Problems

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Solution:

1. \(D\) must be above the centre (of any kind) of the equilateral triangle \(ABC\). Therefore it is a distance \(\frac23 \frac{\sqrt{3}}2 = \frac{\sqrt{3}}3\) from \(A\). \(D\) is \(1\) from \(A\), therefore by Pythagoras \(PD = \sqrt{1-\frac13} = \sqrt{\frac23}\)
2. We can place \(D\) at \(\langle 0,0,\sqrt{\frac23}\rangle\) and \(A'\) (the midpoint of \(BC\)) at \(\langle-\frac{\sqrt{3}}{6},0,0 \rangle\) and we find: \begin{align*} && \cos \theta &= \frac{(\mathbf{a}'-\mathbf{d})\cdot \mathbf{a}'}{|\mathbf{a}'-\mathbf{d}|| \mathbf{a}'|} \\ &&&= \frac{|\mathbf{a}'|}{|\mathbf{a}'-\mathbf{d}|} \\ &&&= \frac{\frac{\sqrt{3}}{6}}{\sqrt{\frac23+\frac{3}{36}}} = \frac13 \end{align*}
3. We have
   

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   And therefore we must have \(\tan \frac{\cos^{-1} \frac13}{2} = \frac{r}{\frac{\sqrt{3}}{6}}\) therefore \begin{align*} && r &= \frac{\sqrt{3}}{6} \tan \left (\frac{\cos^{-1} \frac13}{2} \right) \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{1-\cos(\cos^{-1}\frac13)}{1+\cos(\cos^{-1}\frac13)}} \\ &&&= \frac{\sqrt{3}}6 \sqrt{\frac{\frac23}{\frac43}} \\ &&&= \frac{\sqrt{6}}{12} \end{align*}

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Solution: First let's consider the area of the base. It is an equilateral triangle with side length \(a\), so \(\frac12 a^2 \sin 60^\circ = \frac{\sqrt{3}}4a^2\).

Let's consider the height. The distance to the centre \(\frac23 \frac{\sqrt{3}}2 a = \frac{a}{\sqrt{3}}\) so \(h = \sqrt{b^2 - \frac{a^2}{3}}\) and therefore the volume is: \begin{align*} V &= \frac13 \times \mbox{area of base} \times \mbox {height} \\ &= \frac13 \frac{\sqrt{3}}{4}a^2 \sqrt{\frac{3b^2-a^2}{3}} \\ &= \frac1{12}a^2 (3b^2-a^2)^{\frac12} \end{align*} The area of an isoceles triangle with sides \(a,b,b\) can be found by considering the perpendicular: ie \(\frac{a}{4} \sqrt{b^2-\frac{a^2}{4}} = \frac{a\sqrt{4b^2-a^2}}{8}\). Therefore by considering the volume, we must have \begin{align*} && V &= \frac13 \times \mbox{area of base} \times \mbox {height} \\ \Rightarrow && \frac1{12}a^2 (3b^2-a^2)^{\frac12} &= \frac13 \frac{a\sqrt{4b^2-a^2}}{8} h \\ \Rightarrow && h &= \frac{2a(3b^2-a^2)}{(4b^2-a^2)^{\frac12}} \\ \Rightarrow && h^2 &= \frac{4a^2(3b^2-a^2)}{4b^2-a^2} \end{align*}

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Solution:

The distance to where she enters the water is \(\sqrt{a^2+(b-x)^2}\) and the distance through the water is \(\sqrt{x^2+c^2}\). The total time will be \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{v}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} + \frac{x}{v \sqrt{x^2+c^2}} &= 0 \\ \Rightarrow && v(b-x)(x^2+c^2)^{\frac12} &= xu(a^2+(b-x)^2)^{\frac12} \end{align*} When she is in the water, she can will move with velocity \(\begin{pmatrix} v \cos \theta \\ v \sin \theta -v \end{pmatrix}\). She needs to travel a distance \(\begin{pmatrix} c \\ -x \end{pmatrix}\), so we must have that \begin{align*} && \frac{x}{c} &= \frac{1-\sin \theta}{\cos \theta} \\ \Rightarrow && \sec \theta - \tan \theta &= \frac{x}{c} \\ \Rightarrow && \sec \theta &= \tan \theta + \frac{x}{c} \\ \Rightarrow && \sec^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && 1 + \tan^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && \tan \theta &=\frac{c^2-x^2}{2xc} \\ \Rightarrow && \sin \theta &= \frac{c^2-x^2}{c^2+x^2} \\ && \cos \theta &= \frac{2xc}{c^2+x^2} \\ \end{align*} (where we have taken the positive value for \(\cos \theta\) since we must be heading towards the child). Since \(v \cos \theta t = c\) the time taken to reach the child in the water is \(\frac{c}{v} \frac{c^2+x^2}{2xc} = \frac{c^2+x^2}{2xv}\). So the total time is: \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2xv}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} -\frac{c^2}{2vx^2} + \frac{x^2}{2vx^2}&= 0 \\ \Rightarrow && u(x^2-c^2)\sqrt{a^2+(b-x)^2}&= 2vx^2(b-x) \end{align*} as required. When \(b = c\), the shortest path will be running directly to the bank (there's no quicker way to get to the bank) then swimming directly out (and letting the current take you downstream exactly as far as you need)). Therefore the path will be: If \(b > c\) then she should run a little downstream first. and if \(c > b\) she should actually run a little upstream to take advantage of the current: