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Home » Henry the Stickmins INSANE Struggle That Had You Clicking Nonstop—Heres How! - inBeat

Henry the Stickmins INSANE Struggle That Had You Clicking Nonstop—Heres How! - inBeat

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 y = \frac{-42 \pm \sqrt{1764 - 1512}}{54} = \frac{-42 \pm \sqrt{252}}{54} = \frac{-42 \pm 6\sqrt{7}}{54} = \frac{-7 \pm \sqrt{7}}{9}.

📌 We want positive ratio. Try $ y = \frac{-7 + \sqrt{7}}{9} \approx \frac{-7 + 2.6458}{9} = \frac{-4.3542}{9} \approx -0.484 $, negative. Other root: $ \frac{-7 - 2.6458}{9} < 0 $. Both negative — meaning $ A/d < 0 $, so first and last have opposite signs — impossible for $ A^2 + (A+3d)^2 = (sum)^2 $ unless not ordered.

📌 Wait — the original assumption about ordering may be wrong. Try symmetric AP: let the four terms be $ a - 3d, a - d, a + d, a + 3d $ — symmetric around $ a $. This is a valid arithmetic progression with common difference $ 2d $, but we can scale. Define as AP with common difference $ 2d' $, but to match convention, let common difference be $ 2d $, so terms: $ a - 3d, a - d, a + d, a + 3d $. Then first: $ a - 3d $, last: $ a + 3d $, sum of squares: $ (a-3d)^2 + (a+3d)^2 = 2a^2 + 18d^2 $. Sum of all: $ 4a $, square: $ 16a^2 $. Set equal:

📌 2a^2 + 18d^2 = 16a^2 \Rightarrow 18d^2 = 14a^2 \Rightarrow \frac{d^2}{a^2} = \frac{14}{18} = \frac{7}{9}.

📌 Third term: $ a + d $, first: $ a - 3d $. Ratio:

📌 \frac{a + d}{a - 3d} = \frac{1 + \frac{d}{a}}{1 - 3\frac{d}{a}}.

📌 Let $ x = \frac{d}{a} $, $ x^2 = \frac{7}{9} \Rightarrow x = \pm \frac{\sqrt{7}}{3} $. Try $ x = \frac{\sqrt{7}}{3} $:

📌 \frac{1 + \frac{\sqrt{7}}{3}}{1 - \sqrt{7}} = \frac{\frac{3 + \sqrt{7}}{3}}{1 - \sqrt{7}} = \frac{3 + \sqrt{7}}{3(1 - \sqrt{7})}.

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