Problems

Solution:

1. 

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2. \(2 \textrm{Min}(x^2,2x) = 5x-3\) tells us either \(2x^2 = 5x-3 \Rightarrow 2x^2 - 5x +3 = 0 \Rightarrow (2x-3)(x-1) \Rightarrow x = 1, \frac32\) and \(0 \leq x \leq 2\) or \(4x = 5x-3 \Rightarrow x= 3\) and \(x < 0\) or \(2 > x\), therefore our solutions are \(x = 1, \frac32, 3\)
3. We have different cases based on \(x\) vs \(-2, 0, 2\), ie Case \(x \leq -2\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= 2x + x^3 \end{align*} So \(2x = 2x + x^3 \Rightarrow x^3 = 0\), but \(x \leq -2\) so no solutions. or \(6x = 2x + x^3 \Rightarrow 0 = x(x^2-4) \Rightarrow x = 0, 2, -2\) so \(x = -2\). Case \(-2 < x \leq 0\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= 2x + 4x \end{align*} So \(2x = 2x + 4x\) ie \(x = 0\) which is valid. Or \(6x = 2x + 4x\) ie valid for all values in \(-2 \leq x \leq 0\) Case \(0 < x \leq 2\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= x^2 + x^3 \end{align*} So \(2x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-2) = x(x-1)(x+2)\) so \(x = 0, 1, -2\), but the range means \(x = 0\) or \(x = 1\) Or \(6x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-6) = x(x-2)(x+3)\) so \(x = 0, 2, -3\), but the range means \(x = 0\) or \(x = 2\) Case \(2 \leq x \): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&=2x + 4x \end{align*} So \(2x = 2x + 4x \Rightarrow x = 0\) so no solutions. Or \(6x = 2x + 4x\) so a range of solutions. Therefore the final solutions for \(m = 2\) are \(x = 0, x = 1\) and for \(m = 2\) are \(x \in [-2,0] \cup [2, \infty)\)
4. \(\mathrm{Min}(x^2, x^3)\) switches when \(x = 1\), so we must consider both limits: \begin{align*} && \frac{\d y}{\d x}\vert_{x > 1} &= 4x - 5 \\ \\ && \frac{\d y}{\d x}\vert_{x < 1} &= 6x^2 - 5 \\ \end{align*} so when \(x = 1\) the sign of the derivative changes from positive to negative, hence a local maximum. The other local maxima and minima will be when \(x = \frac54\) or \(x = \pm \sqrt{5/6}\)
   

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Solution:

1. 1. Suppose \(z^2 + |z + b| = a\), then \(z^2 = a- |z + b| \in \mathbb{R}\), since \(a \in \mathbb{R}\). Since the square root of a real number is either purely real or purely imaginary, \(z\) is purely real or purely imaginary.
   2. Suppose \(z = it\) for some \(t \in \mathbb{R}\), then \begin{align*} && \frac72 &= -t^2 +\sqrt{t^2 + \frac{25}{4}} \\ \Rightarrow && \left ( \frac72 + t^2\right)^2 &= t^2 + \frac{25}{4} \\ \Rightarrow && t^4 + 7t^2 + \frac{49}{4} &= t^2 + \frac{25}{4} \\ \Rightarrow && t^4 + 6t^2 + 6 &= 0 \end{align*} but since \(\Delta = 6^2 - 4 \cdot 1 \cdot 6 < 0\) there are no real solutions.
   3. Suppose \(z = t\) for some \(t \in \mathbb{R}\), then either \(t^2 + t + \frac72 = \frac52 \Rightarrow t^2 + t + 1 = 0\) (no solutions) or \(t^2 - t - \frac72 = \frac52 \Rightarrow t^2 - t - 6 = (t-3)(t+2) = 0\). When \(t = 3\) then we must take the positive part for \(|z + \frac72|\) so this cannot work. When \(t = -2\) we also have \(\frac72-2 > 0\) so we are still taking the positive part. Hence no solutions
   4. Suppose \(\frac{1}{2} < b < \frac{3}{4}\), the equation then consider \(z^2 + |z + b| = \frac{1}{2}\). Case 1: \(z = t \in \mathbb{R}\), then we have two cases: Case 1a: \(z+b > 0\). \(z^2 + z + b = \frac12 \Rightarrow z = \frac{-1 \pm \sqrt{1-4b+2}}{2}\) which clearly is a valid real number an \(z + b > 0\). Case 1b: \(z+b < 0\) \(z^2 - z - b = \frac12 \Rightarrow z = \frac{1 \pm \sqrt{1+4b-2}}{2}\)
2. Let \(\omega\) be a (primitive) cube root of unity. \begin{align*} && z^3 &= 4 - |z+2|^2 \\ \Rightarrow && z &\in \mathbb{R} \cup \omega \mathbb{R} \cup \omega^2 \mathbb{R} \end{align*} Case 1:

Solution:

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2. \begin{enumerate}
3. since \(\frac{\ln x}{x}\) is decreasing on \((e, \infty)\) we must have that \(\frac{\ln 3}{3} > \frac{\ln \pi}{\pi} \Rightarrow e^\pi > \pi^3\)
4. similarly, since \(\frac{\ln x}{x}\) is increasing on \((0, e)\) we must have that \(\frac{\ln \sqrt{5}}{\sqrt{5}} < \frac{\ln 9/4}{9/4} \Rightarrow \left(\frac{9}{4}\right)^{\sqrt{5}} > \sqrt{5}^{\frac{9}{4}}\)
5. Since \(2^4 = 4^2\) notice also that:
   

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   from the graph we must have the green area between \(1\) and \(2\) mapping to the (higher) green area between \(3\) and \(4\). Therefore \((x+2)^x > x^{x+2}\) for \(1 < x < 2\)
6. \begin{align*} && 9^{\sqrt 2} & \stackrel{?}{>} \sqrt{2}^9 \\ \Leftrightarrow && (3^2)^{\sqrt2} &\stackrel{?}{>} (\sqrt{2}^3)^3 \\ \Leftrightarrow && 3^{2 \sqrt2} &\stackrel{?}{>} (2\sqrt2)^3 \end{align*} Since \(e^2 < 8 < 9\Rightarrow e < 2\sqrt2 < 3\) therefore: \begin{align*} && \frac{\ln 2 \sqrt2}{2 \sqrt 2} &> \frac{\ln 3}{3} \\ \Leftrightarrow && (2 \sqrt{2})^3 &> 3^{2 \sqrt{2}} \\ \Leftrightarrow && \sqrt{2}^9 &> 9^{\sqrt 2} \\ \end{align*}
7. \begin{align*} && 8^{\sqrt[3]{3}} & \stackrel{?}{>} \sqrt[3]{3}^8 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (\sqrt[3]{3}^4)^2 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (3\sqrt[3]{3})^2 \\ \end{align*} Since \(3\sqrt[3]{3} > 4\) we have \begin{align*} && \frac{\ln (3 \sqrt[3]3)}{3 \sqrt[3]3} &< \frac{\ln 4}{4} \\ &&&= \frac{\ln 2}{2}\\ \Rightarrow && (3 \sqrt[3]{3})^2 &< 2^{3 \sqrt[3]{3}} \\ \Rightarrow && \sqrt[3]3^8 &< 8^{\sqrt[3]3} \end{align*}

Solution:

1. Claim: If \(n \in \mathbb{Z}\) then \(\lfloor x + n \rfloor = \lfloor x \rfloor + n\) Proof: Since \(\lfloor x \rfloor \leq x\) then \(\lfloor x \rfloor + n \leq x + n\) and \(\lfloor x \rfloor + n \in \mathbb{Z}\) we must have that \(\lfloor x \rfloor +n \leq \lfloor x + n \rfloor\). However, since \(\lfloor x \rfloor + 1 > x\) we must also have that \(\lfloor x \rfloor + 1 + n > x + n\), therefore \(\lfloor x \rfloor + n\) is the largest integer less than \(x + n\) as required.
2. 1. Claim: \(f_n\left(x + \frac{1}{n}\right) = f_n(x)\) Proof: \begin{align*} f_n\left(x + \frac{1}{n}\right) &=\left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{1}{n}+ \frac{1}{n} \right\rfloor + \left\lfloor x+ \frac{1}{n} + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x+ \frac{1}{n} + \frac{n-1}{n} \right\rfloor - \left \lfloor n\left ( x + \frac{1}{n} \right) \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \left\lfloor x+ \frac{n}{n} \right\rfloor - \left \lfloor nx + 1 \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \left\lfloor x+ 1 \right\rfloor - \left \lfloor nx + 1 \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \lfloor x \rfloor + 1 - \left ( \lfloor nx \rfloor + 1 \right) \\ &= \lfloor x \rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x + \frac{n-1}{n} \right\rfloor - \lfloor nx \rfloor \\ &= f_n(x) \end{align*}
   2. Suppose \(0 \leq t < \frac1n\), then note that \(\left \lfloor t + \frac{k}{n} \right \rfloor = 0\) for \(0 \leq k \leq n - 1\) and \(\lfloor n t \rfloor = 0\) since \(nt < 1\)
   3. Since \(f_n(x)\) is zero on \([0, \tfrac1n)\) and periodic with period \(\tfrac1n\) it must be constantly zero
3. 1. Claim: \(\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor = \lfloor x \rfloor\) Proof: Suppose \(x = n + \epsilon\) where \(0 \leq \epsilon < 1\), ie \(n = \lfloor x \rfloor\), then consider two cases: Case 1: \(n = 2k\) \begin{align*} \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor &= \left\lfloor \frac{n + \epsilon}{2} \right\rfloor + \left\lfloor \frac{n + \epsilon+1}{2} \right\rfloor \\ &= \left\lfloor \frac{2k + \epsilon}{2} \right\rfloor + \left\lfloor \frac{2k + \epsilon+1}{2} \right\rfloor \\ &= k + \left\lfloor \frac{\epsilon}{2} \right\rfloor + k + \left\lfloor \frac{\epsilon+1}{2} \right\rfloor \\ &= 2k \\ &= n \end{align*} Case 2: \(n = 2k + 1\) \begin{align*} \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor &= \left\lfloor \frac{n + \epsilon}{2} \right\rfloor + \left\lfloor \frac{n + \epsilon+1}{2} \right\rfloor \\ &= \left\lfloor \frac{2k +1+ \epsilon}{2} \right\rfloor + \left\lfloor \frac{2k +1+ \epsilon+1}{2} \right\rfloor \\ &= k + \left\lfloor \frac{\epsilon+1}{2} \right\rfloor + k +1+ \left\lfloor \frac{\epsilon}{2} \right\rfloor \\ &= 2k +1\\ &= n \end{align*} as required.
   2. Since \(\left \lfloor \frac{x+1}{2} \right \rfloor = \lfloor x \rfloor - \lfloor \frac{x}{2} \rfloor\) and in general, \(\left \lfloor \frac{x+2^k}{2^{k+1}} \right \rfloor = \lfloor \frac{x}{2^k} \rfloor - \lfloor \frac{x}{2^{k+1}} \rfloor\) and so in general: \begin{align*} \sum_{k=0}^\infty \left \lfloor \frac{x+2^k}{2^{k+1}} \right \rfloor &= \sum_{k=0}^\infty \left ( \left \lfloor \frac{x}{2^k} \right \rfloor -\left \lfloor \frac{x}{2^{k+1}} \right \rfloor \right) \\ &= \lfloor x \rfloor \end{align*}

Solution:

1. \begin{align*} && I &= \int_0^{\infty} \frac{\sqrt{x}-1}{\sqrt{x(x^3+1)}} \, dx \\ x = u^{-1}, \d x = -u^{-2} \d u: && &= \int_{u=\infty}^{u = 0} \frac{u^{-1/2} - 1}{\sqrt{u^{-1}(u^{-3}+1)}} (-u^{-2}) \d u \\ &&&= \int_0^\infty \frac{u^{-1/2} -1}{\sqrt{1+u^3}} \d u \\ &&&= \int_0^\infty \frac{1-\sqrt{u}}{\sqrt{u(u^3+1)}} \d u \\ &&&= -I \\ \Rightarrow && I &= 0 \end{align*}
2. \begin{align*} && I &= \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx \\ x = u^{-2}, \d x = -2u^{-3} \d u: && &= \int_{u = \infty}^{u = 0} \frac{1}{\sqrt{u^{-6}+1}} (-2u^{-3}) \d u \\ &&&= \int_0^\infty \frac{2}{\sqrt{1+u^6}} \d u \\ &&&= 2\int_0^{\infty} \frac{1}{\sqrt{x^6+1}} \, \d x \end{align*}
3. \begin{align*} && I &= \int_0^{\infty} \frac{x^r - 1}{\sqrt{x^s(x^p+1)}} \, dx \\ x = u^{-1}, \d x = - u^{-2} \d t: &&&= \int_{u=\infty}^{u = 0} \frac{u^{-r}-1}{\sqrt{u^{-s}(u^{-p}+1)}} (- u^{-2}) \d u \\ &&&= \int_0^\infty \frac{u^{-r}-1}{\sqrt{u^{4-s}(u^{-p}+1)}} \d u \\ &&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r}(u^{-p}+1)}} \d u \\ &&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r-p}(1+u^p)}} \d u \\ \end{align*} Therefore if \(4-s+2r-p = s\) or \(r = \frac{p}2+s-2\) we have \(I = -I\), ie \(I = 0\).
4. \begin{align*} && I &= \int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx \\ x = u^{-t}, \d x = -t u^{-t-1}:&&&= \int_{u = -\infty}^{u = 0} \frac{1}{\sqrt{u^{-pt}+1}} (-t u^{-t-1}) \d u \\ &&&= t \int_0^\infty \frac1{\sqrt{u^{-pt+2t+2}+u^{2t+2}}} \d u \end{align*} Therefore if \(\begin{cases} 2t+2 &= q \\ 2t+2 -pt &= 0 \end{cases}\), ie \(q = 2t+2, p = 2 + 2/t\) we will have found the \(p, q\) desired for any \(t\) (or \(k\)). [Alternatively] Let \(\displaystyle I(p) =\int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx\), then clearly \(I(p)\) is decreasing and as \(p \to \infty\) \(I(p) \to 1\), so our integral can take any values on \((1, \infty)\) and so for any positive value we can find two values with a given ratio. In particular given \(k\) and \(p\) we can find a suitable \(q\).

Solution:

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   By symmetry we can observe that the parabola and circle will intersect \(0, 1\) (at the base), \(2, 4\) times. So setting up our system of equations we have: \begin{align*} &&& \begin{cases} x^2 + (y-a)^2 &= r^2 \\ 2ky &= x^2 \end{cases} \\ \Rightarrow && r^2 &= x^2 + \left (\frac{x^2}{2k} - a \right )^2 \\ \Rightarrow &&r^2 &= x^2 + a^2 - \frac{ax^2}{k} + \frac{x^4}{4k^2} \\ \Rightarrow &&0 &= \frac{1}{4k^2} x^4 + \left ( 1 - \frac{a}{k} \right) x^2 + a^2 - r^2 \\ \Rightarrow && \Delta &= \left ( 1 - \frac{a}{k} \right)^2-4 \cdot \frac{1}{4k^2} (a^2 - r^2) \\ &&&= 1 - \frac{2a}{k} + \frac{a^2}{k^2} - \frac{a^2}{k^2} + \frac{r^2}{k^2} \\ &&&= \frac{k^2-2ka+r^2}{k^2} \end{align*} Since there will be (at most) two solutions if \(\Delta = 0\) we must have if the circle and parabola are tangent \(r^2 - 2ka + k^2 = 0 \Rightarrow r^2 = k(2a-k)\). So long as there is a solution \(x^2 > 0\) there will be two tangent points, so if \(-\left(1 - \frac{a}{k}\right) > 0\) or \(a > k > 0\)
2. Since \(y = c \pm x\) are tangent to the circle with radius \(r\) and centre \((0,a)\) we have the following equations: \begin{align*} &&& \begin{cases} x^2 + (y-a)^2 &= r^2 \\ c \pm x &= y \end{cases} \\ \Rightarrow && r^2 &= x^2 + (c -a\pm x)^2 \\ &&&= 2x^2+(c-a)^2 \pm 2x(c-a) \\ \Rightarrow && \Delta &= 4(c-a)^2 -4 \cdot 2 \left ( (c-a)^2 -r^2 \right)\\ &&&= 8r^2-4(c-a)^2 \\ \Rightarrow && x &= \frac{\mp 2(c-a) \pm \sqrt{\Delta}}{4} \\ &&&= \mp \frac12 (c-a) \\ && y &= \pm \frac12 (c+a) \\ && (x,y) &= \left (\frac12 (c-a), \frac12 (c+a)\right), \left (-\frac12 (c-a), -\frac12 (c+a)\right) \end{align*}

Solution: Let \(v = \frac{\d x}{\d t}\) and notice that \(\frac{\d}{\d t} \left ( \frac{\d x}{\d t} \right) = \frac{\d }{\d x} \left ( v \right) \frac{\d x}{\d t} = v \frac{\d v}{\d x}\). Also notice that: \begin{align*} && v \frac{\d v}{\d x} &= 2x v \\ \Rightarrow && \frac{\d v}{\d x} &= 2x \\ \Rightarrow && v &= x^2 + C \\ \Rightarrow && \frac{\d x}{\d t} &= x^2 + C \\ \end{align*}

1. When \(t = 0, \frac{\d x}{\d t} = a^2\) so \(C = 0\), therefore \(\frac{\d x}{\d t} = x^2 \Rightarrow t = -x^{-1} + k\) and so \(k = a^{-1}\) and \(x = \frac{a}{1-at}\). As \(t\) increases from \(0\) the particle heads to infinity at an increasing rate, `reaching' infinity around \(t=\frac{1}{a}\)
2. When \(t = 0, \frac{\d x}{\d t} = a^2 + p\) so \(C = p\). Therefore \(\frac{\d x}{\d t} = x^2 + p \Rightarrow t = \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) + c\). When \(c = - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right)\), so \begin{align*} && t &= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right) \\ &&&= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} \right) \\ \Rightarrow && \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} &= \tan (\sqrt{p} t) \\ \Leftrightarrow && \sqrt{p}(x-a) &= \tan (\sqrt{p} t)(\sqrt{p}-ax) \\ \Leftrightarrow && x(\sqrt{p}+a\tan (\sqrt{p} t)) &= \sqrt{p} (\tan(\sqrt{p}t) + a) \\ \Leftrightarrow && x &= \frac{\sqrt{p} (\tan(\sqrt{p}t) + a)}{\sqrt{p}+a\tan (\sqrt{p} t)} \end{align*} The particle heads to \(\frac{\sqrt{p}}{a}\).
3. When \(t = 0, \frac{\d x}{\d t} = a^2-q^2\) so \(C = -q^2\). Therefore \begin{align*} && \frac{\d x}{\d t} &= x^2 -q^2 \\ \Rightarrow && \int \d t &= \int \frac{1}{(x-q)(x+q)} \d x \\ &&&= \frac{1}{2q} \int \left ( \frac{1}{x-q}- \frac{1}{x+q} \right )\d x \\ &&&= \frac{1}{2q} \left ( \ln (x-q) - \ln(x+q) \right) \\ &&&= \frac{1}{2q} \ln \left ( \frac{x-q}{x+q} \right)\\ \Rightarrow && \frac{x-q}{x+q} &= Ae^{2qt} \\ \underbrace{\Rightarrow}_{t= 0} && A &= \frac{a-q}{a+q} \\ \Rightarrow && x-q &= \frac{a-q}{a+q}e^{2qt}(x+q) \\ \Leftrightarrow && x\left (1-\frac{a-q}{a+q}e^{2qt} \right) &= q\left (1 + \frac{a-q}{a+q}e^{2qt} \right) \\ \Leftrightarrow && x &= q \frac{1 + \frac{a-q}{a+q}e^{2qt}}{1-\frac{a-q}{a+q}e^{2qt}} \end{align*}

Solution:

1. \(A = \{1,4,6,7 \}, B = \{2,3,5,8 \}\)
2. \begin{align*} && \sum_{k=1}^m a_k^3 + \sum_{k=1}^m (c + b_k)^3 &= \sum_{k=1}^m a_k^3 + mc^3 + 3c^2 \sum_{k=1}^m b_k + 3c \sum_{k=1}^m b_k^2 + \sum_{k=1}^m b_k^3 \\ &&&= \sum_{k=1}^m a_k^3 + mc^3 + 3c^2 \sum_{k=1}^m a_k + 3c \sum_{k=1}^m a_k^2 + \sum_{k=1}^m b_k^3 \\ &&&= \sum_{k=1}^m a_k^3 + 3c \sum_{k=1}^m a_k^2+ 3c^2 \sum_{k=1}^m a_k + mc^3 + \sum_{k=1}^m b_k^3 \\ &&&= \sum_{k=1}^m (a_k+c)^3 + \sum_{k=1}^m b_k^3 \end{align*}
3. Note that \begin{align*} && \sum_{k=1}^m a_k^2 + \sum_{k=1}^m (c + b_k)^2 &= \sum_{k=1}^m a_k^2 + mc^2 + 2c \sum_{k=1}^m b_k +\sum_{k=1}^m b_k^2 \\ &&&= \sum_{k=1}^m (a_k+c)^2 + \sum_{k=1}^m b_k^2 \end{align*} Therefore if we take our balanced subsets of \(\{1,2,3,4,5,6,7,8\}\) and take \(A \cup (B+8)\) and \(B \cup (A+8)\) they will be balanced (including the cubes) so: \(A = \{1,4,6,7,10,11,13,16\}, B = \{2,3,5,8,9,12,14,15\}\)
4. Notice that: \begin{align*} \sum_{a \in A} (n+a)^2 &= |A|n^2 + 2n \sum_{a \in A} a + \sum_{a \in A} a^2 \\ &= 6 n^2 + 2n \sum_{b \in B}b + \sum_{b \in B} b^2 \\ &= n^2 + \sum_{b \in B}(n+b)^2 \\ \end{align*} Therefore consider the sets \(C = \{(n+i)^2 : i \in A\}, D = \{n^2\} \cup \{(n+i)^2 : i \in B\}\)

Solution:

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1. Assuming the particles have mass \(m\), and speed \(v\) just before collision, then \begin{align*} \text{COE}: && \underbrace{\frac12 m u^2}_{\text{initial kinetic energy}} + \underbrace{0}_{\text{initial GPE}} &= \underbrace{\frac12m v^2}_{\text{kinetic energy just before collision}} + \underbrace{mgr\left(1-\frac1{\sqrt{2}}\right)}_{\text{GPE just before collision}} \\ \Rightarrow && v^2 &= u^2 - gr(2-\sqrt{2}) \end{align*} Therefore the particles has velocity \(\frac{\sqrt{2}}2v \binom{\pm 1}{1}\) before the collision. By symmetry, the impulse between the particles will be horizontal, so the vertical velocities will be unchanged at \(\frac{\sqrt{2}}{2}v\). By conservation of momentum (or symmetry) the particles will have equal but opposite velocities after the collision (say \(w\)) satisfying: \[ e = \frac{2w}{2\frac{\sqrt{2}}{2}v} \] ie \(w = \frac{\sqrt{2}}2 e v\) as required.
2. Once the particles have rebounded, they will be projectiles whilst the strings are slack. If we consider the left-most point \(A = (0,0)\) then the particles colide at \(\left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\) and the position at time \(t\) after the collision (before the string goes slack) will be: \begin{align*} \mathbf{x}_t &= \frac{\sqrt{2}}{2}r\binom{1}{-1} + \frac{\sqrt{2}}{2} vt \binom{-e}{1} + \frac12 gt^2 \binom{0}{-1} \end{align*} The string will go taught when \(|\mathbf{x}_t|^2 = r^2\), ie \begin{align*} && r^2 &= \left ( \frac{\sqrt{2}}{2} r - \frac{\sqrt{2}}{2}evt \right)^2 + \left (-\frac{\sqrt{2}}{2} r + \frac{\sqrt{2}}{2}vt -\frac12 gt^2 \right)^2 \\ \Rightarrow && r^2 &= \frac12 \left (r - evt \right)^2 + \frac12 \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\ \Rightarrow && 2r^2 &= \left (r - evt \right)^2 + \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\ \end{align*} as required.
3. Suppose \(z = \frac{vt}{r}\), \(c = \frac{\sqrt{2}v^2}{rg}\), then \begin{align*} && 2r^2 &= \left (r - evt \right)^2 + \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\ \Leftrightarrow && 2 &= \left (1 - e\frac{vt}{r} \right)^2 + \left (-1 + \frac{vt}{r}- \frac{\sqrt{2}}{2} \frac{gt^2}{r} \right)^2 \\ \Leftrightarrow && 2 &= \left (1 - ez \right)^2 + \left (-1 +z- \frac{v^2t^2}{r^2} \frac{gr}{\sqrt{2}v^2}\right)^2 \\ \Leftrightarrow && 2 &= \left (1 - ez \right)^2 + \left (-1 +z- \frac{z^2}{c} \right)^2 \\ \Leftrightarrow && 2 &= 1-2ez + e^2z^2 + 1 + z^2 +\frac{z^4}{c^2} - 2z-2\frac{z^3}{c}+2\frac{z^2}{c} \\ \Leftrightarrow && 0 &= z(-2e-2) + z^2(e^2+1 + \frac{2}{c}) + z^3(-\frac{2}{c}) + z^4 \frac{1}{c^2} \\ \underbrace{\Leftrightarrow}_{z \neq 0} && 0 &= \left ( \frac{z}{c} \right)^3 - 2\left ( \frac{z}{c} \right)^2 + \left ( \frac{z}{c} \right) (1 + e^2 + \frac{2}{c} ) - \frac{2}{c}(1+e) \end{align*} as required, (where on the last step we divide by \(z/c\)).
4. If a cubic has \(3\) equal, non-zero roots then it must have the form \((z-a)^3 = z^3 -3az^2 + 3a^2 z -a^3 = 0\), so \(3a = 2\), and so the expansion must be \(\left ( \frac{z}{c} \right)^3 - 2\left ( \frac{z}{c} \right)^2 + \frac{4}{3}\left ( \frac{z}{c} \right) - \frac{8}{27} = 0\) \begin{align*} && \frac{2}{c}(1+e) &= \frac{8}{27} \\ \Rightarrow && \frac{2}{c} &= \frac{8}{27} \frac{1}{1+e} \\ && 1 + e^2 + \frac{2}{c} &= \frac43 \\ \Rightarrow && e^2 + \frac{8}{27(1+e)} &= \frac{1}{3} \\ \Rightarrow && 27(1+e)e^2+8 &= 9(1+e) \\ \Rightarrow && 27e^3 + 27e^2-9e-1 &= 0 \\ \Rightarrow && (3e-1)(9e^2+12e+1) &= 0 \end{align*} The only (positive) root is \(e = \frac13\), therefore \(e = \frac13\). We must also have \begin{align*} && \frac{2}{c} \frac43 &= \frac{8}{27} \\ \Rightarrow && c &= 9 \\ \Rightarrow && \frac{\sqrt{2}v^2}{rg} &= 9 \\ \Rightarrow && v^2 &= \frac{9\sqrt{2}rg}{2} \end{align*} as required. If we consider the path of the particle acting as a projectile, iff the path is tangent to the circle then there will be exactly one solution for \(z/c\) and (importantly) it will be a repeated root. Therefore the particle rejoins the circle at a tangent and the tension is acting perpendicularly to the direction of motion (ie no energy loss).
5. Since the only energy lost is lost in the collision, we can apply conservation of energy again: \begin{align*} \text{COE:} && \frac12 m U^2 &= \frac12 m \frac12v^2(1+e^2) + mgr\left (1 - \frac1{\sqrt{2}} \right) \\ \Rightarrow && U^2 &= \frac12 \frac{9 \sqrt{2}}{2}gr(1+\frac19) + gr(2 - \sqrt{2}) \\ &&&= \left (\frac{5\sqrt{2}}{2}+2 - \sqrt{2} \right)gr \\ &&&= \left (\frac{4+3\sqrt{2}}{2} \right)gr \end{align*}

Solution:

+ - ↺

1. Notice that \(AM = MB = MP\) in particular \(P\) lies on a semi-circle of radius \(a\) and therefore by Thales' theorem \(\angle APB = 90^{\circ}\). Notice that by angles in a triangle and the angles adding to \(90^{\circ}\), \(\angle APM = \theta\). Therefore, \begin{align*} && |PB| &= 2a \sin \theta \\ && |PA| &= 2a \cos \theta \\ && T_A &= \frac{\lambda}{l} \left (2a \cos \theta -l \right) \\ && T_B &= \frac{\lambda}{l} \left (2a \sin\theta -l \right) \\ \end{align*} Since \(T_A\) and \(T_B\) are perpendicular, we can consider the forces as having vector \(\frac{\lambda}{l}\binom{2a\cos \theta-l}{2a\sin \theta - l}\) in this coordinate system, ie the sum of a vector \(\frac{2\lambda a}{l}\binom{\cos \theta}{\sin \theta}\) and \(\displaystyle -\sqrt{2}\lambda \binom{\frac1{\sqrt{2}}}{\frac1{\sqrt{2}}}\) which are unit vectors parallel to the rod and along the bisector of \(APB\) respectively.
2. \begin{align*} \overset{\curvearrowright}{M}: && 0 &= \frac{a}{2} \cdot mg \cos 2 \theta - a\cdot \sqrt{2}\lambda \cos (90-(45-\theta))\\ \Rightarrow && \cos 2 \theta &= \frac{\lambda}{mg} 2 \sqrt{2} \cos (45 + \theta) \\ \Rightarrow && \cos^2 \theta - \sin^2 \theta &= \frac{2\lambda}{mg} (\cos \theta - \sin \theta) \\ \underbrace{\Rightarrow}_{\theta < 45^{\circ}} && \cos \theta + \sin \theta &= \frac{2\lambda}{mg} \end{align*} Over \((0, 45^{\circ})\), \(\cos \theta + \sin \theta\) ranges from \(1\) to \(\sqrt{2}\), therefore \(1 < \frac{2 \lambda}{mg} < \sqrt{2} \Rightarrow \frac12 mg < \lambda < \frac12 \sqrt{2} mg\) as required.
3. \begin{align*} \overset{\curvearrowright}{P}: && 0 &=- \frac{a}{2} \cdot \left ( mg \cos 2\theta \right) - a \cdot F \sin 2 \theta + a \cdot N \cos 2 \theta \\ \Rightarrow && \frac12 mg &= N - F \tan 2 \theta \end{align*} as required.