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

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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:

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


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


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


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


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


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


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.


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:

\(\displaystyle \frac {y - \frac {27} 5} {x - \frac {43} {15} }\) \(=\) \(\displaystyle \frac {\frac {18} 5 - \frac {27} 5} {\frac {92} {15} - \frac {43} {15} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {5 y - 27} {15 x - 43}\) \(=\) \(\displaystyle \frac {18 - 27} {92 - 43}\) 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.

$\blacksquare$


Sources