# 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:

## Proof

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.

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

Thus:

 $\text {(AF)}: \quad$ $\displaystyle \frac {y - 4} {x - 1}$ $=$ $\displaystyle \frac {7 - 4} {5 - 1}$ $\displaystyle$ $=$ $\displaystyle \frac 3 4$ $\displaystyle \leadsto \ \$ $\displaystyle 4 \left({y - 4}\right)$ $=$ $\displaystyle 3 \left({x - 1}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac 3 4 x + \dfrac {13} 4$

 $\text {(BC)}: \quad$ $\displaystyle \frac {y - 8} {x - 2}$ $=$ $\displaystyle \frac {2 - 8} {4 - 2}$ $\displaystyle$ $=$ $\displaystyle -3$ $\displaystyle \leadsto \ \$ $\displaystyle y - 8$ $=$ $\displaystyle -3 \left({x - 2}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -3 x + 14$

 $\text {(BE)}: \quad$ $\displaystyle \frac {y - 8} {x - 2}$ $=$ $\displaystyle \frac {1 - 8} {7 - 2}$ $\displaystyle$ $=$ $\displaystyle -\frac 7 5$ $\displaystyle \leadsto \ \$ $\displaystyle 5 \left({y - 8}\right)$ $=$ $\displaystyle -7 \left({x - 2}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle 5 y - 40$ $=$ $\displaystyle -7 x + 14$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -\frac 7 5 x + \frac {54} 5$

 $\text {(AD)}: \quad$ $\displaystyle \frac {y - 4} {x - 1}$ $=$ $\displaystyle \frac {5 - 4} {8 - 1}$ $\displaystyle$ $=$ $\displaystyle \frac 1 7$ $\displaystyle \leadsto \ \$ $\displaystyle 7 \left({y - 4}\right)$ $=$ $\displaystyle x - 1$ $\displaystyle \leadsto \ \$ $\displaystyle 7 y - 28$ $=$ $\displaystyle x - 1$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \frac x 7 + \frac {27} 7$

 $\text {(CD)}: \quad$ $\displaystyle \frac {y - 2} {x - 4}$ $=$ $\displaystyle \frac {5 - 2} {8 - 4}$ $\displaystyle$ $=$ $\displaystyle \frac 3 4$ $\displaystyle \leadsto \ \$ $\displaystyle 4 \left({y - 2}\right)$ $=$ $\displaystyle 3 \left({x - 4}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle 4 y - 8$ $=$ $\displaystyle 3 x - 12$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \frac 3 4 x - 1$

 $\text {(EF)}: \quad$ $\displaystyle \frac {y - 1} {x - 7}$ $=$ $\displaystyle \frac {7 - 1} {5 - 7}$ $\displaystyle$ $=$ $\displaystyle -3$ $\displaystyle \leadsto \ \$ $\displaystyle y - 1$ $=$ $\displaystyle -3 x + 21$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -3 x + 22$

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

 $\displaystyle y$ $=$ $\displaystyle \dfrac 3 4 x + \dfrac {13} 4$ $\displaystyle y$ $=$ $\displaystyle -3 x + 14$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac 3 4 x + \dfrac {13} 4$ $=$ $\displaystyle -3 x + 14$ $\displaystyle \leadsto \ \$ $\displaystyle 3 x + 13$ $=$ $\displaystyle -12 x + 56$ $\displaystyle \leadsto \ \$ $\displaystyle 15 x$ $=$ $\displaystyle 43$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \dfrac {43} {15}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -3 \left({\dfrac {43} {15} }\right) + 14$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {-129 + 210} {15}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {81} {15}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {27} 5$

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

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

 $\displaystyle y$ $=$ $\displaystyle -\frac 7 5 x + \frac {54} 5$ $\displaystyle y$ $=$ $\displaystyle \frac x 7 + \frac {27} 7$ $\displaystyle \leadsto \ \$ $\displaystyle -\frac 7 5 x + \frac {54} 5$ $=$ $\displaystyle \frac x 7 + \frac {27} 7$ $\displaystyle \leadsto \ \$ $\displaystyle -49 x + 378$ $=$ $\displaystyle 5 x + 135$ $\displaystyle \leadsto \ \$ $\displaystyle 54 x$ $=$ $\displaystyle 243$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \dfrac 9 2$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \frac 1 7 \left({\frac 9 2}\right) + \frac {27} 7$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {9 + 54} {14}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {63} {14}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac 9 2$

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

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

 $\displaystyle y$ $=$ $\displaystyle \frac 3 4 x - 1$ $\displaystyle y$ $=$ $\displaystyle -3 x + 22$ $\displaystyle \leadsto \ \$ $\displaystyle \frac 3 4 x - 1$ $=$ $\displaystyle -3 x + 22$ $\displaystyle \leadsto \ \$ $\displaystyle \frac 3 x - 4$ $=$ $\displaystyle -12 x + 88$ $\displaystyle \leadsto \ \$ $\displaystyle 15 x$ $=$ $\displaystyle 92$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \dfrac {92} {15}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle -3 \left({\dfrac {92} {15} }\right) + 22$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {-92 + 110} 5$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {18} 5$

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.

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

Thus:

 $\displaystyle \frac {y - \frac {27} 5} {x - \frac {43} {15} }$ $=$ $\displaystyle \frac {\frac {18} 5 - \frac {27} 5} {\frac {92} {15} - \frac {43} {15} }$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {5 y - 27} {15 x - 43}$ $=$ $\displaystyle \frac {18 - 27} {92 - 43}$ simplifying $\displaystyle \leadsto \ \$ $\displaystyle \frac {5 y - 27} {15 x - 43}$ $=$ $\displaystyle -\frac 9 {49}$ $\displaystyle \leadsto \ \$ $\displaystyle 245 y - 1323$ $=$ $\displaystyle - 135 x + 387$ $\displaystyle \leadsto \ \$ $\displaystyle 245 y$ $=$ $\displaystyle - 135 x + 1710$ $\displaystyle \leadsto \ \$ $\displaystyle 49 y + 27 x$ $=$ $\displaystyle 342$

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

 $\displaystyle 49 \dfrac 9 2 + 27 \dfrac 9 2$ $=$ $\displaystyle \dfrac {441 + 243} 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {684} 2$ $\displaystyle$ $=$ $\displaystyle 342$

Bingo.

$\blacksquare$