Problems

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Solution:

The orange guard will always see \(2b+b-a = 3b-a\) The blue guard will see \(b + \frac{b(b-a)}{a} = \frac{b^2}{a}\) if \(b < 3a\) and \(3b + \frac{b(b-3a)}{(b-a)} = \frac{2b(2b-3a)}{b-a}\). Therefore the blue guard always sees more if \(b > 3a\). He sees more in the other case if \begin{align*} && \frac{b^2}{a} &> 3b - a \\ \Leftrightarrow && \frac{b^2}{a^2} &> 3\frac{b}{a} - 1 \\ \Leftrightarrow && x^2 - 3x + 1 &> 0\\ \Leftrightarrow && x > \frac{3 + \sqrt{5}}{2} \text{ or } x < \frac{3-\sqrt{5}}{2} \end{align*} Since \(b > a\) we must have \(b > \frac{3+\sqrt{5}}2 a\) There is an alternative interpretation which is that the orange guard is in the top left corner, ie In this case the green guard will always see \(2b + \frac{2b(b-a)}{b+a} = \frac{4b^2}{b+a}\) Comparing \(\frac{4b^2}{b+a}\) with \(\frac{b^2}{a}\) we can see the former is larger if \(3a > b\). Comparing \(\frac{4b^2}{b+a}\) with $$

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Solution: Set up a coordinate frame such that the position of the catch is the origin and the time of the catch is \(t = 0\) . We must have then that the trajectory of the ball is \(\mathbf{s} =\mathbf{u} t + \frac12 \mathbf{g} t^2 = \binom{u_x t}{u_y t - \frac12 gt^2}\). We must then have: \begin{align*} && \tan \theta &= \frac{u_y t - \frac12 gt^2}{u_x t} \\ &&&= \frac{u_y}{u_x} - \frac{g}{2u_x} t \\ \Rightarrow && \frac{\d}{\d \theta} \left ( \tan \theta \right) &= 0 - \frac{g}{2 u_x} \end{align*} which is clearly constant. Set coordinates so \(y\)-axis starts from eye-level and \(t = 0\) the first time the ball reaches that level. (Or move the trajectory backwards if that's not the case). Then the ball has trajectory \(\binom{u_xt}{u_yt - \frac12 gt^2}\). The ball reaches eye level a second time when \(t = \frac{2u_y}{g}\), ie at a point \(\frac{2u_xu_y}{g}\). The fielder therefore needs to have position \(f + (u_x-\frac{g}{2u_y}f)t\) at all times. Therefore \begin{align*} && \tan \theta &= \frac{u_y t - \frac12 gt^2}{f + (u_x-\frac{g}{2u_y}f)t - u_x t} \\ &&&= \frac{u_y t - \frac12 gt^2}{f(1-\frac{g}{2u_y}t)} \\ &&&= \frac{u_yt ( 1- \frac{g}{2u_y}t)}{f( 1- \frac{g}{2u_y}t)} \\ &&&= u_y t \\ \Rightarrow && \frac{\d}{\d \theta} \left ( \tan \theta \right) &= u_y \end{align*} Ie \( \frac{\d}{\d \theta} \left ( \tan \theta \right) \) is constant as required.

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Solution:

The area he cannot see is \begin{align*} &&A &= \underbrace{8^2}_{\text{everywhere above}(4,4)} - \underbrace{4^2}_{\text{inner square}} - \underbrace{\frac12 \cdot 4 \cdot (\frac32(4-p)+p - 4)}_{\text{blue triangle}} - \underbrace{\frac12 \cdot 4 \cdot \frac{4(12-p)}{8-p}}_{\text{green triangle}} \\ &&&= 48 - 3(4-p)-2(p-4) - \frac{8(12-p)}{8-p} \\ &&&= 36-5p-\frac{32}{8-p} \\ \\ \Rightarrow && \frac{\d A}{\d p} &= -5 + \frac{32}{(8-p)^2} \\ &&&> 0 \text{ if } 0 \leq p \leq 4 \end{align*}

1. Since the area not visible is increasing as \(p\) increases, we would like \(p\) to be as large as possible, ie \(p = 4\).
2. Similarly, he can see the most when \(p =0\)