Area of Triangle Inscribed in Parabola

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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 1

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:

Triangles which are on equal bases and in the same parallels are equal to one another.

(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:

$\map \AA {\triangle ABC} = h b = \dfrac 1 8 \paren {u - v}^3$

$\blacksquare$


Proof 2

From Equation of Straight Line in Plane through Two Points, the straight line $AB$ can be expressed as:

\(\ds \dfrac {\paren {y_{AB} - y_1} } {\paren {x - x_1} }\) \(=\) \(\ds \dfrac {\paren {y_2 - y_1} } {\paren {x_2 - x_1} }\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\paren {y_{AB} - v^2} } {\paren {x - v} }\) \(=\) \(\ds \dfrac {\paren {u^2 - v^2} } {\paren {u - v} }\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {u + v} \paren {u - v} } {\paren {u - v} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \paren {u + v}\) dividing top and bottom by $\paren {u - v}$
\(\ds \leadsto \ \ \) \(\ds \paren {y_{AB} - v^2}\) \(=\) \(\ds \paren {u + v} \paren {x - v}\) multiplying both sides by $\paren {x - v}$
\(\ds \) \(=\) \(\ds \paren {u + v} x - u v - v^2\)
\(\ds \leadsto \ \ \) \(\ds y_{AB}\) \(=\) \(\ds \paren {u + v} x - u v\)

The coordinates of $C$ are:

$\tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 4 \paren {u + v}^2 } $

Let $P$ be the midpoint of $AB$, such that $AP = BP$.

We can use Equation of Straight Line in Plane through Two Points above to determine the coordinates of $P$:

\(\ds y_{AB}\) \(=\) \(\ds \paren {u + v} x - u v\) from above
\(\ds \) \(=\) \(\ds \paren {u + v} \paren {\dfrac 1 2 \paren {u + v} } - u v\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \dfrac 1 2 \paren {u^2 + v^2}\)

Therefore, the coordinates of $P$ are:

$\tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 2 \paren {u^2 + v^2} } $
Inscribed Triangle.JPG

We now connect points $P$ and $C$ to create line segment $PC$

This in turn creates two triangles.

We see from the diagram above:

$\map \AA {\triangle ABC} = \map \AA {\triangle APC} + \map \AA {\triangle CPB}$

From Area of Triangle in Terms of Two Sides and Angle, we have:

$\map \AA {\triangle APC} = \dfrac 1 2 \paren {AP} \paren {PC} \map \sin {\angle APC} $

We now notice that:

\(\ds \paren {AP} \map \sin {\angle APC}\) \(=\) \(\ds \dfrac {\paren {u + v} } 2 - v\) from above
\(\ds \) \(=\) \(\ds \dfrac {\paren {u - v} } 2\)

Therefore:

$\map \AA {\triangle APC} = \dfrac 1 2 \paren {PC} \dfrac {\paren {u - v} } 2$

We also notice:

\(\ds \map \sin {\angle CPB}\) \(=\) \(\ds \map \sin {\pi - \angle APC}\) $\angle CPB$ is supplementary to $\angle APC$
\(\ds \) \(=\) \(\ds \map \sin {\angle APC}\) Sine of Supplementary Angle

Since $AP = PB$, $PC = PC$ and $\map \sin {\angle CPB} = \map \sin {\angle APC}$, then by Area of Triangle in Terms of Two Sides and Angle we have:

$\map \AA {\triangle APC} = \map \AA {\triangle CPB}$

Therefore:

\(\ds \map \AA {\triangle ABC}\) \(=\) \(\ds \map \AA {\triangle APC} + \map \AA {\triangle CPB}\)
\(\ds \) \(=\) \(\ds 2 \map \AA {\triangle APC}\) since $\map \AA {\triangle APC} = \map \AA {\triangle CPB}$
\(\ds \) \(=\) \(\ds 2 \dfrac 1 2 \paren {PC} \dfrac {\paren {u - v} } 2\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {u - v} P C\)


All that remains is to determine the length of $CP$:

\(\ds CP\) \(=\) \(\ds P_y - C_y\) Definition of Length of Line
\(\ds CP\) \(=\) \(\ds \dfrac 1 2 \paren {u^2 + v^2} - \dfrac 1 4 \paren {u + v}^2\) substitution
\(\ds CP\) \(=\) \(\ds \dfrac 1 4 \paren {2 u^2 + 2 v^2 - \paren {u + v}^2}\) factor out $\dfrac 1 2$
\(\ds CP\) \(=\) \(\ds \dfrac 1 4 \paren {u^2 - 2 u v + v^2}\) expansion and subtraction
\(\ds CP\) \(=\) \(\ds \dfrac 1 4 \paren {u - v}^2\) factor

Substituting from above:

$\map \AA {\triangle ABC} = \dfrac 1 2 \paren {u - v} \dfrac 1 4 \paren {u - v}^2$

Therefore:

$\map \AA {\triangle ABC} = \dfrac 1 8 \paren {u - v}^3$

$\blacksquare$

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