Problems

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Solution:

1. Claim: \(S_n \leq 2\sqrt{n} -1\). Proof: (By induction) (Base case: \(n = 1\)). \(\frac{1}{\sqrt{1}} \leq 1 = 2 \cdot \sqrt1 - 1\). Therefore the base case is true. (Inductive step): Suppose our result is true for \(n = k\). Then consider \(n = k+1\). \begin{align*} && \sum_{r=1}^{k+1} \frac{1}{\sqrt{r}} &=\sum_{r=1}^{k} \frac{1}{\sqrt{r}} + \frac{1}{\sqrt{k+1}} \\ &&&\leq 2\sqrt{k} - 1 + \frac{1}{\sqrt{k+1}} \\ &&&= \frac{2 \sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}} - 1 \\ &&&\underbrace{\leq}_{AM-GM} \frac{(k+k+1)+1}{\sqrt{k+1}} - 1 \\ &&&=\frac{2(k+1)}{\sqrt{k+1}} - 1 \\ &&&= 2\sqrt{k+1}-1 \end{align*} Therefore, since if our statement is true for \(n = k\), it is also true for \(n = k+1\). By the principle of mathematical induction we can say that it is true for all \(n \geq 1, n \in \mathbb{Z}\)
2. Claim: \((4k+1)\sqrt{k+1} > (4k+3)\sqrt k\,\) for \(k\ge0\,\) Proof: \begin{align*} && (4k+1)\sqrt{k+1} &> (4k+3)\sqrt k \\ \Leftrightarrow && (4k+1)^2(k+1) &> (4k+3)^2k \\ \Leftrightarrow && (16k^2+8k+1)(k+1) &> (16k^2 + 24k+9)k \\ \Leftrightarrow && 16 k^3 + 24 k^2 + 9 k +1&> 16k^3 + 24k^2+9k \end{align*} But this last inequality is clearly true, hence our original inequality is true. Suppose \(S_n \geq 2\sqrt{n} + \frac{1}{2 \sqrt{n}} - C\), then adding \(\frac{1}{\sqrt{n+1}}\) to both sides we have: \begin{align*} S_{n+1} &\geq 2\sqrt{n} + \frac{1}{2 \sqrt{n}} - C + \frac{1}{\sqrt{n+1}} \\ &= 2\sqrt{n+1} + \frac{1}{2\sqrt{n+1}} - C + \frac{1}{2\sqrt{n+1}} +\frac{1}{2 \sqrt{n}} +2(\sqrt{n} - \sqrt{n+1})\\ &= 2\sqrt{n+1} + \frac{1}{2\sqrt{n+1}} - C + \frac{1}{2\sqrt{n+1}} +\frac{1}{2 \sqrt{n}} -\frac{2}{\sqrt{n+1} + \sqrt{n}}\\ \end{align*} Therefore as long as the inequality is satisfied for \(n=1\), ie \(1 \geq 2\sqrt{1} + \frac{1}{2 \sqrt{1}} - C = \frac52 - C \Rightarrow C \geq \frac32\)

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Solution: \begin{align*} \rightarrow: && x &= u \cos \alpha t\\ \Rightarrow && t &= \frac{x}{u \cos \alpha}\\ \uparrow: && -h &= u\sin \alpha t- \frac12gt^2 \\ && - h &= x\tan \alpha - \frac12 g \frac{x^2}{u^2}\sec^2 \alpha \\ \Rightarrow && \frac{gx^2}{u^2} &= h(2\cos^2 \alpha) + x2 \tan \alpha \cos^2 \alpha \\ &&&= h(1 + \cos 2 \alpha) + x \sin 2\alpha \\ \frac{\d}{\d \alpha}: && \frac{g}{u^2} 2 x \frac{\d x}{\d \alpha} &= -2h \sin 2 \alpha + 2x \cos 2 \alpha +\frac{\d x}{\d \alpha} \sin 2 \alpha \\ \Rightarrow && \frac{\d x}{\d \alpha} \left ( \frac{2xg}{u^2} - \sin 2 \alpha \right) &= 2\cos 2 \alpha (x -h \tan 2 \alpha) \end{align*} Since the turning point will be a maximum must be \(x = h \tan 2 \alpha\). Therefore, let \(c = \cos 2 \alpha\) \begin{align*} && \frac{gh^2}{u^2} \tan^2 2 \alpha &= h(1 + \cos 2 \alpha) + h \tan 2 \alpha \sin 2 \alpha \\ \Rightarrow && \frac{gh}{u^2}(c^{-2}-1) &= 1+c+\frac{1-c^2}{c} \\ \Rightarrow && \frac{gh(1-c^2)}{u^2c^2} &= \frac{c+c^2+1-c^2}{c}\\ &&&= \frac{1+c}{c} \\ \Rightarrow && \frac{gh(1-c)}{u^2c} &= 1 \\ \Rightarrow && u^2c &= gh(1-c) \\ \Rightarrow && c(u^2+gh) &= gh \\ \Rightarrow && \cos 2 \alpha &= \frac{gh}{u^2+gh} \\ \\ \Rightarrow && d_{max}^2 &= h^2 + h^2 \tan^2 2 \alpha \\ &&&= h^2\sec^2 2 \alpha \\ &&&= h^2 \frac{(u^2+gh)^2}{g^2h^2} \\ &&&= \frac{(u^2+gh)^2}{g^2} \\ &&&= \left (\frac{u^2}{g}+h \right)^2 \\ \Rightarrow && d_{max} &= \frac{u^2}{g}+h \end{align*}

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Solution:

1. \(\,\) \begin{align*} && y&=\frac{x^2+x\sin\theta+1}{x^2+x\cos\theta+1} \\ \Leftrightarrow && 0 &= x^2(y-1) + x(y \cos \theta - \sin \theta) + y-1 \\ \Leftrightarrow && 0 &\leq \Delta = (y\cos \theta - \sin \theta)^2 - 4(y-1)^2 \\ \Leftrightarrow && (y\cos \theta - \sin \theta)^2 &\geq 4(y-1)^2 \end{align*} [Assuming that \(y \neq 1\), if \(y = 1\) then the RHS is \(0\) and it is automatically satisfied]. Notice that \((y\cos \theta - \sin \theta)^2 \leq (y^2+1)(\cos^2 \theta + \sin^2 \theta)\) by Cauchy-Schwarz, so \(y^2 + 1 \geq 4(y-1)^2\). \begin{align*} && y^2 + 1 &\geq 4(y-1)^2 \\ \Leftrightarrow && 0 &\geq 3y^2-8y+3 \\ \text{c.v.} && y&= \frac{8 \pm \sqrt{64-4\cdot3 \cdot 3}}{6} \\ &&&= \frac{4 \pm \sqrt{16-9}}{3} = \frac{4 \pm \sqrt{7}}3 \end{align*} so \(\frac{4-\sqrt{7}}3 \leq y \leq \frac{4+\sqrt7}3\).
2. If \(y = \frac{4+\sqrt7}3\) then \(y - 1 = \frac{1+\sqrt7}3\) and since \(y^2+1 = 4(y-1)^2\) taking square roots we obtain \(\sqrt{y^2+1} = 2(y-1)\). Since equality must hold in our C-S identity, we must have \(\langle y, -1 \rangle\) parallel to \( \langle \cos \theta , \sin \theta \rangle\), ie \(\tan \theta = -\frac{3}{4+\sqrt{7}}\) and \begin{align*} && x & = \frac{-(y \cos \theta - \sin \theta) \pm \sqrt{\Delta}}{2(y-1)} \\ &&&= \frac{\pm2(y-1)}{2(y-1)} \\ &&&= \pm1 \end{align*}

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Solution: \begin{align*} && R\cosh(x + \gamma) &=R \cosh x \cosh \gamma + R \sinh x \sinh \gamma \\ \Rightarrow && R \cosh \gamma &= B \\ && R \sinh \gamma &= A \\ \Rightarrow && R^2 &= B^2 - A^2 \\ \Rightarrow && \tanh \gamma &= \frac{A}{B} \\ \end{align*} Therefore it is possible, by writing \(R = \sqrt{B^2-A^2}\) and \(\gamma = \textrm{artanh} \left ( \frac{A}{B} \right)\). This works as long as \(|B| > A > 0\). Supposing \(A >|B| \), try \(S \sinh (x + \delta) = S \sinh x \cosh \delta +S \cosh x \sinh \delta\) \begin{align*} && S \cosh \delta &= A \\ && -S \sinh \delta &= B \\ \Rightarrow && S^2 &= A^2 - B^2 \\ \Rightarrow && \tanh \delta &= \frac{B}{A} \\ \end{align*} Therefore in this case we can write \(\sqrt{A^2-B^2} \sinh \left (x + \tanh^{-1} \left ( \frac{B}{A} \right) \right)\) If \(A = \pm B > 0\) we can we have \(A \sinh x + B \cosh x = \pm Ae^{\pm x}\)

1. Suppose \(y \cosh x = 1\) and \(y \cosh x = a \sinh x +b \cosh x\) so \begin{align*} && 1 & = a \sinh x + b \cosh x \\ &&&= \sqrt{b^2-a^2} \cosh(x + \textrm{artanh} \frac{a}{b} ) \\ \Rightarrow && x + \textrm{artanh} \frac{a}{b} &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \\ \Rightarrow && x &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) -\textrm{artanh} \frac{a}{b} \end{align*}
2. If \( a > b > 0\) we have \begin{align*} && 1 & = \sqrt{a^2-b^2} \sinh \left ( x - \textrm{artanh} \frac{b}{a} \right) \\ \Rightarrow && x &= \textrm{arsinh} \left ( \frac{1}{\sqrt{a^2-b^2}} \right) + \textrm{artanh} \left ( \frac{b}{a} \right) \end{align*}
3. To intersect at distinct points we must have \(b > a\) and \(\textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \neq 0\) which is always true.
4. For the curves to touch, we need them to intersect and have matching derivatives, ie \begin{align*} && -\tanh x \cdot \textrm{sech}x &= a\textrm{sech}^2 x \\ \Rightarrow && 0 &= \textrm{sech}^2 x (a + \sinh x) \\ \Rightarrow && x &= -\textrm{arsinh} \, a\\ \Rightarrow && \sinh x &= - a\\ \Rightarrow &&\cosh x &= \sqrt{1 + a^2} \\ \end{align*} So if the curves touch, we must have \(1 = -a^2+b\sqrt{1+a^2} \Rightarrow b = \sqrt{1+a^2}\) and since this does work it is a necessary and sufficient condition. We will also have the \(y\) coordinate is \(\frac{1}{\sqrt{1+a^2}}\)

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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: Let \(O\) be the origin, and let \(\mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}'\) be the four points. The conditions give us \begin{align*} && \mathbf{a} \cdot \mathbf{b} &= 0 \\ && |\mathbf{a}| &= |\mathbf{b}| \\ && \mathbf{a}' \cdot \mathbf{b}' &= 0 \\ && |\mathbf{a}'| &= |\mathbf{b}'| \\ \end{align*} So \begin{align*} \text{midpoint }AB \text{ to midpoint } BA' &= (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}'))\cdot (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}')) \\ &= \frac12(\mathbf{a}-\mathbf{a}')\cdot \frac12(\mathbf{a} - \mathbf{a}') \\ \text{midpoint }BA' \text{ to midpoint } A'B' &= (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}')) \cdot (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}'))\\ &= \frac12(\mathbf{b}-\mathbf{b}')\cdot \frac12(\mathbf{b} - \mathbf{b}') \\ &= \frac14 (|\mathbf{b}|^2 + |\mathbf{b}'|^2 - 2\mathbf{b}\cdot\mathbf{b}')\\ &= \frac14(|\mathbf{a}|^2 + |\mathbf{a}'|^2 - 2\mathbf{b}\cdot\mathbf{b}') \\ \text{midpoint }A'B' \text{ to midpoint } B'A &= (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a})) \cdot (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a}))\\ &= \frac12(\mathbf{a}'-\mathbf{a})\cdot \frac12(\mathbf{a}' - \mathbf{a}) \\ \text{midpoint }B'A \text{ to midpoint } AB &= (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b})) \cdot (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b}))\\ &= \frac12(\mathbf{b}'-\mathbf{b})\cdot \frac12(\mathbf{b}' - \mathbf{b}) \\ \end{align*} So it's sufficient to prove \(\mathbf{a}\cdot \mathbf{a}' = \mathbf{b}\cdot \mathbf{b}'\) but this is clear from looking at a diagram for 1 second. Given the length of the square is what it is, we want to minimise \(\mathbf{b}\cdot \mathbf{b}'\) which is when they are vertically opposite each other, ie \(\angle BOA' = 90^\circ\)

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Solution: \(f'(x) = 6ax-18x^2\), therefore \(f\) has turning points at \(0\) and \(\frac{a}3\) (ie decreasing for \(x \leq 0\) and \(x \geq \frac{a}{3}\) and increasing otherwise). Therefore possible maxima are \(f(-\tfrac13), f(\frac{a}{3}), f(1)\) where we consider \(\frac{a}{3}\) if \(a \leq 3\) and \(1\) otherwise. \(f(-\frac13) = \frac{a}{3} + \frac{2}{9} = \frac{3a+2}{9}\) \(f(\frac{a}{3}) = \frac{a^3}{3} - \frac{2a^3}{9} = \frac{a^3}{9}\) \(f(1) = 3(a-2)\) Comparing \(\frac{a^3}{9}\) to \(\frac{3a+2}{9}\) we have a double root at \(a = -1\) and a single root at \(a = 2\), therefore \(\frac{a^3}9\) is larger if \(a \geq 2\) Comparing \(3(a-2)\) to \(\frac{3a+2}9\) we have a cross-over at \(a = \frac{7}3\). Therefore we have: \begin{align*} M(a) &= \begin{cases} \frac{3a+2}{9} & 0 \leq a \leq 2 \\ \frac{a^3}{9} & 2 \leq a \leq 3 \\ 3(a-2) & 3 \leq a \end{cases} \end{align*}

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Solution: The bullet fired at time \(t\) will hit the ground at time \(t+\frac{2u \sin (\frac13\pi - \lambda t)}{g}\). To find the last time a bullet hits the ground, we can differentiate, noting that \begin{align*} && T(t) &= t + \frac{2u \sin \alpha}{g} \\ \Rightarrow && T'(t) &= 1 - \frac{2u\lambda}{g} \cos \alpha \\ && T''(t) &= \frac{2u \lambda^2}{g} \sin \alpha > 0 \end{align*} If \(k = \frac{g}{2\lambda u} \in [\frac12, \frac12\sqrt{3}]\) then notice that this turning point is always achieved, and will be a maximum. It will be when \(\cos \alpha = k, \sin \alpha = \sqrt{1-k^2}\). The distance will be \(u \cos \alpha \cdot \frac{2 u \sin \alpha}{g} = \frac{2ku^2\sqrt{1-k^2}}{g}\). If \(k < \frac12\) then the last bullet to hit the ground will be the last bullet fired, ie \(\frac{2u^2 \sin \frac16\pi \cos \frac16\pi}{g} = \frac{u^2 \sin \frac13 \pi}{g} = \frac{\sqrt{3}u^2}{2g}\)

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Solution:

1. \(\,\) \begin{align*} && \mathbb{E}(C) &= \int_0^\infty \text{cost}(x) f(x) \d x \\ &&&= ky + \int_y^{\infty} a(x-y) \lambda e^{-\lambda x} \d x\\ &&&= ky + \int_0^{\infty} a u \lambda e^{-\lambda u -\lambda y} \d x \\ &&&= ky + ae^{-\lambda y} \left( \left [ -ue^{-\lambda u} \right]_0^\infty -\int_0^\infty e^{-\lambda u} \d u\right) \\ &&&= ky + \frac{a}{\lambda}e^{-\lambda y} \\ \\ && \frac{\d \mathbb{E}(C)}{\d y} &= k - ae^{-\lambda y} \\ \Rightarrow && y &= \frac{1}{\lambda}\ln \left ( \frac{a}{k} \right) \end{align*} Since \(\mathbb{E}(C)\) is clearly increasing when \(y\) is very large, the optimal value will be \(\frac{1}{\lambda}\ln \left ( \frac{a}{k} \right)\), if \(\frac{a}{k} > 1\), otherwise you should spend nothing on flood defenses.
2. \begin{align*} && \mathbb{E}(C^2) &= \int_0^{\infty} \text{cost}(x)^2 f(x) \d x \\ &&&= \int_0^{\infty}(ky + a(x-y)\mathbb{1}_{x > y})^2 f(x) \d x \\ &&&= k^2y^2 + \int_y^{\infty}2kya(x-y)f(x)\d x + \int_y^{\infty}a^2 (x-y)^2 f(x) \d x \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{- \lambda y}+a^2e^{-\lambda y}\int_{u=0}^\infty u^2 \lambda e^{-\lambda u} \d u \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y}+a^2e^{-\lambda y}(\textrm{Var}(Exp(\lambda)) + \mathbb{E}(Exp(\lambda))^2\\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} \\ && \textrm{Var}(C) &= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} - \left ( ky + \frac{a}{\lambda} e^{-\lambda y}\right)^2 \\ &&&= a^2e^{-\lambda y} \frac{2}{\lambda^2} - a^2 e^{-2\lambda y}\frac{1}{\lambda^2} \\ &&&= \frac{a^2}{\lambda^2} e^{-\lambda y}\left (2 - e^{-\lambda y} \right) \\ \\ && \frac{\d \textrm{Var}(C)}{\d y} &= \frac{a^2}{\lambda^2} \left (-2\lambda e^{-\lambda y} +2\lambda e^{-2\lambda y} \right) \\ &&&= \frac{2a^2}{\lambda} e^{-\lambda y}\left (e^{-\lambda y}-1 \right) \leq 0 \end{align*} so \(\textrm{Var}(C)\) is decreasing in \(y\).

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Solution:

1. \(\,\) \begin{align*} && \int_0^b x^2 \d x &= \left ( \int_0^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} &= \left ( \frac{b^2}{2} \right)^2 \\ \Rightarrow && b &= \frac{4}{3} \end{align*}
2. \(\,\) \begin{align*} && \int_1^b x^2 \d x &= \left ( \int_1^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{1}{3} &= \left ( \frac{b^2}{2} - \frac{1}{2} \right)^2 \\ \Rightarrow && 4(b^3 - 1) &= 3(b^2-1)^2 \\ \Rightarrow && 4(b^3-1) &= 3(b^4-2b^2+1) \\ \Rightarrow && 0 &= 3b^4-4b^3-6b^2+7 \\ &&&= (b-1)(3b^3-b^2-7b-7) \\ \Rightarrow && 0 &= 3b^3-b^2-7b-7 \end{align*}
   

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   Let \(f(x) = 3x^3-x^2-7x-7\) then \(f(2) = 3 \cdot 8 - 4 - 14 - 7 = -1 < 0\), \(f(3) = 3 \cdot 27 - 9 - 21 - 7 = 44 > 0\) therefore the root must lie between \(2\) and \(3\).
3. \(,\) \begin{align*} && \int_a^b x^2 \d x &= \left ( \int_a^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{a^3}{3} &= \left ( \frac{b^2}{2} - \frac{a^2}{2} \right)^2 \\ \Rightarrow && 4(b^3 - a^3) &= 3(b^2-a^2)^2 \\ \Rightarrow && 4(b^2+ab+a^2) &= 3(b-a)(b+a)^2 \\ \Rightarrow && 4 \left ( \left ( \frac{p+q}{2}\right)^2+\left ( \frac{p+q}{2}\right)\left ( \frac{p-q}{2}\right)+\left ( \frac{p-q}{2}\right)^2\right) &= 3qp^2 \\ \Rightarrow && 3p^2 + q^2 &= 3qp^2 \\ \Rightarrow && 3p^2(q-1) &= q^2 \\ \Rightarrow && p^2 &= \frac{q^2}{3(q-1)} \\ \Rightarrow && 1 &\leq \frac{1}{3(q-1)} \\ \Rightarrow && 3(q-1) &\leq 1 \\ \Rightarrow && q & \leq \frac{4}{3} \\ \end{align*}

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Solution: The position of the hands are \(\begin{pmatrix} a\sin(-t) \\ a \cos(-t) \end{pmatrix}\) and \(\begin{pmatrix} b\sin(-60t) \\ b \cos(-60t) \end{pmatrix}\), the distance between the hands is \begin{align*} x &= \sqrt{\left ( a \sin t - b \sin 60t\right)^2+\left ( a \cos t - b \cos 60t\right)^2} \\ &= \sqrt{a^2+b^2-2ab\left (\sin t \sin 60t+\cos t \cos 60t \right)} \\ &= \sqrt{a^2+b^2-2ab \cos(59t)} = \sqrt{a^2+b^2-2ab \cos \theta} \\ \\ \frac{\d x}{\d \theta} &= \frac{ab \sin \theta}{ \sqrt{a^2+b^2-2ab \cos \theta}} \\ \frac{\d^2 x}{\d \theta^2} &= \frac{ab \cos \theta\sqrt{a^2+b^2-2ab \cos \theta} - \frac{a^2b^2 \sin^2 \theta}{\sqrt{a^2+b^2-2ab \cos \theta}} }{a^2+b^2-2ab \cos \theta} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2 \sin^2 \theta }{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2(1-\cos^2 \theta)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab(a^2+b^2) \cos \theta-a^2b^2 \cos \theta- a^2b^2}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{-ab(a\cos \theta -b)(b \cos \theta - a)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ \end{align*} So the rate of increase is largest when \(\cos \theta = \frac{a}{b}\) (since \(\frac{b}{a}\) is impossible. Therefore when \(x = \sqrt{a^2+b^2-2ab \frac{a}{b}} = \sqrt{a^2+b^2-2a^2} = \sqrt{b^2-a^2}\) If \(b = 2a\) then \(\cos \theta = \frac{a}{2a} = \frac12 = \frac{\pi}{3} = 60^\circ\) The relative speed of the hands is \(5.5^\circ\) per minute, so \(\frac{60}{5.5} = \frac{120}{11} \approx 11\) but clearly also less than since \(121 = 11^2\).

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Solution:

1. Note that \(f(x) = (x+2a)^3 - 27a^2x\) so \(f'(x) = 3(x+2a)^2-27a^2 = 3x^2+12ax-15a^2 = (x-a)(3x+15a)\) so the turning points are at \(x = a, x = -5a\). But \(f(a) = 0\), so the curve just touches the \(x\)-axis. Note also that \(f(0) = 8a^3 \geq 0\) so:
   

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2. \(\,\) \begin{align*} && 0 &\leq (x+2y)^3 - 27y^2 x \\ &&& \leq 3^3 - 27 y^2 x \\ &&&= 27(1-y^2x) \\ \Rightarrow && xy^2 &\leq1 \end{align*} with equality when \(x = y = 1\)
3. Notice that \begin{align*} && 0 &\leq (p+q+r)^3 - 27\left (\frac{q+r}{2}\right)^2 p \\ \Rightarrow && (p+q+r)^3 &\geq 27\left (\frac{q+r}{2}\right)^2 p \\ &&&\underbrace{\geq}_{AM-GM} 27\left (\sqrt{qr}\right)^2 p \\ &&&= 27pqr \end{align*} Equality can only hold if \(p = \frac{q+r}{2}\), but by symmetry we must also have \(q = \frac{r+p}{2}, r = \frac{p+q}{2}\) ie \(p = q = r\). And indeed equality does hold in this case.

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Solution: \begin{align*} && \mathbb{E}(X^2) &= \frac12 \left (\frac16 \left ( 3(k -1)^2+2(2k-1)^2+(3k-1)^2 \right) +\frac16 \left ( 3(k -1)^2+2(k-2)^2+(k-3)^2 \right) \right) \\ &&&= \frac12 \left (\frac16 \left (20k^2-20k+6 \right) + \frac16 \left ( 6k^2-20k+20\right) \right) \\ &&&= \frac1{12} \left (26k^2-40k+ 26\right) \\ &&&= \frac{13}{6} (k^2+1) - \frac{10}{3}k \\ &&&= \frac{13}{6}(k-1)^2+k \end{align*} Since \(k\) a single digit positive number and \(\mathbb{E}(X^2)\) is an integer, \(6 \mid k-1 \Rightarrow k = 1, 7\). \begin{align*} \mathbb{E}(X | k=1) &= \frac12 \left (\frac16 \left ( 2+2 \right) +\frac16 \left ( 2+2 \right) \right) = \frac23 \not \in \mathbb{Z}\\ \mathbb{E}(X | k=7) &= \frac12 \left (\frac16 \left ( 3\cdot6+2\cdot13+20 \right) +\frac16 \left ( 3\cdot6+2\cdot5+4 \right) \right) = 8 \end{align*} Therefore \(k = 7\) The probability distribution is \begin{align*} && \mathbb{P}(X=4) = \frac1{12} \\ && \mathbb{P}(X=5) = \frac1{6} \\ && \mathbb{P}(X=6) = \frac12 \\ && \mathbb{P}(X=13) = \frac1{6} \\ && \mathbb{P}(X=20)= \frac1{12} \\ \end{align*} The only ways to score more than \(25\) are: \(20+6, 20+13, 20+20, 13+13\) The only ways to score exactly \(25\) are \(20+5\) \begin{align*} \mathbb{P}(>25) &= \frac1{12} \cdot\left(2\cdot \frac12+2\cdot\frac16+\frac1{12}\right) + \frac{1}{6^2} \\ &= \frac{7}{48} \\ \mathbb{P}(=25) &= \frac{2}{12 \cdot 6} = \frac{1}{36} \\ \\ \mathbb{E}(\text{payout}) &= \frac{7}{48}w + \frac{1}{36} = \frac{21w+4}{144} \end{align*} The casino needs \(\frac{21w+4}{144} < 1 \Rightarrow 21w< 140 \Rightarrow w < \frac{20}{3}\)

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Solution:

1. If \(f(x) = \sin nx\) then \(f'(x) = n \cos n x\) and so \begin{align*} && LHS &= \int_0^\pi \sin^2 n x \d x \\ &&&= \left [ \frac{x+\frac1{2n}\sin 2n x}{2} \right ]_0^{\pi} \\ &&&= \frac{\pi}{2} \\ \\ && RHS &= \int_0^{\pi} n^2 \cos^2 n x \d x \\ &&&= n^2 \left [ \frac{\frac{1}{2n}\sin 2n x + x}{2} \right]_0^{\pi} \\ &&&= n^2\frac{\pi}{2} \geq LHS \end{align*} [\(f(0) = 0, f(\pi) \neq 0\)] Suppose \(f(x) = x\) then \(f'(x) = 1\) and \(LHS = \frac{\pi^3}{3} > \pi = RHS\). [\(f(0) \neq 0, f(\pi) = 0\)] Suppose \(f(x) = \pi - x\) then \(f'(x) = -1\) and \(LHS = \frac{\pi^3}{3} > \pi = RHS\)
2. Suppose \(f(x) = x(\pi - x)\) then \(f'(x) = \pi - 2x\) and so \begin{align*} && \int_0^\pi x^2(\pi-x)^2 \d x &\leq \int_0^\pi (\pi-2x)^2 \d x \\ \Leftrightarrow && \left [\pi^2 \frac{x^3}{3} - 2\pi \frac{x^4}{4} + \frac{x^5}{5} \right]_0^{\pi} &\leq \left [ \pi^2x - 4\pi \frac{x^2}{2} + \frac{4x^3}{3} \right]_0^{\pi} \\ \Leftrightarrow && \pi^5 \left (\frac13 - \frac12+\frac15 \right) &\leq \pi^3 \left ( 1 - 2+\frac43 \right) \\ \Leftrightarrow && \pi^2 \frac{1}{30} &\leq \frac13 \\ \Leftrightarrow && \pi^2 &\leq 10 \end{align*} Suppose \(f(x) = p\sin \tfrac12 x + q \cos \tfrac12 x + r\), so \(f(0) = q + r\) and \(f(\pi) = p + r\), so say \(p = q = 1, r = -1\) \begin{align*} && LHS &= \int_0^{\pi} \left ( \sin \tfrac12 x + \cos \tfrac12 x-1\right)^2 \d x \\ &&&=\int_0^\pi \left ( \sin^2 \tfrac12 x + \cos^2 \tfrac12 x+1-2\sin \tfrac12 x - 2\cos \tfrac12 x+ \sin x \right)\\ &&&= \left [2x + 4\cos \tfrac12 x - 4\sin \tfrac12 x - \cos x \right]_0^{\pi} \\ &&&= \left ( 2\pi -4+1 \right) - \left ( 4-1 \right) \\ &&&= 2\pi -6\\ \\ && RHS&= \int_0^{\pi} \left ( \tfrac12 \cos \tfrac12 x -\tfrac12 \sin \tfrac12 x\right)^2 \d x \\ &&&= \int_0^{\pi} \left ( \tfrac14 \cos^2 \tfrac12 x +\tfrac14 \sin^2 \tfrac12 x-\tfrac14 \sin x\right) \d x \\ &&&= \frac{\pi}{4} - \frac12 \\ \Rightarrow && 2\pi -6 &\leq \frac{\pi}{4} - \frac12 \\ \Rightarrow && \frac{7\pi}{4} &\leq \frac{11}{2} \\ \Rightarrow && \pi &\leq \frac{22}{7} \end{align*} \(22^2/7^2 = 484/49 < 10\) therefore \(\pi \leq \frac{22}{7}\) is the better estimate.