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Home » Can we have a non-path configuration with exactly 2 close pairs? - RTA

Can we have a non-path configuration with exactly 2 close pairs? - RTA

📅 May 10, 2026 👤 scraface

May 10, 2026

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📌 Since $ g''(x) = \frac{4}{x^3} > 0 $ for $ x > 0 $, this is a minimum. Thus, the minimum value of $ f(n) $ is $ \sqrt{2} + \frac{2}{\sqrt{2}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2} $.

📌 \boxed{2\sqrt{2}}

📌 Question: A paleobotanist is studying the symmetry of a fossilized flower with radial structure. If the flower has 7 equally spaced petals, and a vector $ \vec{v} $ from the center to a petal tip has components $ ( \cos \theta, \sin \theta ) $, and a second vector to an adjacent petal is $ \vec{w} = (\cos(\theta + \frac{2\pi}{7}), \sin(\theta + \frac{2\pi}{7})) $, find the angle between $ \vec{v} $ and $ \vec{w} $.

📌 Solution: The angle between two unit vectors $ \vec{v} $ and $ \vec{w} $ is given by the cosine of the difference of their angles:

📌 \cos \phi = \vec{v} \cdot \vec{w} = \cos \theta \cos\left(\theta + \frac{2\pi}{7}\right) + \sin \theta \sin\left(\theta + \frac{2\pi}{7}\right).

📌 Using the cosine of difference identity:

📌 \cos\left( \theta - \left(\theta + \frac{2\pi}{7} \right) \right) = \cos\left( -\frac{2\pi}{7} \right) = \cos\left( \frac{2\pi}{7} \right).

📌 Therefore, the angle between $ \vec{v} $ and $ \vec{w} $ is $ \frac{2\pi}{7} $.

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