The Golden Ratio
By definition, the Golden Ratio is a ratio involving overlapping line segments. Given collinear points A, B, and C, such that B is between A and C, if the ratio between the two subsegments is the same as the ratio between the entire segment and the longer segment, then that ratio is \(\frac{1 + \sqrt{5}}{2} \approx 1.618\). For ease of reference, this constant is called \(\phi\) (“phi”).
The Golden Ratio is said to appear in all manner of artworks and constructions, although others have claimed that many of its appearances are exaggerated. It has also been found throughout nature. It is fundamental to the construction of the pentagram. For a detailed treatment of this constant, I recommend Mario Livio’s The Golden Ratio.
The Golden Ratio can fairly easily be calculated. Let \(AB = 1\) and \(BC = x\). Since we’re dealing with ratios, our answer can be generalized across all lengths of \(\overline{AB}\). Since \(AC = AB + BC \), \(AC = x + 1\). This gives us: \[\frac{BC}{AB} = \frac{AC}{BC} \Rightarrow \frac{x}{1} = \frac{x + 1}{x} \]
We then cross-multiply: \[x \cdot x = x + 1 \Rightarrow x^2 – x – 1 = 0 \]
Use the Quadratic Formula to find the roots: \[\frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\ \frac{-(-1) \pm \sqrt{(-1)^2 – 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2} \]
This has a positive root and a negative root, but since \(x\) is a segment measurement, it has to be positive, hence \(\frac{1 + \sqrt{5}}{2} \).
The Power of a Point
Consider this problem: Draw a tangent line and a secant line from an external point to a circle such that the chord and the tangent segment are the same length. What is the ratio of the chord to the external segment of the secant line?
Said another way, with regards to the figure shown, if \(AD\) is tangent to \(\odot P\) and \(\overline{BC} \cong \overline{AD}\), then calculate \(\frac{BC}{AB}\).
Let’s call \(AB = 1\) and \(BC = x\). As above, because we’re finding the ratio between these lengths, we can generalize our results. The Power of a Point Theorem, as it applies to tangents and secants, says: \[AB \cdot AC = AD^2 \]
Furthermore, we know that \(AC = AB + BC = 1 + x\), which means that \(1 \cdot (1 + x) = x^2 \Rightarrow x^2 – x – 1 = 0\).
This might look familiar: It’s the same equation we have above for the Golden Ratio. The ratio of the chord to the external secant segment (or, if you prefer, the ratio of the tangent segment to the external secant segment) is ϕ.
Note that, since the collinear segments are in a golden ratio relationship, the ratio of the entire secant segment (AC) to the chord segment (BC) is likewise ϕ.
For me, this is an unexpected appearance of ϕ. This is the beauty of mathematics: Finding familiar things in unfamiliar places.