it spins on an axis. The x^2 + y^2 = r^2, is combined with y = x^2, with a bouncing and rotating
cube. The number of dimensions, is correct, as you can draw the path of the centre of the cube, and its corners, on an xy
plane.
A rotating circular swingset.