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Home » \boxed{-\frac{\sqrt2}2} - Imagemakers

\boxed{-\frac{\sqrt2}2} - Imagemakers

📅 April 29, 2026 👤 scraface

Apr 29, 2026

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📌 Question:** A cultural anthropologist observes a ceremonial circle divided into 12 equal segments, each representing a lunar phase. If a participant starts at segment 1 and jumps \( n \) segments clockwise each day, where \( n \equiv 7 \pmod{12} \), after how many days will the participant return to segment 1?

📌 The participant jumps \( n = 7 \) segments clockwise each day modulo 12. We seek the smallest positive integer \( d \) such that:

📌 7d \equiv 0 \pmod{12}

📌 This means \( 12 \mid 7d \). Since \( \gcd(7, 12) = 1 \), 12 must divide \( d \). Thus, the smallest such \( d \) is:

📌 After 12 days, the total displacement is \( 7 \times 12 = 84 \equiv 0 \pmod{12} \), returning to the starting segment.

📌 Therefore, the participant returns to segment 1 after:

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S = rac{2(1 + (-1))}{1 - (-1)} = 0.
But earlier steps suggest $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $. For $...
S = rac{2(e^{2i heta} + e^{2i\phi})}{e^{2i heta} - e^{2i\phi}} = 2 \c...
Multiply numerator and denominator by $ e^{-i( heta + \phi)/2} $:
S = 2i \cdot rac{e^{i( heta - \phi)/2} + e^{-i( heta - \phi)/2}}{e^{i...
ight)}{2i\sin\left(rac{ heta - \phi}{2}
ight)} = -2 \cot\left(rac{ heta - \phi}{2}
eq b $ implies $ heta
eq \phi $, so $ S $ is purely imaginary. However, the original questio...
Let $ S = rac{(a + b)^2 + (a - b)^2}{a^2 - b^2} = rac{2a^2 + 2b^2}{a...

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