Area of Triangle Inscribed in Parabola/Proof 1
Theorem
Let $T$ be the parabola which is the locus of points $\tuple {x, y}$ satisfying $y = x^2$.
Let $A$ and $B$ be two arbitrary points on $T$ with coordinates:
- $A = \tuple {u, u^2}$
- $B = \tuple {v, v^2}$, with $u > v$
Let $C$ be a third point on $T$ whose $x$-coordinate is the average of those of $A$ and $B$.
The area of $\triangle ABC$ is:
- $\map \AA {\triangle ABC} = \dfrac 1 8 \paren {u - v}^3$
Proof
The point $\tuple {x = 0, y = 0}$ satisfies $y = x^2$.
Since $x^2 \ge 0$, $0$ is the minimum value for $y$.
Thus, the vertex of $T$ lies at the origin.
The coordinates of $C$ are given as the average those for $A$ and $B$.
Given:
- $A = \tuple {u, u^2}$
- $B = \tuple {v, v^2}$, with $u > v$
By the definition of average:
- $C = \tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 4 \paren {u + v}^2 } $
Let $P$ be the midpoint of $AB$, such that $AP = BP$.
Let the coordinates of $P$ be $\tuple {P_x, P_y}$.
By the Midline Theorem:
- $P_x = \dfrac 1 2 \paren {u + v}$
Also by the Midline Theorem:
- $P_y = \dfrac 1 2 \paren {u^2 + v^2}$
Let $h$ be one-half the distance between the $x$-coordinates of $A$ and $B$:
- $h = \dfrac 1 2 \paren {u - v}$
By construction:
- the height of both $\triangle APC$ and $\triangle BPC$ is equal to $h$.
Let $b$ be the length of $PC$, the shared base of $\triangle APC$ and $\triangle BPC$.
| \(\ds b\) | \(=\) | \(\ds P_y - C_y\) | Definition above | |||||||||||
| \(\ds b\) | \(=\) | \(\ds \dfrac 1 2 \paren {u^2 + v^2} - \dfrac 1 4 \paren {u + v}^2\) | Substitution | |||||||||||
| \(\ds b\) | \(=\) | \(\ds \dfrac 1 4 \paren {2 u^2 + 2 v^2 - \paren {u + v}^2}\) | Factor out $\dfrac 1 2$ | |||||||||||
| \(\ds b\) | \(=\) | \(\ds \dfrac 1 4 \paren {u^2 - 2 u v + v^2}\) | Expansion and subtraction | |||||||||||
| \(\ds b\) | \(=\) | \(\ds \dfrac 1 4 \paren {u - v}^2\) | Factor |
In the words of Euclid:
(The Elements: Book $\text{I}$: Proposition $38$)
By Area of Triangle:
- $\map \AA {\triangle APC} = \dfrac 1 2 h b$
Substituting from above:
- $\map \AA {\triangle APC} = \dfrac 1 {16} \paren {u - v}^3$
Since they have the same height and a shared base:
- $\map \AA {\triangle APC} = \map \AA {\triangle BPC}$
But:
- $\map \AA {\triangle ABC} = \map \AA {\triangle APC} + \map \AA {\triangle BPC}$
By addition:
- $\map \AA {\triangle ABC} = h b = \dfrac 1 8 \paren {u - v}^3$
$\blacksquare$