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:
Proof
Let the points be labelled to simplify:
- $A := \tuple {1, 4}$
- $B := \tuple {2, 8}$
- $C := \tuple {4, 2}$
- $D := \tuple {8, 5}$
- $E := \tuple {7, 1}$
- $F := \tuple {5, 7}$
Let $ABCDEF$ be considered as a hexagon.
We join the opposite points of $ABCDEF$:
- $AF: \tuple {1, 4} \to \tuple {5, 7}$
- $BC: \tuple {2, 8} \to \tuple {4, 2}$
- $BE: \tuple {2, 8} \to \tuple {7, 1}$
- $AD: \tuple {1, 4} \to \tuple {8, 5}$
- $CD: \tuple {4, 2} \to \tuple {8, 5}$
- $EF: \tuple {7, 1} \to \tuple {5, 7}$
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 in Plane through Two Points:
- $\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$
Thus:
| \(\text {(AF)}: \quad\) | \(\ds \frac {y - 4} {x - 1}\) | \(=\) | \(\ds \frac {7 - 4} {5 - 1}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 3 4\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 4 \paren {y - 4}\) | \(=\) | \(\ds 3 \paren {x - 1}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac 3 4 x + \dfrac {13} 4\) |
| \(\text {(BC)}: \quad\) | \(\ds \frac {y - 8} {x - 2}\) | \(=\) | \(\ds \frac {2 - 8} {4 - 2}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds -3\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y - 8\) | \(=\) | \(\ds -3 \paren {x - 2}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds -3 x + 14\) |
| \(\text {(BE)}: \quad\) | \(\ds \frac {y - 8} {x - 2}\) | \(=\) | \(\ds \frac {1 - 8} {7 - 2}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac 7 5\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 5 \paren {y - 8}\) | \(=\) | \(\ds -7 \paren {x - 2}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 5 y - 40\) | \(=\) | \(\ds -7 x + 14\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds -\frac 7 5 x + \frac {54} 5\) |
| \(\text {(AD)}: \quad\) | \(\ds \frac {y - 4} {x - 1}\) | \(=\) | \(\ds \frac {5 - 4} {8 - 1}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 7\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 7 \paren {y - 4}\) | \(=\) | \(\ds x - 1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 7 y - 28\) | \(=\) | \(\ds x - 1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac x 7 + \frac {27} 7\) |
| \(\text {(CD)}: \quad\) | \(\ds \frac {y - 2} {x - 4}\) | \(=\) | \(\ds \frac {5 - 2} {8 - 4}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 3 4\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 4 \paren {y - 2}\) | \(=\) | \(\ds 3 \paren {x - 4}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 4 y - 8\) | \(=\) | \(\ds 3 x - 12\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac 3 4 x - 1\) |
| \(\text {(EF)}: \quad\) | \(\ds \frac {y - 1} {x - 7}\) | \(=\) | \(\ds \frac {7 - 1} {5 - 7}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds -3\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y - 1\) | \(=\) | \(\ds -3 x + 21\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds -3 x + 22\) |
Evaluate the intersection of $AF$ and $BC$:
| \(\ds y\) | \(=\) | \(\ds \dfrac 3 4 x + \dfrac {13} 4\) | ||||||||||||
| \(\ds y\) | \(=\) | \(\ds -3 x + 14\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac 3 4 x + \dfrac {13} 4\) | \(=\) | \(\ds -3 x + 14\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 3 x + 13\) | \(=\) | \(\ds -12 x + 56\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 15 x\) | \(=\) | \(\ds 43\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \dfrac {43} {15}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds -3 \paren {\dfrac {43} {15} } + 14\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {-129 + 210} {15}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {81} {15}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {27} 5\) |
So $AF$ and $BC$ intersect at $\paren {\dfrac {43} {15}, \dfrac {27} 5}$.
Evaluate the intersection of $BE$ and $AD$:
| \(\ds y\) | \(=\) | \(\ds -\frac 7 5 x + \frac {54} 5\) | ||||||||||||
| \(\ds y\) | \(=\) | \(\ds \frac x 7 + \frac {27} 7\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -\frac 7 5 x + \frac {54} 5\) | \(=\) | \(\ds \frac x 7 + \frac {27} 7\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -49 x + 378\) | \(=\) | \(\ds 5 x + 135\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 54 x\) | \(=\) | \(\ds 243\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \dfrac 9 2\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac 1 7 \paren {\frac 9 2} + \frac {27} 7\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {9 + 54} {14}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {63} {14}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac 9 2\) |
So $BE$ and $AD$ intersect at $\paren {\dfrac 9 2, \dfrac 9 2}$.
Evaluate the intersection of $CD$ and $EF$:
| \(\ds y\) | \(=\) | \(\ds \frac 3 4 x - 1\) | ||||||||||||
| \(\ds y\) | \(=\) | \(\ds -3 x + 22\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 3 4 x - 1\) | \(=\) | \(\ds -3 x + 22\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 3 x - 4\) | \(=\) | \(\ds -12 x + 88\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 15 x\) | \(=\) | \(\ds 92\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \dfrac {92} {15}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds -3 \paren {\dfrac {92} {15} } + 22\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {-92 + 110} 5\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {18} 5\) |
So $CD$ and $EF$ intersect at $\paren {\dfrac {92} {15}, \dfrac {18} 5}$.
It remains to be shown that those points of intersection:
- $\paren {\dfrac {43} {15}, \dfrac {27} 5}$, $\paren {\dfrac 9 2, \dfrac 9 2}$, $\paren {\dfrac {92} {15}, \dfrac {18} 5}$
all lie on the same straight line.
From Equation of Straight Line in Plane through Two Points:
- $\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$
Thus:
| \(\ds \frac {y - \frac {27} 5} {x - \frac {43} {15} }\) | \(=\) | \(\ds \frac {\frac {18} 5 - \frac {27} 5} {\frac {92} {15} - \frac {43} {15} }\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {5 y - 27} {15 x - 43}\) | \(=\) | \(\ds \frac {18 - 27} {92 - 43}\) | simplifying | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {5 y - 27} {15 x - 43}\) | \(=\) | \(\ds -\frac 9 {49}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 245 y - 1323\) | \(=\) | \(\ds - 135 x + 387\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 245 y\) | \(=\) | \(\ds - 135 x + 1710\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 49 y + 27 x\) | \(=\) | \(\ds 342\) |
It remains to demonstrate that $\paren {\dfrac 9 2, \dfrac 9 2}$ lies on this line:
| \(\ds 49 \dfrac 9 2 + 27 \dfrac 9 2\) | \(=\) | \(\ds \dfrac {441 + 243} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {684} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 342\) |
Bingo.
$\blacksquare$
Sources
- Oct. 1987: Edward Kitchen: Problem 1248: A Curious Property of 1/7: Solution (Math. Mag. Vol. 60, no. 4: p. 245) www.jstor.org/stable/2689350
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $142,857$