avewright/proofwiki-math huggingface dataset:
Theorem: Length of Inradius of Triangle

Let $\triangle ABC$ be a triangle whose sides are of lengths $a, b, c$.
Then the length of the inradius $r$ of $\triangle ABC$ is given by:
:$r = \dfrac {\sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } } s$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.

Proof:
:410px
Let $\AA$ be the area of $\triangle ABC$.
From Area of Triangle in Terms of Inradius:
:$\AA = r s$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
From Heron's Formula:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
Hence the result:
:$r = \dfrac {\sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } } s$
{{qed}}
Theorem: Length of Circumradius of Triangle

Let $\triangle ABC$ be a triangle whose sides are of lengths $a, b, c$.
Then the length of the circumradius $R$ of $\triangle ABC$ is given by:
:$R = \dfrac {abc} {4 \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.

Proof:
:320px
Let $\AA$ be the area of $\triangle ABC$.
From Area of Triangle in Terms of Circumradius:
:$\AA = \dfrac {a b c} {4 R}$
From Heron's Formula:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
Hence the result:
:$R = \dfrac {abc} {4 \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } }$
{{qed}}
Theorem: Area of Regular Polygon by Circumradius

Let $P$ be a regular $n$-gon.
Let the circumradius of $P$ be $r$.
Then the area $\AA$ of $P$ is given by:
:$\AA = \dfrac 1 2 n r^2 \sin \dfrac {2 \pi} n$

Proof:
:400px
From Regular Polygon is composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$.
Then $\AA$ is equal to $n$ times the area of $\triangle OAB$.
Let $d$ be the length of one side of $P$.
Then $d$ is the length of the base of $\triangle OAB$.
Let $h$ be the altitude of $\triangle OAB$.
The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.
Then:
:$(1): \quad h = r \cos \dfrac \pi n$
:$(2): \quad d = 2 r \sin \dfrac \pi n$
So:
{{begin-eqn}}
{{eqn | l = \AA
      | r = n \frac {h d} 2
      | c = Area of Triangle in Terms of Side and Altitude
}}
{{eqn | r = \frac n 2 \paren {r \cos \frac \pi n} \paren {2 r \sin \dfrac \pi n}
      | c = substituting from $(1)$ and $(2)$ above
}}
{{eqn | r = \frac 1 2 n r^2 2 \paren {\cos \frac \pi n} \paren {\sin \dfrac \pi n}
      | c = rearranging
}}
{{eqn | r = \frac 1 2 n r^2 \paren {\sin \frac {2 \pi} n}
      | c = Double Angle Formula for Sine
}}
{{eqn | r = \frac 1 2 n r^2 \sin \frac {2 \pi} n
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Perimeter of Regular Polygon by Circumradius

Let $P$ be a regular $n$-gon.
Let $C$ be a circumcircle of $P$.
Let the radius of $C$ be $r$.
Then the perimeter $\PP$ of $P$ is given by:
:$\PP = 2 n r \sin \dfrac \pi n$

Proof:
:400px
From Regular Polygon is composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$.
Then $\PP$ is equal to $n$ times the base of $\triangle OAB$.
Let $d$ be the length of one side of $P$.
Then $d$ is the length of the base of $\triangle OAB$.
The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.
Then:
:$d = 2 r \sin \dfrac \pi n$
So:
{{begin-eqn}}
{{eqn | l = \PP
      | r = n d
      | c = 
}}
{{eqn | r = n \paren {2 r \sin \dfrac \pi n}
      | c = substituting from above
}}
{{eqn | r = 2 n r \sin \dfrac \pi n
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Regular Polygon by Inradius

Let $P$ be a regular $n$-gon.
Let $C$ be an incircle of $P$.
Let the radius of $C$ be $r$.
Then the area $\AA$ of $P$ is given by:
:$\AA = n r^2 \tan \dfrac \pi n$

Proof:
:400px
From Regular Polygon is composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$.
Then $\AA$ is equal to $n$ times the area of $\triangle OAB$.
Also, $r$ is the length of the altitude of $\triangle OAB$.
Let $d$ be the length of one side of $P$.
Then $d$ is the length of the base of $\triangle OAB$.
The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.
Then $d = 2 r \tan \dfrac \pi n$
So:
{{begin-eqn}}
{{eqn | l = \AA
      | r = n \frac {r d} 2
      | c = Area of Triangle in Terms of Side and Altitude
}}
{{eqn | r = \frac n 2 r \paren {2 r \tan \dfrac \pi n}
      | c = substituting from above
}}
{{eqn | r = n r^2 \tan \dfrac \pi n
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Perimeter of Regular Polygon by Inradius

Let $P$ be a regular $n$-gon.
Let $C$ be an incircle of $P$.
Let the radius of $C$ be $r$.
Then the perimeter $\PP$ of $P$ is given by:
:$\PP = 2 n r \tan \dfrac \pi n$

Proof:
:400px
From Regular Polygon is composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$.
Then $\AA$ is equal to $n$ times the area of $\triangle OAB$.
Also, $r$ is the length of the altitude of $\triangle OAB$.
Let $d$ be the length of one side of $P$.
Then $d$ is the length of the base of $\triangle OAB$.
The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.
Then:
:$d = 2 r \tan \dfrac \pi n$
So:
{{begin-eqn}}
{{eqn | l = \PP
      | r = n d
      | c = Perimeter of Regular Polygon
}}
{{eqn | r = 2 n r \tan \dfrac \pi n
      | c = substituting from above
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Isosceles Triangle

Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.
Let $\theta$ be the angle of the apex $A$.
Let $r$ be the length of a leg of $\triangle ABC$.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = \dfrac 1 2 r^2 \sin \theta$

Proof:
:300px
{{begin-eqn}}
{{eqn | l = \AA
      | r = \frac 1 2 b h
      | c = Area of Triangle in Terms of Side and Altitude
}}
{{eqn | r = \frac 1 2 b \paren {r \cos \dfrac \theta 2}
      | c = {{Defof|Cosine of Angle|Cosine}}
}}
{{eqn | r = \frac 1 2 2 \paren {r \sin \dfrac \theta 2} \paren {r \cos \dfrac \theta 2}
      | c = {{Defof|Sine of Angle|Sine}}
}}
{{eqn | r = \frac 1 2 r^2 \sin \theta
      | c = Double Angle Formula for Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Isosceles Triangle

Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.
Let $\theta$ be the angle of the apex $A$.
Let $r$ be the length of a leg of $\triangle ABC$.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = \dfrac 1 2 r^2 \sin \theta$

Proof:
A direct application of Area of Triangle in Terms of Two Sides and Angle:
:$\AA = \dfrac 1 2 a b \sin \theta$
where $a = b = r$.
{{qed}}
Theorem: Area of Segment of Circle

Let $C$ be a circle of radius $r$.
Let $S$ be a segment of $C$ such that its base subtends an angle of $\theta$ at the center of the circle.
Then the area $\AA$ of $S$ is given by:
:$\AA = \dfrac 1 2 r^2 \paren {\theta - \sin \theta}$
where $\theta$ is measured in radians.

Proof:
:350px
Let $BDCE$ be the segment $S$.
Let $b$ be the length of the base of $S$.
Let $BACE$ be the sector of $C$ whose angle is $\theta$.
The $\AA$ is equal to the area of $BACE$ minus the area of the isosceles triangle $\triangle ABC$ whose base is $b$.
Let $h$ be the altitude of $\triangle ABC$.
From Area of Sector, the area of sector $BACE$ is $\dfrac 1 2 r^2 \theta$.
From Area of Isosceles Triangle, the area of $\triangle ABC$ is $\dfrac 1 2 r^2 \sin \theta$.
Thus:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \frac 1 2 r^2 \theta - \frac 1 2 r^2 \sin \theta
      | c = 
}}
{{eqn | r = \frac 1 2 r^2 \paren {\theta - \sin \theta}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Mills' Theorem

There exists a real number $A$ such that $\floor {A^{3^n} }$ is a prime number for all $n \in \N_{>0}$, where:
:$\floor x$ denotes the floor function of $x$
:$\N$ denotes the set of all natural numbers.

Proof:
{{refactor|level = basic|extract the definition below into its own page}}
We define $\map f x$ as a '''prime-representing function''' {{iff}}:
:$\forall x \in \N: \map f x \in \Bbb P$
where:
:$\N$ denotes the set of all natural numbers
:$\Bbb P$ denotes the set of all prime numbers.
Let $p_n$ be the $n$th prime number.
From Difference between Consecutive Primes:
:$p_{n + 1} - p_n < K {p_n}^{5 / 8}$
where $K$ is an unknown but fixed positive integer.
Theorem: Equation of Ellipse in Reduced Form

Let $K$ be an ellipse aligned in a cartesian plane in reduced form.
Let:
:the major axis of $K$ have length $2 a$
:the minor axis of $K$ have length $2 b$.

Proof:
:500px
By definition, the foci $F_1$ and $F_2$ of $K$ are located at $\tuple {-c, 0}$ and $\tuple {c, 0}$ respectively.
Let the vertices of $K$ be $V_1$ and $V_2$.
By definition, these are located at $\tuple {-a, 0}$ and $\tuple {a, 0}$.
Let the covertices of $K$ be $C_1$ and $C_2$.
By definition, these are located at $\tuple {0, -b}$ and $\tuple {0, b}$.
Let $P = \tuple {x, y}$ be an arbitrary point on the locus of $K$.
From the equidistance property of $K$ we have that:
:$F_1 P + F_2 P = d$
where $d$ is a constant for this particular ellipse.
From Equidistance of Ellipse equals Major Axis:
:$(1): \quad d = 2 a$
Also, from Linear Eccentricity of Ellipse from Major and Minor Axis:
:$(2): \quad a^2 - c^2 = b^2$
Then:
{{begin-eqn}}
{{eqn | l = \sqrt {\paren {x - c}^2 + y^2} + \sqrt {\paren {x + c}^2 + y^2}
      | r = d
      | c = Pythagoras's Theorem
}}
{{eqn | r = 2 a
      | c = from $(1)$
}}
{{eqn | ll= \leadsto
      | l = \sqrt {\paren {x + c}^2 + y^2}
      | r = 2 a - \sqrt {\paren {x - c}^2 + y^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x + c}^2 + y^2
      | r = \paren {2 a - \sqrt {\paren {x - c}^2 + y^2} }^2
      | c = squaring both sides
}}
{{eqn | ll= \leadsto
      | l = x^2 + 2 c x + c^2 + y^2
      | r = 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2
      | c = Square of Difference
}}
{{eqn | ll= \leadsto
      | l = x^2 + 2 c x + c^2 + y^2
      | r = 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2
      | c = Square of Difference
}}
{{eqn | ll= \leadsto
      | l = a^2 - c x
      | r = a \sqrt {\paren {x - c}^2 + y^2}
      | c = gathering terms and simplifying
}}
{{eqn | ll= \leadsto
      | l = \paren {a^2 - c x}^2
      | r = a^2 \paren {\paren {x - c}^2 + y^2}
      | c = squaring both sides
}}
{{eqn | ll= \leadsto
      | l = c^2 x^2 - 2 c x a^2 + a^4
      | r = a^2 \paren {x^2 - 2 c x + c^2 + y^2}
      | c = Square of Difference
}}
{{eqn | r = a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2
      | c = Distributive Laws of Arithmetic
}}
{{eqn | ll= \leadsto
      | l = c^2 x^2 + a^4
      | r = a^2 x^2 + a^2 c^2 + a^2 y^2
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = a^4 - a^2 c^2
      | r = a^2 x^2 - c^2 x^2 + a^2 y^2
      | c = gathering terms
}}
{{eqn | ll= \leadsto
      | l = a^2 \paren {a^2 - c^2}
      | r = \paren {a^2 - c^2} x^2 + a^2 y^2
      | c = Distributive Laws of Arithmetic
}}
{{eqn | ll= \leadsto
      | l = a^2 b^2
      | r = b^2 x^2 + a^2 y^2
      | c = substituting $a^2 - c^2 = b^2$ from $(2)$
}}
{{eqn | ll= \leadsto
      | l = 1
      | r = \frac {x^2} {a^2} + \frac {y^2} {b^2}
      | c = dividing by $a^2 b^2$
}}
{{end-eqn}}
{{qed}}
Theorem: Equation of Ellipse in Reduced Form

Let $K$ be an ellipse aligned in a cartesian plane in reduced form.
Let:
:the major axis of $K$ have length $2 a$
:the minor axis of $K$ have length $2 b$.

Proof:
:500px
Let $P$ be an arbitrary point in the plane.
Let $PM$ be dropped perpendicular to $V_1 V_2$.
Hence $M = \tuple {x, 0}$.
From Intersecting Chord Theorem for Conic Sections:
:$PM^2 = k V_1 M \times M V_2$
for some constant $k$.
Hence:
{{begin-eqn}}
{{eqn | l = y^2
      | r = k \paren {a + x} \paren {a - x}
      | c = 
}}
{{eqn | r = k \paren {a^2 - x^2}
      | c = 
}}
{{end-eqn}}
Putting $x = 0$, we find the points where $K$ crosses the $y$-axis are defined by:
:$y^2 = k a^2$
This gives us:
:$k a^2 = b^2$
and so:
{{begin-eqn}}
{{eqn | l = y^2
      | r = \dfrac {b^2} {a^2} \paren {a^2 - x^2}
      | c = 
}}
{{eqn | r = b^2 - x^2 \dfrac {b^2} {a^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {y^2} {b^2}
      | r = 1 - \dfrac {x^2} {a^2}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Area of Ellipse

Let $K$ be an ellipse whose major axis is of length $2 a$ and whose minor axis is of length $2 b$.
The area $\AA$ of $K$ is given by:
:$\AA = \pi a b$

Proof:
Let $K$ be an ellipse aligned in a cartesian plane in reduced form.
Then from Equation of Ellipse in Reduced Form:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
Thus:
:$y = \pm b \sqrt {1 - \dfrac {x^2} {a^2} }$
From the geometric interpretation of the definite integral:
{{begin-eqn}}
{{eqn | l = \AA
      | r = b \int_{-a}^a \paren {\sqrt {1 - \dfrac {x^2} {a^2} } - \paren {-\sqrt {1 - \dfrac {x^2} {a^2} } } } \rd x
}}
{{eqn | r = b \int_{-a}^a 2 \sqrt {1 - \dfrac {x^2} {a^2} } \rd x
}}
{{end-eqn}}
Let $x = a \sin \theta$ (note that we can do this because $-a \le x \le a$).
Thus:
:$\theta = \map \arcsin {\dfrac x a}$
and:
:$\d x = a \cos \theta \rd \theta$
Then:
{{begin-eqn}}
{{eqn | l = \AA
      | r = b \int_{\map \arcsin {\frac {-a} a} }^{\map \arcsin {\frac a a} } 2 a \sqrt {1 - \frac {\paren {a \sin \theta}^2} {a^2} } \cos \theta \rd \theta
      | c = Integration by Substitution
}}
{{eqn | r = b \int_{-\frac \pi 2}^{\frac \pi 2} 2 a \sqrt {1 - \sin^2 \theta} \cos \theta \rd \theta
}}
{{eqn | r = b \int_{-\frac \pi 2}^{\frac \pi 2} 2 a \sqrt {\cos^2 \theta} \cos \theta \rd \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = a b \int_{-\frac \pi 2}^{\frac \pi 2} 2 \cos^2 \theta \rd \theta
}}
{{eqn | r = a b \int_{-\frac \pi 2}^{\frac \pi 2} \paren {1 + \map \cos {2 \theta} } \rd \theta
      | c = {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{eqn | r = a b \intlimits {\theta + \frac 1 2 \map \sin {2 \theta} } {-\frac \pi 2} {\frac \pi 2}
      | c = Definite Integral of Constant and Primitive of Cosine Function
}}
{{eqn | r = a b \paren {\frac {\pi} 2 + \frac 1 2 \map \sin {2 \cdot \frac {-\pi} 2} - \frac {-\pi} 2 - \frac 1 2 \map \sin {2 \cdot \frac \pi 2} }
}}
{{eqn | r = a b \paren {2 \cdot \frac \pi 2 + 2 \cdot \frac 1 2 \cdot 0}
}}
{{eqn | r = \pi a b
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Ellipse

Let $K$ be an ellipse whose major axis is of length $2 a$ and whose minor axis is of length $2 b$.
The area $\AA$ of $K$ is given by:
:$\AA = \pi a b$

Proof:
Let $K$ be an ellipse aligned in a cartesian plane in reduced form.
Then from Equation of Ellipse in Reduced Form:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
Thus:
:$y = \pm \dfrac b a \sqrt {a^2 - x^2}$
Consider a circle of radius $a$ whose center is at the origin.
From Equation of Circle center Origin, its equation is given by:
:$x^2 + y^2 = a^2$
and so:
:$y = \pm \sqrt {a^2 - x^2}$
The formulas show that each ordinate of the ellipse is $\dfrac b a$ the ordinate of the circle.
Since the same thing is true of the vertical chords:
{{begin-eqn}}
{{eqn | l = \AA_E
      | r = \dfrac b a \AA_C
      | c = where $\AA_E$ and $\AA_C$ are the areas of the ellipse and the circle respectively
}}
{{eqn | r = \dfrac b a \paren {\pi a^2}
      | c = Area of Circle
}}
{{eqn | r = \pi a b
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Equation of Ellipse in Reduced Form/Cartesian Frame

The equation of $K$ is:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$

Proof:
:500px
By definition, the foci $F_1$ and $F_2$ of $K$ are located at $\tuple {-c, 0}$ and $\tuple {c, 0}$ respectively.
Let the vertices of $K$ be $V_1$ and $V_2$.
By definition, these are located at $\tuple {-a, 0}$ and $\tuple {a, 0}$.
Let the covertices of $K$ be $C_1$ and $C_2$.
By definition, these are located at $\tuple {0, -b}$ and $\tuple {0, b}$.
Let $P = \tuple {x, y}$ be an arbitrary point on the locus of $K$.
From the equidistance property of $K$ we have that:
:$F_1 P + F_2 P = d$
where $d$ is a constant for this particular ellipse.
From Equidistance of Ellipse equals Major Axis:
:$(1): \quad d = 2 a$
Also, from Linear Eccentricity of Ellipse from Major and Minor Axis:
:$(2): \quad a^2 - c^2 = b^2$
Then:
{{begin-eqn}}
{{eqn | l = \sqrt {\paren {x - c}^2 + y^2} + \sqrt {\paren {x + c}^2 + y^2}
      | r = d
      | c = Pythagoras's Theorem
}}
{{eqn | r = 2 a
      | c = from $(1)$
}}
{{eqn | ll= \leadsto
      | l = \sqrt {\paren {x + c}^2 + y^2}
      | r = 2 a - \sqrt {\paren {x - c}^2 + y^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x + c}^2 + y^2
      | r = \paren {2 a - \sqrt {\paren {x - c}^2 + y^2} }^2
      | c = squaring both sides
}}
{{eqn | ll= \leadsto
      | l = x^2 + 2 c x + c^2 + y^2
      | r = 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2
      | c = Square of Difference
}}
{{eqn | ll= \leadsto
      | l = x^2 + 2 c x + c^2 + y^2
      | r = 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2
      | c = Square of Difference
}}
{{eqn | ll= \leadsto
      | l = a^2 - c x
      | r = a \sqrt {\paren {x - c}^2 + y^2}
      | c = gathering terms and simplifying
}}
{{eqn | ll= \leadsto
      | l = \paren {a^2 - c x}^2
      | r = a^2 \paren {\paren {x - c}^2 + y^2}
      | c = squaring both sides
}}
{{eqn | ll= \leadsto
      | l = c^2 x^2 - 2 c x a^2 + a^4
      | r = a^2 \paren {x^2 - 2 c x + c^2 + y^2}
      | c = Square of Difference
}}
{{eqn | r = a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2
      | c = Distributive Laws of Arithmetic
}}
{{eqn | ll= \leadsto
      | l = c^2 x^2 + a^4
      | r = a^2 x^2 + a^2 c^2 + a^2 y^2
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = a^4 - a^2 c^2
      | r = a^2 x^2 - c^2 x^2 + a^2 y^2
      | c = gathering terms
}}
{{eqn | ll= \leadsto
      | l = a^2 \paren {a^2 - c^2}
      | r = \paren {a^2 - c^2} x^2 + a^2 y^2
      | c = Distributive Laws of Arithmetic
}}
{{eqn | ll= \leadsto
      | l = a^2 b^2
      | r = b^2 x^2 + a^2 y^2
      | c = substituting $a^2 - c^2 = b^2$ from $(2)$
}}
{{eqn | ll= \leadsto
      | l = 1
      | r = \frac {x^2} {a^2} + \frac {y^2} {b^2}
      | c = dividing by $a^2 b^2$
}}
{{end-eqn}}
{{qed}}
Theorem: Equation of Ellipse in Reduced Form/Cartesian Frame

The equation of $K$ is:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$

Proof:
:500px
Let $P$ be an arbitrary point in the plane.
Let $PM$ be dropped perpendicular to $V_1 V_2$.
Hence $M = \tuple {x, 0}$.
From Intersecting Chord Theorem for Conic Sections:
:$PM^2 = k V_1 M \times M V_2$
for some constant $k$.
Hence:
{{begin-eqn}}
{{eqn | l = y^2
      | r = k \paren {a + x} \paren {a - x}
      | c = 
}}
{{eqn | r = k \paren {a^2 - x^2}
      | c = 
}}
{{end-eqn}}
Putting $x = 0$, we find the points where $K$ crosses the $y$-axis are defined by:
:$y^2 = k a^2$
This gives us:
:$k a^2 = b^2$
and so:
{{begin-eqn}}
{{eqn | l = y^2
      | r = \dfrac {b^2} {a^2} \paren {a^2 - x^2}
      | c = 
}}
{{eqn | r = b^2 - x^2 \dfrac {b^2} {a^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {y^2} {b^2}
      | r = 1 - \dfrac {x^2} {a^2}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Equation of Ellipse in Reduced Form/Cartesian Frame/Parametric Form

The equation of $K$ in parametric form is:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cos \theta
}}
{{eqn | l = y
      | r = b \sin \theta
}}
{{end-eqn}}
where $\theta$ is the eccentric angle of the point $P = \tuple {x, y}$ {{WRT}} $K$.

Proof:
Let the point $\tuple {x, y}$ satisfy the equations:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cos \theta
}}
{{eqn | l = y
      | r = b \sin \theta
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \frac {x^2} {a^2} + \frac {y^2} {b^2}
      | r = \frac {\paren {a \cos \theta}^2} {a^2} + \frac {\paren {b \sin \theta}^2} {b^2}
      | c = 
}}
{{eqn | r =  \frac {a^2} {a^2} \cos^2 \theta + \frac {b^2} {b^2} \sin^2 \theta
      | c = 
}}
{{eqn | r = \cos^2 \theta + \sin^2 \theta
      | c = 
}}
{{eqn | r = 1
      | c = Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
{{qed}}
Theorem: Perimeter of Ellipse

Let $K$ be an ellipse whose major axis is of length $2 a$ and whose minor axis is of length $2 b$.
The perimeter $\PP$ of $K$ is given by:
:$\PP = 4 a \map E e$
where:
:$\ds \map E e = \int_0^{\pi / 2} \sqrt{1 - e^2 \sin^2 \theta} \rd \theta$ is the complete elliptic integral of the second kind
:$e = \dfrac {\sqrt {a^2 - b^2} } a$ is the eccentricity of $K$.

Proof:
Let $K$ be aligned in a cartesian plane such that:
:the major axis of $K$ is aligned with the $x$-axis
:the minor axis of $K$ is aligned with the $y$-axis.
Then from Equation of Ellipse in Reduced Form: parametric form:
:$x = a \cos \theta, y = b \sin \theta$
Thus:
{{begin-eqn}}
{{eqn | l = \frac {\d x} {\d \theta}
      | r = -a \sin \theta
      | c = Derivative of Cosine Function
}}
{{eqn | l = \frac {\d y} {\d \theta}
      | r = b \cos \theta
      | c = Derivative of Sine Function
}}
{{end-eqn}}
From Arc Length for Parametric Equations, the length of one quarter of the perimeter of $K$ is given by:
{{begin-eqn}}
{{eqn | l = \frac {\PP} 4
      | r = \int_0^{\pi / 2} \sqrt {\paren {-a \sin \theta}^2 + \paren {b \cos \theta}^2} \rd \theta
      | c = 
}}
{{eqn | r = \int_0^{\pi / 2} \sqrt {a^2 \paren {1 - \cos^2 \theta} + b^2 \cos^2 \theta} \rd \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \int_0^{\pi / 2} \sqrt {a^2 - \paren {a^2 - b^2} \cos^2 \theta} \rd \theta
      | c = simplifying
}}
{{eqn | r = a \int_0^{\pi / 2} \sqrt {1 - \paren {1 - \frac {b^2} {a^2} } \cos^2 \theta} \rd \theta
      | c = extracting $a$ as a factor
}}
{{eqn | n = 1
      | r = a \int_0^{\pi / 2} \sqrt {1 - k^2 \cos^2 \theta} \rd \theta
      | c = setting $k^2 = 1 - \dfrac {b^2} {a^2} = \dfrac {a^2 - b^2} {a^2}$
}}
{{end-eqn}}
Since $\cos \theta = \map \sin {\dfrac \pi 2 - \theta}$ we can write for any real function $\map f x$:
:$\ds \int_0^{\pi / 2} \map f {\cos \theta} \rd \theta = \int_0^{\pi / 2} \map f {\map \sin {\frac \pi 2 - \theta} } \rd \theta$
So substituting $t = \dfrac \pi 2 - \theta$ this can be converted to:
{{begin-eqn}}
{{eqn | l = \int_0^{\pi / 2} \map f {\cos \theta} \rd \theta
      | r = -\int_{\pi / 2}^0 \map f {\sin t} \rd t
      | c = 
}}
{{eqn | r = \int_0^{\pi / 2} \map f {\sin t} \rd t
      | c = 
}}
{{end-eqn}}
justifying the fact that $\cos$ can be replaced with $\sin$ in $(1)$ above, giving:
:$\ds \PP = 4 a \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \theta} \rd \theta$
{{qed}}
Theorem: Rational Numbers under Multiplication form Commutative Monoid

The set of rational numbers under multiplication $\struct {\Q, \times}$ forms a countably infinite commutative monoid.

Proof:
From Rational Numbers under Multiplication form Monoid, $\struct {\Q, \times}$ is a monoid.
Then:
:from Rational Multiplication is Commutative we have that $\times$ is commutative on $\Q$
:from Rational Numbers are Countably Infinite we have that $\Q$ is a countably infinite set.
{{Qed}}
Theorem: Distance of Point from Origin in Cartesian Coordinates

Let $P = \tuple {x, y}$ be a point in the cartesian plane.
Then $P$ is at a distance of $\sqrt {x^2 + y^2}$ from the origin.

Proof:
:300px
By definition of the cartesian plane, the point $P$ is $x$ units from the $y$-axis and $y$ units from the $x$-axis.
The $y$-axis and $x$-axis are perpendicular to each other, also by definition.
Thus $x$, $y$ and $OP$ form a right-angled triangle.
By Pythagoras' Theorem:
:$OP^2 = x^2 + y^2$
Hence the result.
{{qed}}
Theorem: Composite of Monomorphisms is Monomorphism

Let:
:$\struct {S_1, \circ_1, \circ_2, \ldots, \circ_n}$
:$\struct {S_2, *_1, *_2, \ldots, *_n}$
:$\struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$
be algebraic structures.
Let:
:$\phi: \struct {S_1, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {S_2, *_1, *_2, \ldots, *_n}$
:$\psi: \struct {S_2, *_1, *_2, \ldots, *_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$
be monomorphisms.
Then the composite of $\phi$ and $\psi$ is also a monomorphism.

Proof:
From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism.
From Composite of Injections is Injection, $\psi \circ \phi$ is an injection.
A monomorphism is an injective homorphism.
Hence $\psi \circ \phi$ is a monomorphism.
{{qed}}