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--- math54-f22:s:bryanli [2022/08/26 23:32]
alphyte [Homework]
+++ math54-f22:s:bryanli [2026/02/21 14:41] (current)
@@ Line 20: / Line 20: @@
         <li> $3\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} + \overrightarrow{OB} + \overrightarrow{BM} + \overrightarrow{OC} + \overrightarrow{CM} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$ by Problem 5.
         <li> $\overrightarrow{AM} = \frac{2}{3}\overrightarrow{AA'}$ with similar formulas for the other vertices by geometry. By Problem 7, we can then write $\overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC}$ as $-\frac{1}{3}(\overrightarrow{AB} + \overrightarrow{AC}) + \frac{1}{3}(\overrightarrow{BA} + \overrightarrow{BC}) + \frac{1}{3}(\overrightarrow{CA} + \overrightarrow{CB})$ which clearly sums to $0$.
-        <li>
+        <li> If $z_1, z_2, z_3$ are the centers of the vertices' circles, then $v_i = z_i + c_ie^{i(\omega t + \phi_1)} = z_i + k_ie^{i\omega t}$, then the barycenter follows the motion $M = \frac{1}{3}(v_1 + v_2 + v_3 + (k_1 + k_2 + k_3)e^{i\omega t}$
         <li> This is by definition of a median. The vector $\overrightarrow{AB} + \overrightarrow{AC}$ passes between $B$ and $C$, but is twice the length, so we scale it by $1/2$.
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Lecture Notes