Week 1
- The gravitational force is equal to $10m$ downwards, and so the normal and frictional forces must sum to the opposite the gravitational force $(0, -10m)$. This gives us the equation $N \cdot (1/2, \sqrt{3}/2) + \mu \cdot (-\sqrt{3}/2, 1/2) = (0, 10m)$, where $N$ and $\mu$ are the magnitude of the normal and frictional forces respectively. Solving this system of linear equations gives us $\mu = 5m$ and $N = 5\sqrt{3}m$, implying that the normal and frictional forces are $F_N = (5\sqrt{3}m/2, 15m/2)$ and $F_\mu = (-5\sqrt{3}m/2, 5m/2)$ respectively.
- Assume the river flows horizontally, then the horizontal component of the velocity vector of the ship has magnitude $3$, which means the vertical component must have magnitude $4$ by the Pythagorean theorem. To travel $1.2$ miles in a round trip with a velocity of $4$ mph would take $0.3$ hours.
- We proceed by induction on the number of breaks in the line segment. For 0 breaks, this is trivially true. Now, suppose we have a broken segment $A_1\cdots A_{n + 1}$. Then, by the induction hypothesis we have $\overrightarrow{A_1A_2} + \cdots + \overrightarrow{A_{n - 1}A_n} + \overrightarrow{A_nA_1} = 0$, thus $\overrightarrow{A_1A_2} + \cdots + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = -\overrightarrow{A_nA_1} + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = \overrightarrow{A_1A_n} + \overrightarrow{A_nA_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = \overrightarrow{A_1A_{n + 1}} + \overrightarrow{A_{n + 1}A_1} = 0$.
- $3\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} + \overrightarrow{OB} + \overrightarrow{BM} + \overrightarrow{OC} + \overrightarrow{CM} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$ by Problem 5.
- $\overrightarrow{AM} = \frac{2}{3}\overrightarrow{AA'}$ 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 $0$.
- 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 $M = \frac{1}{3}(v_1 + v_2 + v_3 + (k_1 + k_2 + k_3)e^{i\omega t}$
- 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 $1/2$.