Problems

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Solution:

1. \(\,\) \begin{align*} && E(x) &= \left ( \frac{\d y}{\d x} \right)^2 + \frac12 y^4 \\ \Rightarrow && E'(x) &= 2 \frac{\d y}{\d x} \frac{\d^2 y}{\d x^2} + 2y^3 \frac{\d y}{\d x} \\ &&&= 2\frac{\d y}{\d x} \left ( \frac{\d^2 y}{\d x^2} + y^3 \right) \\ &&&= 0 \end{align*} Therefore \(E\) is constant. \(E(0) = \frac12\) and \begin{align*} &&\frac12 &= \left ( \frac{\d y}{\d x} \right)^2 + \frac12 y^4 \\ &&&\geq \frac12 y^4 \\ \Rightarrow && |y| &\leq 1 \end{align*}
2. \(\,\) \begin{align*} && E(x) &= \left ( \frac{\d v}{\d x} \right)^2 + 2 \cosh v \\ \Rightarrow && E'(x) &= 2 \frac{\d v}{\d x}\frac{\d^2 v}{\d^2 x} + 2 \sinh v \frac{\d v}{\d x} \\ &&&= 2 \frac{\d v}{\d x} \left ( \frac{\d^2 v}{\d^2 x} + \sinh v\right) \\ &&&= 2 \frac{\d v}{\d x} \left ( -x \frac{\d v}{\d x}\right) \\ &&&= -2x \left ( \frac{\d v}{\d x} \right)^2 \leq 0 \tag{\(x \geq 0\)} \\ \\ && E(0) &= 0^2 + 2 \cosh \ln 3 \\ &&&= 3 + \frac13 = \frac{10}{3} \\ \Rightarrow && \frac{10}{3} &\geq E(x) \\ &&&= \left ( \frac{\d v}{\d x} \right)^2 + 2 \cosh v \\ &&&\geq 2 \cosh v \\ \Rightarrow && \cosh v &\leq \frac53 \end{align*}
3. \(\,\) \begin{align*} && E(x) &= \left ( \frac{\d w}{\d x} \right)^2 + 2(w \sinh w + \cosh w) \\ && E'(x) &= 2 \frac{\d w}{\d x}\frac{\d^2 w}{\d^2 x} + 2(w \cosh w + 2 \sinh w) \frac{\d w}{\d x} \\ &&&= 2 \frac{\d w}{\d x} \left ( \frac{\d^2 w}{\d^2 x}+(w \cosh w + 2 \sinh w)\right) \\ &&&= -2 \left ( \frac{\d w}{\d x} \right)^2 \left (\underbrace{5\cosh x - 4 \sinh x -3}_{\geq0} \right) \\ &&&\leq 0 \\ && E(0) &= \frac12 + 2 = \frac52 \\ \Rightarrow && \frac52 &\geq E(x) \\ &&&=\underbrace{ \left ( \frac{\d w}{\d x} \right)^2}_{\geq0} + 2(\underbrace{w \sinh w}_{\geq 0} + \cosh w) \\ &&&\geq2\cosh w \\ \Rightarrow && \cosh w &\leq \frac54 \end{align*}

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Solution: \begin{align*} \sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)} &= e^{2i\alpha} \sum_{r=0}^{n-1} \left (\e^{2i\pi/n} \right)^r \\ &= e^{2i\alpha} \frac{1-\left (\e^{2i\pi/n} \right)^n}{1-\e^{2i\pi/n} } \\ &= 0 \end{align*} \(d = s + r \cos \theta\) ie \(s = d - r \cos \theta\) Therefore \(d = \frac{r}{k} + r \cos \theta \Rightarrow r = \frac{kd}{1+k \cos \theta}\). The \(l_j\) will come from \(r(\alpha + \frac{j \pi}{n} )+r(\alpha + \pi + \frac{j \pi}{n} )\) \begin{align*} && l_j &= r(\alpha + \frac{(j-1) \pi}{n} )+r(\alpha + \pi + \frac{(j-1) \pi}{n} ) \\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1+k \cos \left ( \alpha+\pi+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1-k \cos \left ( \alpha+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{2kd}{1-k^2 \cos^2 \left ( \alpha + \frac{(j-1) \pi}{n}\right)}\\ \Rightarrow && \sum_{j=1}^n \frac 1 {l_j} &= \sum_{j=0}^{n-1} \frac{1-k^2 \cos^2 \left ( \alpha + \frac{j \pi}{n}\right)}{2kd} \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \cos^2 \left ( \alpha + \frac{j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \frac{1+ \cos \left ( 2\alpha + \frac{2j \pi}{n}\right)}{2} \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \sum_{j=0}^{n-1}\cos \left ( 2\alpha + \frac{2j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \underbrace{\textrm{Re} \left ( \sum_{j=0}^{n-1}e^{ 2i(\alpha + \frac{j \pi}{n})} \right)}_{=0} \\ &&&= \frac{n}{2kd} - \frac{nk^2}{4kd} \\ &&&= \frac{n(2-k^2)}{4kd} \end{align*}

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Solution:

1. Notice that \(\E[X_1] = \frac{a}{n}\) and consider \(\E[X_i]\) with \(i > 1\). the probability that this is \(1\) is \(\frac{b}{n} \cdot \frac{a}{n-1}\). So \begin{align*} && \E[S] &= \E[X_1] + \sum_{i=2}^n \E[X_i] \\ &&&= \frac{a}{n} + (n-1) \frac{ab}{n(n-1)} \\ &&&= \frac{a(b+1)}{n} \end{align*}
2. 1. The probability \(X_1X_j = 1\) is \(\frac{a}{n} \cdot \frac{b}{n-1} \cdot \frac{a-1}{n-2} = \frac{a(a-1)b}{n(n-1)(n-2)}\) since there is nothing special about the order, and the first is an \(A\) with probability \(\frac{a}{n}\) and given this occurs there are now \(a-1\) \(A\) and \(n-1\) letters left etc... Therefore \(\E[X_1X_j] = \frac{a(a-1)b}{n(n-1)(n-2)}\)
   2. \(\E[X_iX_j]\) when the pairs don't overlap is \(\frac{a}{n} \frac{b}{n-1} \frac{a-1}{n-2} \frac{b-1}{n-3}\), and so \begin{align*} && \sum\limits_{i=2}^{n-2} \bigg( \sum\limits_{j=i+2}^n \E(X_iX_j)\bigg) &= \sum\limits_{i=2}^{n-2} \bigg( \sum\limits_{j=i+2}^n \frac{a(a-1)b(b-1)}{n(n-1)(n-2)(n-3)}\bigg) \\ &&&= \frac{a(a-1)b(b-1)}{n(n-1)(n-2)(n-3)}\sum\limits_{i=2}^{n-2} \bigg( \sum\limits_{j=i+2}^n 1\bigg) \\ &&&= \frac{a(a-1)b(b-1)}{n(n-1)(n-2)(n-3)}\sum\limits_{i=2}^{n-2} (n-(i+1)) \\ &&&= \frac{a(a-1)b(b-1)}{n(n-1)(n-2)(n-3)} \left ((n-1)(n-3)-\frac{(n-2)(n-1)}{2}+1 \right) \\ &&&= \frac{a(a-1)b(b-1)}{n(n-1)(n-2)(n-3)} \left ( \frac{2n^2-8n-6-n^2+3n-2+2}{2}\right) \\ &&&= \frac{a(a-1)b(b-1)}{n(n-1)(n-2)(n-3)} \left ( \frac{n^2-5n-6}{2}\right) \\ &&&= \frac{a(a-1)b(b-1)}{2n(n-1)} \end{align*}
   3. We also need to consider the other cross terms. \(X_iX_{i+1}=0\). (Since \(X_i = 1\) means the \(i\)th letter is \(A\) and \(X_{i+1} = 1\) means the \(i\)th letter is \(B\)). It's the same story for \(X_1X_2\), and so all the cross terms are accounted for. Therefore \begin{align*} && \E[S^2] &= \E \left [\sum X_i^2 + 2\sum_{i \neq j} X_i X_j \right] \\ &&&= \frac{a(b+1)}{n} +2(n-2)\frac{a(a-1)b}{n(n-1)(n-2)}+ 2 \frac{a(a-1)b(b-1)}{2n(n-1)} \\ &&&= \frac{a(b+1)}{n} +\frac{2a(a-1)b}{n(n-1)} + \frac{a(a-1)b(b-1)}{n(n-1)} \\ &&&= \frac{a(b+1)}{n} +\frac{a(a-1)b(b+1)}{n(n-1)} \\ && \var[S] &= \E[S^2] - \left ( \E[S] \right)^2 \\ &&&= \frac{a(b+1)}{n} + \frac{a(a-1)b(b+1)}{n(n-1)} - \frac{a^2(b+1)^2}{n^2} \\ &&&= \frac{a(b+1) \left (n(n-1) + (a-1)b n -a(b+1)(n-1) \right)}{n^2(n-1)} \\ &&&= \frac{a(b+1) \left ( (n-a)(n-b-1) \right)}{n^2(n-1)} \\ &&&= \frac{a(b+1) \left ( b(a-1) \right)}{n^2(n-1)} \\ \end{align*}

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Solution:

1. 1. \(\,\) \begin{align*} && 0 &\leq f(t) &\leq M \\ \Rightarrow && \int_0^x 0 \d t &\leq \int_0^x f(t) \d t & \leq \int_0^x M \d x \\ \Rightarrow && 0 &\leq F(x) &\leq Mx \end{align*}
   2. \(\,\) \begin{align*} && \int_0^1 2g(x)F(x)f(x) \d x &= \left [ g(x) F(x)^2 \right] - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \\ &&&= g(1) - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \end{align*}
2. 1. \(\,\) \begin{align*} && 1 &= \int_0^1 kF(y)f(y) \d y \\ &&&= k\left [ \frac12 F(y)^2\right]_0^1 \\ &&&= \frac{k}{2} \\ \Rightarrow && k &= 2 \end{align*}
   2. \(\,\) \begin{align*} \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &\geq \int_0^1 y^n 2My f(y) \d y \\ &= 2M\int_0^1 y^{n+1} f(y) \d y \\ &= 2M \E[X^{n+1}] = 2M\mu_{n+1} \\ \\ \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &= 1 - \int_0^1 ny^{n-1} F(y)^2 \d y \\ &\geq 1 - \int_0^1 ny^{n-1}My F(y) \d y \\ &= 1 - M\int_0^1 ny^n F(y) \d y \\ &= 1 - M[\frac{n}{n+1}y^{n+1} F(y)]_0^1 + M\int_0^1\frac{n}{n+1} y^{n+1} f(y) \d y \\ &= 1 - \frac{nM}{n+1} + \frac{nM}{n+1} \mu_{n+1} \end{align*}
   3. Since \(\E[Y^{n-1}] \geq 0\) we must have \begin{align*} && 2M\mu_n \geq 1 + \frac{(n-1)M}{n}\mu_n - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \left (2M + \frac{(n-1)M}{n} \right) \geq 1 - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \frac{3Mn-M}{n} & \geq \frac{n-(n-1)M}{n} \\ \Rightarrow && \mu_n & \geq \frac{n-(n-1)M}{3Mn-M} \end{align*}

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Solution: \begin{align*} && b &= c - ma \\ \Rightarrow && c &= b+ma \\ \Rightarrow && y &= m(a-x)+b \\ \Rightarrow && q &= ma+b \\ && p &= \frac{ma+b}{m} \\ \\ && d &= p+q \\ &&&= a + \frac{b}{m} + ma + b \\ \Rightarrow && d' &= -bm^{-2}+a \\ \Rightarrow && m &= \sqrt{b/a} \\ \\ \Rightarrow &&d &= a + \sqrt{ba}+\sqrt{ba} + b \\ &&&= (\sqrt{a}+\sqrt{b})^2 \\ \\ && |PQ|^2 &= p^2 + q^2 \\ &&&= a^2 + \frac{2ab}{m} + \frac{b^2}{m^2} + m^2a^2 + 2mab + b^2 \\ &&&= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ && \frac{\d}{\d m}&= -2b^2m^{-3}-2abm^{-2}+2ab + 2a^2m \\ && 0 &=2a^2m^4+2abm^3-2abm-2b^2 \\ &&&= 2(am^3-b)(am+b) \\ \Rightarrow && m &= \sqrt[3]{\frac{b}{a}} \\ \\ &&|PQ|^2 &= \left[ a^{1/3}(a^{2/3} + b^{2/3}) \right]^2 + \left[ b^{1/3}(a^{2/3} + b^{2/3}) \right]^2 \\ &&&= a^{2/3}(a^{2/3} + b^{2/3})^2 + b^{2/3}(a^{2/3} + b^{2/3})^2 \\ &&&= (a^{2/3} + b^{2/3})^2 \cdot (a^{2/3} + b^{2/3}) \\ &&&= (a^{2/3} + b^{2/3})^3 \\ \Rightarrow && |PQ| &= (a^{2/3} + b^{2/3})^{3/2} \end{align*} We can also do this with AM-GM instead: \begin{align*} && d &= a + b + \frac{b}{m} + am \\ &&&\geq a+b + 2 \sqrt{\frac{b}{m} \cdot am} \\ &&&= a+2\sqrt{ab}+b \\ \\ && |PQ|^2 &= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ &&&= a^2+b^2 + \frac{b^2}{m} + abm + abm + a^2m^2 + \frac{ab}{m} + \frac{ab}{m} \\ &&&= a^2+b^2 + 3\sqrt[3]{ \frac{b^2}{m} \cdot abm \cdot abm} + 3 \sqrt[3]{ a^2m^2 \cdot \frac{ab}{m} \cdot \frac{ab}{m} } \\ &&&= a^2 + 3b^{4/3}a^{2/3}+3b^{2/3}a^{4/3}+b^2 \\ &&&= (a^{2/3}+b^{2/3})^3 \end{align*}

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Solution:

1. \(\,\)
   

   + - ↺
   1. \(n = 0\) if \(b > 9\)
   2. \(n = 1\) is not possible, since by symmetry if \(x\) is a root, so is \(-x\), and \(0\) can never be the only root.
   3. \(n = 2\) if \(b < 0\) or \(b = 9\)
   4. \(n = 3\) if \(b = 0\)
   5. \(n = 4\) if \(0 < b < 9\)
2. \(\,\) \begin{align*} && y' &= 4x^3-12x+a \\ && y'' &= 12x^2-12 \\ \Rightarrow && x &= \pm 1 \\ \Rightarrow && 0 &= 4(\pm 1) - 12 (\pm 1) + a \\ &&&= a \mp 8 \\ \Rightarrow && a &= \pm 8 \end{align*} When \(a = 8\), we have \(y = x^4-6x^2+8x\) and \begin{align*} &&y' &= 4x^3-12x+8 \\ &&&= 4(x^3-3x+2) \\ &&&= 4(x-1)^2(x+2) \\ \Rightarrow && y(1) &= 3\\ && y(-2) &= -24 \end{align*}
   

   + - ↺
   Therefore there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise. Similarly, if \(a = -8\), we have \(y = x^4 - 6x^2-8x\) \begin{align*} && y' &= 4x^3-12x-8 \\ &&&= 4(x^3-3x-2) \\ &&&= 4(x-2)(x+1)^2 \end{align*} So we have stationary points at \(x = 2\) and \(x = -1\) (which is also a inflection point) and at \(x = 2\) \(y = -24\), so we have the same story: there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise.
3. \(\,\)
   

   + - ↺

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Solution:

1. \(\,\)
   

   + - ↺
   The area under the blue curve is \(1-\cos b\). The area under the green line is \(\frac12 b \sin b\) The area under the red line is \(\frac12 b^2\) Therefore \(\frac12 b \sin b < 1- \cos b < \frac12 b^2 \Rightarrow 1- \frac12 b^2 < \cos b < 1 - \frac12 b \sin b\)
2. \(\,\)
   

   + - ↺
   \begin{align*} &&\text{Area under blue curve}: &= \int_0^1 a^x \d x\\ &&&= \left [ \frac{1}{\ln a}e^{x \ln a} \right]_0^1 \\ &&&= \frac{a-1}{\ln a} \\ \\ &&\text{Area under green line}: &=\frac12 \cdot 1 \cdot (a + 1)\\ &&&= \frac{a+1}{2} \\ \\ &&\text{Area under tangent}: &=\frac12 \cdot 1 \cdot (1+\ln a + 1)\\ &&&= \frac{\ln a+2}{2} \\ \\ \Rightarrow && \frac{a+1}{2} & > \frac{a-1}{\ln a} \\ \Rightarrow && \ln a& > \frac{2(a-1)}{a+1} \\ \\ \Rightarrow && \frac{a-1}{\ln a} &> \frac{\ln a +2}{2} \\ \Rightarrow && 2(a-1) -2\ln a - (\ln a)^2 &> 0 \\ \Rightarrow && \ln a & < -1 + \sqrt{2a-1} \end{align*}

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Solution: \begin{align*} && \frac{\d y}{\d x} &= -\frac1{2x^2} \\ \Rightarrow && \frac{y - \frac{1}{2p}}{x - p} &= - \frac{1}{2p^2} \\ \Rightarrow && y - \frac1{2p} &= -\frac{1}{2p^2}x +\frac1{2p} \\ \Rightarrow && y +\frac{1}{2p^2}x &= \frac1p \\ \Rightarrow && 2p^2 y + x &= 2p\\ \Rightarrow && 2q^2 y + x &= 2q \\ \Rightarrow && (p^2-q^2)y &= p-q \\ \Rightarrow && y &= \frac{1}{p+q} \\ && x &= \frac{2pq}{p+q} \end{align*} \begin{align*} \text{normal}: && \frac{y-\frac1{2p}}{x-p} &= 2p^2 \\ \Rightarrow && y - \frac1{2p} &= 2p^2x - 2p^3 \\ \Rightarrow && 2py -4p^3x &= 1-4p^4 \\ \Rightarrow && 2qy -4q^3x &= 1-4q^4 \\ pq = \tfrac12: && y - 2p^2 x &= q-2p^3 \\ && y - 2q^2 x &= p-2q^3 \\ \Rightarrow && (2q^2-2p^2)x &= q-p +2q^3-2p^3 \\ &&&= (q-p)(q+p+2q^2+1+2p^2) \\ \Rightarrow && x &= \frac{1+2(p^2+q^2)+1}{2(p+q)} \\ &&&= \frac{1+2(p^2+q^2+2pq-1)+1}{2(p+q)} \\ &&&= p+q\\ && y &= 2p^2 \left ( p+q \right) + q - 2p^3 \\ &&&= p+q \end{align*} So \(N(p+q, p+q)\) and \(T\left (\frac{1}{p+q}, \frac{1}{p+q} \right)\), so both points lie on \(y = x\). \[ OT \cdot ON = \frac{\sqrt{2}}{p+q} \cdot (p+q)\sqrt{2} = 2 \] which is clearly constant.

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Solution: \begin{align*} &&\int_0^{\frac14\pi} \sin (2x) \ln(\cos x)\, \d x &= \int_0^{\frac14 \pi} 2 \sin x \cos x \ln (\cos x) \d x \\ u = \cos \theta :&&&= \int_{u=1}^{u=\frac1{\sqrt2}} -2u \ln u \d u \\ &&&= \int_{\frac1{\sqrt{2}}}^1 2u \ln u \d u \\ &&&= \left [u^2 \ln u \right]_{\frac1{\sqrt{2}}}^1-\int_{\frac1{\sqrt{2}}}^1 u \d u \\ &&&= -\frac12 \ln \frac{1}{\sqrt{2}} - \l\frac12 - \frac14 \r \\ &&&= \frac14 (\ln 2 - 1) \end{align*} \begin{align*} && \int_0^{\frac14\pi} \cos (2x) \ln(\cos x)\, \d x &= \left [ \frac12 \sin 2x \ln (\cos x) \right]_0^{\frac14\pi}- \int_0^{\frac14\pi} \frac12 \sin 2x \frac{-\sin x}{\cos x} \d x \\ &&&=\frac12 \ln \frac{1}{\sqrt{2}}+\int_0^{\frac14\pi} \sin^2 x \d x \\ &&&= -\frac14 \ln 2 + \int_0^{\frac14\pi} \frac{1-\cos 2x }{2} \d x \\ &&&= -\frac14 \ln 2 +\frac{\pi}{8} -\frac{1}{4} \\ &&&= \frac18 (\pi - 2\ln 2 - 2) \\ &&&= \frac18 (\pi - \ln 4 - 2) \\ \end{align*} Notice that \(\cos x + \sin x = \sqrt{2} \cos (x -\frac{\pi}{4})\), so: \begin{align*} &&\int_{\frac14\pi}^{\frac12\pi} \big ( \cos(2x) + \sin (2x)\big) \ln \big( \cos x + \sin x\big)\d x &= \int_{\frac14\pi}^{\frac12\pi} \big ( \cos(2x) + \sin (2x)\big) \ln (\sqrt{2} \cos ( x - \frac{\pi}{4}) ) \d x \\ &&&= \int_{u=0}^{u=\frac{\pi}{4}} \l \cos(2u+\frac{\pi}{2})+\sin(2u+\frac{\pi}{2}) \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\ &&&= \int_{0}^{\frac{\pi}{4}} \l -\sin 2u+\cos 2u \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\ &&&= \frac14 \ln 2\left [ \cos 2u + \sin 2u \ \right]_{0}^{\frac{\pi}{4}} - \frac14(\ln2 - 1) + \frac18\pi - \frac14(\ln 2 +1) \\ &&&= \frac{\pi}{8}-\frac12 \ln 2 \end{align*}

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Solution:

\begin{align*} && \cot \alpha &= \frac{AD}{h} \\ && \cot \beta &= \frac{BD}{h} \\ && \cot \phi &= \frac{DC}h \\ && CD^2 &= AB \cdot BD \tag{intersecting chords} \\ \Rightarrow && \cot^2 \phi &= \cot \alpha \cot \beta \end{align*} \begin{align*} && LHS &= \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) \\ &&&= \cos p \cos q \left ( \frac{1-\cos(p+q)}{2} \right) - \sin p\sin q \left (\frac{1+\cos(p+q)}{2} \right) \\ &&&= \frac12 \left (\cos p \cos q(1-\cos(p+q)) - \sin p\sin q (1+\cos(p+q)) \right) \\ &&&= \frac12 \left ((\cos p \cos q- \sin p\sin q) - (\cos p \cos q+ \sin p\sin q)\cos(p+q) \right) \\ &&&= \frac12 \left (\cos(p+q) - \cos (p-q)\cos(p+q) \right) \\ &&&= RHS \end{align*} Therefore \begin{align*} \cot p \cot q -\cot^2 \tfrac12 (p+q) &= \frac{\tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)}{\sin p \sin q \sin^2 \tfrac12(p+q)} \\ &=\frac{\cos(p+q)(1-\cos(p-q))}{\sin p \sin q \sin^2 \tfrac12(p+q)} \end{align*} Since \(p,q\) are acute, the denominator is positive. Since \(p+q \leq \frac{\pi}{2}\), we have \(\cos(p+q) \geq 0\). Also \((1-\cos(p-q)) \geq 0\). Thus, the expression is \(\geq 0\). So we must have \begin{align*} && \cot^2 \phi &= \cot \alpha \cot \beta \\ &&&\geq \cot^2 \tfrac12(\alpha+\beta) \end{align*} Since \(\cot\) is decreasing on \((0, \tfrac12 \pi)\) we can deduce \(\phi \leq \tfrac12 (\alpha+\beta)\)

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Solution:

1. Suppose \(t_n = x\) for all \(n\), then we must have \begin{align*} && x &= p x + q x \\ \Leftrightarrow && 1 &= p+q \end{align*} and this clearly works for any value of \(x\).
2. Suppose \(t_{2n} = x, t_{2n+1} = y\) for all \(n\), then \begin{align*} && x &= py + q x \\ && y &= px + q y \\ \Rightarrow && 0 &= py + (q-1) x \\ && 0 &= px + (q-1) y \\ \Rightarrow && p &= (q-1) \frac{x}{y} = (q-1) \frac{y}{x} \\ \Rightarrow && \frac{y}{x} = \pm 1 & \text{ or } q = 1, p = 0 \\ \Rightarrow && y = \pm x & \text{ or } (p,q) = (0,1) \\ \end{align*}
3. Suppose \(t_{3n} = x\), \(t_{3n+1}=y\) and \(t_{3n+2}=z\) , so \begin{align*} && x &= pz + qy \\ && y & = px + qz \\ && z &= py + qx \\ \\ && z &= p(px+qz) + q(pz + qy) \\ &&&= p^2x + 2pqz + q^2 y \\ &&&= p^2(pz+qy) + 2pqz + q^2(px+qz) \\ &&&= p^3 z + p^2qy + 2pqz + q^2p x + q^3 z \\ &&&= (p^3+q^3+2pq)z + pq(py+qx) \\ &&&= (p^3 + q^3 + 2pq)z + pq z \\ &&&= (p^3 + q^3 + 3pq)z \\ \Rightarrow && 0 &= p^3 + q^3 + 3pq- 1 \\ &&&= (p+q-1)(p^2+q^2+1+p+q-pq) \\ &&&= \tfrac12(p+q-1)((p-q)^2+(p+1)^2+(q+1)^2) \end{align*} Therefore \(p+q = 1\) or \((p-q)^2+(p+1)^2+(q+1)^2 = 0 \Rightarrow p = q = -1\). If \(p+q = 1\), then \(z = py + (1-p)x\) and \(x = p(py+(1-p)x) + (1-p)y \Rightarrow (1-p+p^2)x = (1-p+p^2)y \Rightarrow x = y \Rightarrow x= y = z\). If \(p = q = -1\) then adding all the equations we get \(x + y + z = -2(x+y+z) \Rightarrow x + y + z = 0\)

Note that what is actually going on here is that solutions must be of the form \(t_n = \lambda^n\) so the only way to be constant is for \(\lambda = 1\) to be a root, the only way for it to be \(2\)-periodic is for \(\lambda = -1\) to be a root, and the only way for it to be \(3\)-periodic is for \(\lambda = 1, \omega, \omega^2\) to be the roots (although we see this via the classic \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x + \omega y + \omega^2 z)(x+\omega^2 y +\omega z)\) which is because of the real constraint in the question.

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Solution:

1. \(\,\) \begin{align*} && y &= xv \\ && y' &= v + xv' \\ \Rightarrow && 0 &= x^2 v \left ( v + x\frac{\d v}{\d x} \right) +(x^2v^2) - 2x^2 \\ &&&= 2x^2v^2 + x^3 v \frac{\d v}{\d x} - 2x^2 \\ \Rightarrow && 0 &= xv \frac{\d v}{\d x} + 2v^2-2 \\ \\ \Rightarrow && \frac{v}{1-v^2} \frac{\d v}{\d x} &= \frac{2}{x} \\ \Rightarrow && \int \frac{v}{1-v^2} \d v &=2 \ln |x| \\ \Rightarrow && -\frac12\ln |1-v^2| &= 2\ln |x| + C \\ \Rightarrow && 4\ln |x| + \ln |1-v^2| &= K \\ \Rightarrow && x^4(1-v^2) &= K \\ \Rightarrow && x^2(x^2-y^2) &= K \end{align*}
2. \(\,\) \begin{align*} && 0 &= xv \left (v +x \frac{\d v}{\d x} \right) + 6x + 5xv \\ &&&= x^2 v \frac{\d v}{\d x} +xv^2 + 6x+5xv \\ \Rightarrow && 0 &= xv\frac{\d v}{\d x} +v^2 +5v+6 \\ \Rightarrow && -\int \frac{1}{x} \d x &=\int \frac{v}{v^2+5v+6} \d v \\ \Rightarrow && -\ln |x| &= \int \frac{v}{(v+2)(v+3)} \d v \\ &&&= \int \left (\frac{3}{v+3} - \frac{2}{v+2} \right) \d v \\ \Rightarrow && -\ln |x| &= 3\ln |v+3| - 2 \ln |v+2| + C\\ \Rightarrow && -\ln |x| &= \ln \frac{|v+3|^3}{|v+2|^2} + C \\ \Rightarrow && \frac{1}{|x|}|v+2|^2 &= A|v+3|^3 \\ \Rightarrow && \frac{1}{|x|}|\frac{y}{x} + 2|^2&= A|\frac{y}{x} + 3|^3 \\ \Rightarrow && \frac{1}{|x|^3} |y +2x|^2 &= \frac{A}{|x|^3}|y + 3x|^3 \\ \Rightarrow && (y+2x)^2 &= A|y+3x|^3 \end{align*}