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Home » Kentlands Conspiracy: Was This Hidden Gem Disguised as a Ghost Town?" - Network4

Kentlands Conspiracy: Was This Hidden Gem Disguised as a Ghost Town?" - Network4

📅 May 19, 2026 👤 scraface

May 19, 2026

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📌 x^2 + y^2 = c^2 \sin^4\phi, \quad z = c \sin\phi \cos\phi

📌 Let \(r = \sqrt{x^2 + y^2} = c \sin^2\phi\)

📌 But from \(z = c \sin\phi \cos\phi = c \sin\phi \sqrt{1 - \sin^2\phi}\)

📌 Set \(u = \sin\phi\), then:

📌 r = c u^2, \quad z = c u \sqrt{1 - u^2}

📌 u = rac{r^{1/2}}{\sqrt{c}}, \quad z = c \cdot rac{r^{1/2}}{\sqrt{c}} \cdot \sqrt{1 - rac{r^2}{c^2}} = \sqrt{c} \cdot r^{1/2} \sqrt{1 - rac{r^2}{c^2}}

📌 z^2 = c \cdot r \left(1 - rac{r^2}{c^2}

📌 ight) = c r - rac{r^3}{c}

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Only way is if the added value pushes the middle up, but 18 is large.
Perhaps original has values above 16.
The a5 in a 9-list needs to be at least 16 to have average 16 with a6.
Since sum a5+a6=32, minimum a5 is 16 (if a6=16).
So if a5=16, a6=16, then both are 16.
Possible: 14,14,15,16,15,16,16,16 — too messy.
But if original has a value >16, say 20, but median still 15.
But the problem likely assumes real numbers or allows generality.
Given the complexity and olympiad style, likely the intended answer is...
values greater than the previous median (15) but less than the new med...

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