Points Defined by Adjacent Pairs of Digits of Reciprocal of 7 lie on Ellipse

Theorem
Consider the digits that form the recurring part of the reciprocal of $7$:
 * $\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$

Take the digits in ordered pairs, and treat them as coordinates of a Cartesian plane.

It will be found that they all lie on an ellipse:


 * EllipseFromSeventh.png

Proof

 * EllipseFromSeventhSolution.png

Let the points be labelled to simplify:
 * $A := \left({1, 4}\right)$
 * $B := \left({2, 8}\right)$
 * $C := \left({4, 2}\right)$
 * $D := \left({8, 5}\right)$
 * $E := \left({7, 1}\right)$
 * $F := \left({5, 7}\right)$

Let $ABCDEF$ be considered as a hexagon.

We join the opposite points of $ABCDEF$:
 * $AF: \left({1, 4}\right) \to \left({5, 7}\right)$
 * $BC: \left({2, 8}\right) \to \left({4, 2}\right)$
 * $BE: \left({2, 8}\right) \to \left({7, 1}\right)$
 * $AD: \left({1, 4}\right) \to \left({8, 5}\right)$
 * $CD: \left({4, 2}\right) \to \left({8, 5}\right)$
 * $EF: \left({7, 1}\right) \to \left({5, 7}\right)$

It is to be shown that the intersections of:
 * $AF$ and $BC$
 * $BE$ and $AD$
 * $CD$ and $EF$

all lie on the same straight line.

The result then follows from Pascal's Mystic Hexagram.

From Equation of Straight Line through Two Points:


 * $\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$

Thus:

Evaluate the intersection of $AF$ and $BC$:

So $AF$ and $BC$ intersect at $\left({\dfrac {43} {15}, \dfrac {27} 5}\right)$.

Evaluate the intersection of $BE$ and $AD$:

So $BE$ and $AD$ intersect at $\left({\dfrac 9 2, \dfrac 9 2}\right)$.

Evaluate the intersection of $CD$ and $EF$:

So $CD$ and $EF$ intersect at $\left({\dfrac {92} {15}, \dfrac {18} 5}\right)$.

It remains to be shown that those points of intersection:
 * $\left({\dfrac {43} {15}, \dfrac {27} 5}\right)$, $\left({\dfrac 9 2, \dfrac 9 2}\right)$, $\left({\dfrac {92} {15}, \dfrac {18} 5}\right)$

all lie on the same straight line.

From Equation of Straight Line through Two Points:


 * $\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$

Thus:

It remains to demonstrate that $\left({\dfrac 9 2, \dfrac 9 2}\right)$ lies on this line:

Bingo.