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Home » A vector \(\mathbfv = eginpmatrix x \ y \endpmatrix\) is invariant under \(R\) if: - go-checkin.com

A vector \(\mathbfv = eginpmatrix x \ y \endpmatrix\) is invariant under \(R\) if: - go-checkin.com

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Solution: A regular 12-sided polygon (dodecagon) has rotational symmetry of order 12. The full set of rotational symmetries consists of rotations by multiples of $ \frac{360^\circ}{12} = 30^\circ $. These rotations are: $ 0^\circ, 30^\circ, 60^\circ, \dots, 330^\circ $. This forms an arithmetic sequence with first term $ 0^\circ $, common difference $ 30^\circ $, and 12 terms. The sum of an arithmetic sequence is given by:

📌 S = \frac{n}{2}(a + l) = \frac{12}{2}(0^\circ + 330^\circ) = 6 \cdot 330^\circ = 1980^\circ

📌 Thus, the sum of all distinct rotational symmetries is $ \boxed{1980^\circ} $.

📌 Question: A middle school student is building a robot with gears. One gear rotates 48 times per minute, and another rotates 72 times per minute. If both gears start aligned, after how many minutes will they align again?

📌 Solution: To find when the gears align again, we compute the least common multiple (LCM) of their rotation periods. Since they rotate at 48 and 72 rpm (rotations per minute), the time until alignment is the time it takes for each to complete a whole number of rotations such that both return to start simultaneously. This is equivalent to the LCM of the number of rotations per minute in terms of cycle time. First, find the LCM of the rotation counts over time or convert to cycle periods: The time for one rotation is $ \frac{1}{48} $ minutes and $ \frac{1}{72} $ minutes. So we find $ \mathrm{LCM}\left(\frac{1}{48}, \frac{1}{72}\right) = \frac{1}{\mathrm{GCD}(48, 72)} $. Compute $ \mathrm{GCD}(48, 72) $:

📌 Prime factorization: $ 48 = 2^4 \cdot 3 $, $ 72 = 2^3 \cdot 3^2 $, so $ \mathrm{GCD} = 2^3 \cdot 3 = 24 $.

📌 Thus, the LCM of the periods is $ \frac{1}{24} $ minutes? No — correct interpretation: The time until alignment is the least $ t $ such that $ 48t $ and $ 72t $ are both integers and the angular positions coincide. Actually, the alignment occurs at $ t $ where $ 48t \equiv 0 \pmod{360} $ and $ 72t \equiv 0 \pmod{360} $ in degrees per rotation. Since each full rotation is 360°, we want smallest $ t $ such that $ 48t \cdot \frac{360}{360} = 48t $ is multiple of 360 and same for 72? No — better: The number of rotations completed must be integer, and the alignment occurs when both complete a number of rotations differing by full cycles. The time until both complete whole rotations and are aligned again is $ \frac{360}{\mathrm{GCD}(48, 72)} $ minutes? No — correct formula: For two periodic events with periods $ T_1, T_2 $, time until alignment is $ \mathrm{LCM}(T_1, T_2) $, where $ T_1 = 1/48 $, $ T_2 = 1/72 $. But in terms of complete rotations: Let $ t $ be time. Then $ 48t $ rows per minute — better: Let angular speed be $ 48 \cdot \frac{360}{60} = 288^\circ/\text{sec} $? No — $ 48 $ rpm means 48 full rotations per minute → period per rotation: $ \frac{60}{48} = \frac{5}{4} = 1.25 $ seconds. Similarly, 72 rpm → period $ \frac{5}{12} $ minutes = 25 seconds. Find LCM of 1.25 and 25/12. Write as fractions: $ 1.25 = \frac{5}{4} $, $ \frac{25}{12} $. LCM of fractions: $ \mathrm{LCM}(\frac{a}{b}, \frac{c}{d}) = \frac{\mathrm{LCM}(a, c)}{\mathrm{GCD}(b, d)} $? No — standard: $ \mathrm{LCM}(\frac{m}{n}, \frac{p}{q}) = \frac{\mathrm{LCM}(m, p)}{\mathrm{GCD}(n, q)} $ only in specific cases. Better: time until alignment is $ \frac{\mathrm{LCM}(48, 72)}{48 \cdot 72 / \mathrm{GCD}(48,72)} $? No.

📌 Correct approach: The gear with 48 rotations/min makes a rotation every $ \frac{1}{48} $ minutes. The other every $ \frac{1}{72} $ minutes. They align when both complete integer numbers of rotations and the total time is the same. So $ t $ must satisfy $ t = 48 a = 72 b $ for integers $ a, b $. So $ t = \mathrm{LCM}(48, 72) $.

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