Problems

---

Solution: Since \(\omega^3 = 1\) and \(\omega \neq 1\) we must have that \((\omega-1)(1 + \omega + \omega^2) = 0\) but by dividing by \(\omega - 1\) we obtain the desired result. \(1+\omega^2 = -\omega\) so \(|1 + \omega^2| = |-\omega| = 1\) and \(\arg ( 1 + \omega^2) = \arg(-\omega) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}\) \begin{align*} && (1 + 1)^n &= \sum_{k=0}^n \binom{n}{k}\\ && (1+ \omega)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{k} \\ && (1+ \omega^2)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{2k} \\ \Rightarrow && 2^n+(-\omega^2)^n + (-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n (1+1+1)\binom{n}{k} + \sum_{k=0, k \equiv 1 \pmod{3}}^n (1 + \omega + \omega^2) \binom{n}{k} + \sum_{k=0, k \equiv 2 \pmod{3}}^n (1 + \omega^2 + \omega) \binom{n}{k} \\ \Rightarrow && 2^n +((-\omega)^n)^{-1}+(-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n \binom{n}{k} \end{align*} \(2^n +((-\omega)^n)^{-1}+(-\omega)^n = 2^n + 2 \textrm{Re}(-\omega^n) = 2^n + 2 \cos \frac{n\pi}{3}\) Therefore our answer follows. \begin{align*} \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} &= \frac13 \l 2^{25} + 2\cos (\frac{25 \pi}{3}) \r \\ &= \frac13 \l 2^{25} + 2 \cos \frac{\pi}{3} \r \\ &= \frac13 \l 2^{25} + 1 \r \\ &= \frac13 \l (4096 \cdot 4096 \cdot 2) + 1 \r \\ &= 11\,184\,811 \end{align*} Notice that \(S_2 = \binom{24}{2} + \cdots +\binom{24}{23} = \binom{24}{1} + \cdots + \binom{24}{22} = S_1\) and \(S_0 = \binom{24}0 + \cdots + \binom{24}{21} = \frac13 \l 2^{24} + 2 \r\) Therefore since \(S_0 + 2 \cdot S_2 = 2^{24}\) we must have \begin{align*} S_2 &= \frac12 \l 2^{24} - \frac13 \l 2^{24} + 2 \r \r \\ &= \frac13 \l 2^{24} - 1 \r \\ &= \frac13 \l 16777216- 1 \r \\ &= \frac13 \cdot 16777215 \\ &= 5\,592\,405 \end{align*}

---

Solution:

1. Suppose we have \(y = x^k\), then the tangent at \((t,t^k)\) has gradient \(\frac{\d y}{\d x} = kx^{k-1} = kt^{k-1}\) and the normal has gradient \(-\frac1k x^{1-k}\). For the push-off to be the positive \(x\)-axis, we need \(p(x)\) to be the length of the line. The line will have the equation: \begin{align*} && \frac{y - t^k}{x - t} &= -\frac1k t^{1-k} \\ y = 0: && x-t &= \frac{kt^k}{t^{1-k}} \\ && x& =t + kt^{2k-1} \end{align*} The distance from \((t,t^k)\) to \((t+kt^{2k-1},0)\) is \(\sqrt{t^{2k} + k^2t^{4k-2}} = t^k \sqrt{1+k^2t^{2k-2}} = y \sqrt{1+k^2 \frac{y^2}{x^2}} = \frac{y}{x} \sqrt{x^2 + k^2y^2}\), ie \(p(x) = \frac{y}{x}\sqrt{x^2+k^2y^2}\) If \(\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|\), then we need \(\sqrt{x^2+y^2} = \frac{y}{x}\sqrt{x^2+ky^2}\), but clearly this is satisfied when \(k = 1\).
2. The points are \((t, t^2)\) and the normal has graident \(-\frac{1}{2t}\), the normal vector is \(\frac{1}{\sqrt{1+\frac{1}{4t^2}}}\binom{1}{-\frac1{2t}} = \frac{1}{\sqrt{4t^2+1}} \binom{2t}{-1}\). If we write \(p(x) = q(x)\sqrt{4x^2+1}\) then the the new points are at \(\binom{t+q(t)2t}{t^2-\q(t)}\) an the gradient will be \(\frac{2t-q'(t)}{1+q'(t)2t+2q(t)}\). We need it to be the case that \begin{align*} && 2t &= \frac{2t-q'(t)}{1+2q(t)+2tq'(t)} \\ \Rightarrow && 4tq(t) &= -q'(t)(1+4t^2) \\ \Rightarrow && \frac{q'(t)}{q(t)} &= -\frac{4t}{1+4t^2} \\ \Rightarrow && \ln q(t) &= -\frac12 \ln(1+4t^2)+C \\ \Rightarrow && q(t) &= A(1+4t^2)^{-1/2} \\ \Rightarrow && p(x) &= A \end{align*} So the push-off's are constants.

---

Solution: Since the common difference is \(a_2 - a_1\) we can find that \(a_n = a_1 + (n-1)(a_2-a_1)\), then \begin{align*} && a_1 &&+&& a_2 &&+&& \cdots &&+&& (a_1 + (n-2)(a_2 - a_1) && + && (a_1 + (n-1) (a_2 - a_1)) \\ + && (a_1 + (n-1) (a_2 - a_1))&&+&& (a_1 + (n-2)(a_2 - a_1)&&+&& \cdots &&+&& a_2 && + && a_1 \\ \hline \\ = && 2a_1 + (n-1)(a_2 - a_1) && + && 2a_1 + (n-1)(a_2 - a_1) && + && \cdots && + 2a_1 + (n-1)(a_2 - a_1) && + 2a_1 + (n-1)(a_2 - a_1) \\ = && n(2a_1 + (n-1) (a_2 - a_1)) \end{align*} Therefore the sum is \(a_1 n + \frac{n(n-1)}{2} (a_2 - a_1)\). Since \(b_n - b_1 = (b_n - b_{n-1}) + (b_{n-1} - b_{n-2}) + \cdots (b_2 - b_1)\), \(b_n - b_1 = a_1 (n-1) + \frac{(n-1)(n-2)}{2}(a_2 - a_1) = (b_2-b_1)(n-1) + \frac{(n-1)(n-2)}{2}(b_3 -2b_2 +b_1)\). So \(b_n = b_1 + (b_2 - b_1)(n-1) + \frac{(n-1)(n-2)}{2} (b_3 - 2b_2 + b_1)\). In particular \begin{align*} \sum_{i=1}^n b_i &= \sum_{i=1}^n \l b_1 + (b_2 - b_1)(n-1) + \frac{(n-1)(n-2)}{2} (b_3 - 2b_2 + b_1)\r \\ &= nb_1 + (b_2-b_1) \frac{n(n-1)}{2} + \frac{n(n-1)(n-2)}{6}(b_3-2b_2+b_1) \end{align*} Let \(b_2 - b_1 = x\) and \(b_3 - 2b_2+b_1 = y\), then \begin{align*} b_4 - b_2 &= d &= &2x + 3y \\ b_6-b_4 &= 2d &=& 2x +(10-3)y \\ &&=&2x + 7y \\ 220-b_6&=6d &=& 220-(1 + 5x + 10y) \\ \end{align*} \begin{align*} && 4x + 6y &= 2x + 7y \\ && 6x+21y &= 219-5x-10y \\ \Rightarrow && 2x - y &= 0 \\ && 11x + 31y &= 219 \\ \Rightarrow && x &= 3 \\ && y &= 6 \end{align*} Therefore the final sum is \begin{align*} n + 3 \frac{n(n-1)}{2} + 6 \frac{n(n-1)(n-2)}{6} &= n^3-\frac32n^2+\frac32n \end{align*}

---

Solution:

1. \begin{align*} \int_{1}^{3}\frac{1}{6x^{2}+19x+15}\,\mathrm{d}x &= \int_1^3 \frac1{(2x+3)(3x+5)} \d x \\ &= \int_1^3 \l \frac{2}{2x+3} - \frac{3}{3x+5} \r \d x \\ &= \left [\ln(2x+3) - \ln(3x+5) \right ]_1^3 \\ &= \l \ln9 - \ln14 \r - \l \ln 5 - \ln 8 \r \\ &= \ln \frac{72}{70} \\ &= \ln \frac{36}{35} \end{align*}
2. 

   + - ↺
   When \(q = 0\) the roots are \(-1, 0, 1\) There can be \(0, 2, 3, 4\) roots. There will be no roots if \(q > -\min (x^{1760} - x^{220})\) since the whole graph will be above the axis. There will be \(2\) roots if \(q = -\min (x^{1760} - x^{220})\) or \(q > 0\) There will be \(4\) roots if \(0 > q > -\min (x^{1760} - x^{220})\). There will be \(3\) roots if \(q =0\)

---

Solution: Set up coordinates such that \(A\) at the origin and \(\vec{AB} = \mathbf{x}\) and \(\vec{AD} = \mathbf{y}\) and so \(\vec{AC} = \mathbf{x}+\mathbf{y}\)

1. \begin{align*} AC^2 + BD^2 &= (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) + (\mathbf{y}-\mathbf{x})\cdot(\mathbf{y}-\mathbf{x}) \\ &= 2\mathbf{x}\cdot\mathbf{x} + 2\mathbf{y}\cdot\mathbf{y} \\ &= AB^2 + CD^2 +AD^2 + BC^2 \end{align*}
2. \begin{align*} AC^2 -BD^2 &= (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) - (\mathbf{y}-\mathbf{x})\cdot(\mathbf{y}-\mathbf{x}) \\ &= 4 \mathbf{x}\cdot \mathbf{y} \end{align*} \(\mathbf{x}\cdot\mathbf{y} = |\mathbf{x}||\mathbf{y}|\cos \theta\) which is exactly the lenth of one side mutliplied by the length of the projection to that same side.

\begin{align*} AB^2 - AD^2 &= \mathbf{x}\cdot\mathbf{x} - \mathbf{y}\cdot \mathbf{y} \\ &= (\mathbf{x}+\mathbf{y})\cdot(\mathbf{x}-\mathbf{y}) \\ &= AC \cdot BD \end{align*} So this is the area of the rectangle formed by the length of one diagonal and the projection of the other diagonal onto it.

---

Solution: Suppose \(y = uv\) then and suppose \(\frac{\d u}{\d x} + P u = 0\) then \begin{align*} && \frac{\d y}{\d x} + Py &= Q \\ && uv' + u'v + Puv &= Q \\ && uv' &= Q \\ && \frac{\d v}{\d x} &= \frac{Q}{u} \end{align*} Consider \begin{align*} && 0 &= \frac{\d u}{\d x} - \frac{2u}{x+1} \\ \Rightarrow && \ln u &= 2\ln (1 + x) + C \\ \Rightarrow && u &= A(1+x)^2 \end{align*} and \begin{align*} && \frac{\d v}{\d x} &= \frac1{A}(x+1)^{\frac12} \\ \Rightarrow && v &= \frac2{3A}(x+1)^{\frac32} + k \\ \Rightarrow && y &= \frac23(x+1)^\frac72 + k(x+1)^2 \\ && 0 &= y(1) \\ &&&= \frac23 2^{7/2}+k2^2 \\ \Rightarrow && k &= -\frac{2^{5/2}}{3} \\ \Rightarrow && y &= \frac23 (x+1)^{7/2} - \frac{2^{5/2}}{3}(x+1)^2 \end{align*}

---

Solution: \begin{align*} &&\sin \alpha &= 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \\ &&&= 4 \sin \frac{\alpha}{4} \cos \frac{\alpha}{4} \cos \frac{\alpha}{2} \\ \Rightarrow && \frac{\sin \alpha}{4 \sin \frac{\alpha}{4}} &= \cos \frac{\alpha}{2} \cos \frac{\alpha}{4} \end{align*} We proceed by induction on \(n\). Clearly this is true for \(n = 1\) (as we just established). Assume it is true for \(n=k\). Then: \begin{align*} && \frac{\sin \alpha}{2^n \sin \frac{\alpha}{2^n}} &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\ \Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} \cos \frac{\alpha}{2^{n+1}}} &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\ \Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} } &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)\cos \left ( \frac{\alpha}{2^{n+1}} \right) \\ \end{align*} Therefore it is true for \(n=k+1\) Therefore since it is true for \(n=1\) and if it is true for \(n=k\) it is also true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 1\) \begin{align*} \lim_{n \to \infty} \frac{\sin \alpha}{\cos\left(\frac{\alpha}{2}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)} &= \lim_{n \to \infty} 2^n \sin \frac{\alpha}{2^n} \\ &= \lim_{n \to \infty} \alpha \frac{\sin \frac{\alpha}{2^n}}{\frac{\alpha}{2^n}} \\ &= \alpha \lim_{t \to 0} \frac{\sin t}{t} \\ &= \alpha \end{align*} When \(\alpha = \frac{\pi}{2}\) notice that \(\sin \alpha =1\), \(\cos \frac{\alpha}{2} = \sqrt{\frac12}\) and \(2\cos^2 \frac{\alpha}{2^{n+1}}-1 = \cos \frac{\alpha}{2} \Rightarrow \cos \frac{\alpha}{2^{n+1}} = \sqrt{\frac12 + \cos \frac{\alpha}{2^n}}\) exactly the series we see.

---

Solution:

\begin{align*} && y &= 1/x \\ \Rightarrow && \frac{\d y}{\d x} &= -1/x^2 \end{align*} Therefore the tangent at \((1,1)\) will be \(\frac{y - 1}{x-1} = -1 \Rightarrow y = -x + 2\) and at \((b, 1/b)\) will be \(\frac{y-1/b}{x-b} = -\frac{1}{b^2} \Rightarrow y = -\frac{x}{b^2} + \frac{2}{b}\) The intersection will be at \begin{align*} && x + y & = 2 \\ && x + b^2 y &= 2b \\ \Rightarrow && (b^2-1)y &= 2(b-1) \\ \Rightarrow && y &= \frac{2}{b+1} \\ && x &= \frac{2b}{b+1} \end{align*} Therefore \(\displaystyle C = \left (\frac{2b}{b+1}, \frac{2}{b+1} \right)\). The areas of the two trapeziums will be: \begin{align*} [ACC'A'] &= \frac12 \left (1 + \frac{2}{b+1} \right) \left (\frac{2b}{b+1} - 1 \right) \\ &= \frac12 \cdot \frac{b+3}{b+1} \cdot \frac{2b - b - 1}{b+1} \\ &= \frac 12 \frac{(b+3)(b-1)}{(b+1)^2} \end{align*} \begin{align*} [CBB'C'] &= \frac12 \left (\frac{2}{b+1} + \frac{1}{b} \right) \left (b- \frac{2b}{b+1} \right) \\ &= \frac12 \cdot \frac{3b+1}{b(b+1)} \cdot \frac{b^2+b-2b}{b+1} \\ &= \frac 12 \frac{(3b+1)b(b-1)}{b(b+1)^2} \\ &= \frac12 \frac{(3b+1)(b-1)}{(b+1)^2} \end{align*} The area under the curve between \(A\) and \(B\) will be: \begin{align*} \int_1^b \frac{1}{x} \d x &= \left [\ln x \right]_1^b \\ &= \ln b \end{align*} The area of a rectangle of height \(1\) from \(A\) will clearly be above the curve and will have area \(b-1\). The area of \(ACBB'C'A'\) will be: \begin{align*} [ACBB'C'A'] &= [ACC'A']+[CBB'C'] \\ &=\frac 12 \frac{(b+3)(b-1)}{(b+1)^2}+ \frac12 \frac{(3b+1)(b-1)}{(b+1)^2} \\ &= \frac12 \frac{(b-1)(4b+4)}{(b+1)^2} \\ &= \frac{2(b-1)}{b+1} \end{align*} By comparing areas, we must have: \(\frac{2(b-1)}{b+1} \leq \ln b \leq b-1\) and since \(b > 1\) we can write it as \(1 + z\) for \(z >0\), ie: \(\displaystyle \frac{2z}{2+z} \leq \ln (1 + z) \leq z\). [By considering the area of \(ABB'A'\) which is \begin{align*} [ABB'A'] &= \frac12 \left (1 + \frac{1}{b} \right) \left ( b- 1 \right) \\ &= \frac12 \frac{(b+1)(b-1)}{b} \end{align*} we can tighten the right hand bound to \(\displaystyle \frac{(2+z)z}{2(z+1)} = \left (1 - \frac{z}{2z+2} \right)z\)

---

Solution: Note that \(P = Fv\). Since he is running at a steady pace, we can say that \(F\) must be equal to the resistive forces (as net force is \(0\)). Therefore \(F = 50 + 2(v+3)\) on the way out. ie, \(300 = (2v + 56)v \Rightarrow 150 = v^2 + 28v \Rightarrow v = \sqrt{346}-14\) On the way back, \(F = 50 + 2(v-3)\), ie \(300 = (2v+44)v \Rightarrow 150 = v^2 +22v \Rightarrow v = \sqrt{271}-11\) Therefore the total time will be \(\frac{5000}{\sqrt{346}-150} + \frac{5000}{\sqrt{271}-11} \approx 2002\), or 33 minutes, 22 seconds. Very respectable! The total energy in this first run is \(E = Pt = 2002 \cdot 300\). Now suppose we apply two different powers as in the question, then we must have: \begin{align*} && x_1 &= 2v_1^2 + 56v_1 \\ && x_2 &= 2v_2^2 + 44v_2 \\ && E &= x_1 \frac{5000}{v_1} + x_2 \frac{5000}{v_2} \\ &&&= 5000 \left ( \frac{x_1}{v_1} + \frac{x_2}{v_2} \right) \\ \Rightarrow && \frac{x_1}{v_1} &= 2v_1 + 56 \\ && \frac{x_2}{v_2} &= 2v_2 + 44 \\ \Rightarrow && \frac{E}{5000} &= 2(v_1+v_2) + 100 \\ \Rightarrow && v_1+v_2 &\text{ is independent of the choices for }x_i \end{align*} We wish to minimize \begin{align*} && \frac{5000}{v_1} + \frac{5000}{v_2} &\underbrace{\geq}_{AM-HM} 10\,000 \cdot \frac{2}{v_1+v_2} \\ &&&= 10\,000 \cdot \frac{2}{\sqrt{346}-14+\sqrt{271}-11} \\ &&&\approx 1987 \end{align*} ie they can go 15 seconds quicker with better strategy.

---

Solution: The particle has initial velocity \(\displaystyle \binom{v \cos \frac{\pi}{3}}{v \sin \frac{\pi}{3}}\) and acceleration \(\displaystyle \binom{0}{-g}\). It will have zero vertical speed (ie be at the top of its trajectory) when \(t = \frac{\sqrt{3}v}{2g}\). Since \(0 = v^2-u^2 + 2as\) the height achieved will be \(\frac{3v^2}{8g}\) At this point it will need to travel the same distance again, but this time the initial speed is \(\frac{3v}{2}\) so: \begin{align*} && \frac{3v^2}{8g} &= \frac{3v}{2} t + \frac12 g t^2 \\ \Rightarrow && 0 &= 4g^2t^2+12vgt - 3v^2 \\ \Rightarrow && t &= \l \frac{-3+2\sqrt{3}}{2} \r \frac{v}{g} \end{align*} Therefore the total time is: \begin{align*} \l \frac{\sqrt{3}}{2} - \frac32 + \sqrt{3} \r \frac{v}{g} &= \frac{3\sqrt{3}-3}{2}\frac{v}{g} \end{align*} \begin{align*} COM(\uparrow): && 0 &= 2m v_y - m \frac{3}{2}v \\ \Rightarrow &&v_y &= \frac34 v \\ COM(\rightarrow): && 3mV &= 2m (3v) +m \frac{v}{2} \\ \Rightarrow && V &= \frac{13}6\\ \end{align*} Therefore superman is now travelling at a vector of \(\displaystyle \binom{\frac{13}6}{\frac34}v\) ie an angle of \(\tan^{-1} \frac 9{26}\) to the horizontal, approximately \(19^\circ\)