Problems

Solution:

1. \begin{align*} (x+y+z)^3 &= x^3+y^3+z^3+ \\ &\quad 3xy^2 + 3xz^2 + 3yx^2 + \cdots + 3zy^2 \\ &\quad\quad + 6xyz \\ (x+y+z)(xy+yz+zx) &= x^2y+x^2z + \cdots + z^2 x + 3xyz \\ x^3+y^3+z^3 &= (x+y+z)^3 - 3(xy^2 + \cdots + zy^2) - 6xyz \\ &= \alpha^3 - 3(\alpha \beta - 3\gamma)-6\gamma \\ &= \alpha^3-3\alpha \beta+3\gamma \end{align*} Since \(4 = 1^3-3\cdot1\cdot(-1) + 3 \gamma \Rightarrow \gamma = 0\), therefore one of \(x,y,z = 0\). WLOG \(x = 0\), so \(y+z = 1, y^2 + z^2 = 3 \Rightarrow y^2 + (1-y)^2 = 3 \Rightarrow y^2 -y -1 = 0 \Rightarrow y = \frac{1 \pm \sqrt{5}}{2}\), so we have \((x,y,z) = (0, \frac{1 +\sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2})\) and permutations.
2. \begin{align*} A^2 &= s(s-a)(s-b)(s-c) \\ \end{align*} Notice the second part is the same as plugging \(s= 16/2 = 8\) into our polynomial Therefore \begin{align*} A^2 &= 8 \cdot (8^3 - 16 \cdot 8^2 + 81 \cdot 8 - 128) \\ &= 8 \cdot 8 (8^2 - 16 \cdot 8 + 81- 16) \\ &= 64 (-64+81-16) \\ &= 64 \end{align*} Therefore \(A = 8\)

Solution: \(p = a+b+c, q = ab+bc+ca, r = abc\)

1. Suppose two roots are equal to each other, this means that one of the roots is also a root of the derivative. ie \begin{align*} && 0 &= x^3+qx - r \\ && 0 &= 3x^2+q \end{align*} have a common root, but this root must satisfy \(x^2 = -\frac{q}{3}\). Then \begin{align*} &&0 &= x^3 + qx - r \\ &&&= x^3 -3x^3 - r \\ &&&= -2x^3 -r \\ \Rightarrow && r^2 &= 4x^6 \\ &&&= 4 \left ( -\frac{q}{3}\right)^3 \\ \Rightarrow && 0 &= 27r^2+4q^3 \end{align*}
2. Consider \(x = z + \frac{p}{3}\), then the equation is: \begin{align*} x^{3}-px^{2}+qx-r &= (z + \frac{p}{3})^3 - p(z + \frac{p}{3})^2 + q(z + \frac{p}{3}) - r \\ &= z^3 + pz^2 + \frac{p^2}{3}z + \frac{p^3}{27} - \\ &\quad -pz^2-\frac{2p^2}{3}z-\frac{p^3}{9} + \\ &\quad\quad qz + \frac{pq}{3} - r \\ &= z^3+\left (\frac{p^2}{3}-\frac{2p^2}{3}+q \right)z + \left (\frac{p^3}{27}-\frac{p^3}{9}+\frac{pq}{3}-r \right) \\ &= z^3+\left (-\frac{p^2}{3}+q \right)z + \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right) \\ \end{align*} Since this equation must also have repeated roots we must have: \begin{align*} 4\left (-\frac{p^2}{3}+q \right)^3 + 27 \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right)^2 = 0 \end{align*} which is exactly our desired result

Solution: \begin{align*} && x-y &= \frac1z - \frac1y \\ && x-z &= \frac1x - \frac1y \\ && y-z &= \frac1x - \frac1z \\ \Rightarrow && (x-y)(x-z)(y-z) &= \frac{(y-z)(y-x)(z-x)}{x^2y^2z^2} \\ \Rightarrow && x^2y^2 z^2 &= 1 \\ \end{align*} Suppose \(x + \frac1{y} =k \Rightarrow xy + 1 = ky\) Therefore \(y + \frac{1}{z} = y \pm xy = k\) Therefore \(1 \mp y = k(y \mp 1) \Rightarrow k = \pm 1\)

Solution: Claim: \(S_p = q^p\) for \(1 \leq p \leq n-1\) Proof: When \(p = 1\), \(S_p = \Sigma_1 = q\) as expected. Note that \(\Sigma_i = 0\) for \(i = 2, \cdots, n-1\). Using \(S_p = S_{p-1}\Sigma_{1}-S_{p-2}\Sigma_{2}+\cdots+(-1)^{p-1+1}S_{1}\Sigma_{p-1}+(-1)^{p+1}p\Sigma_{p}\), we can see that \(S_p = qS_{p-q}\) when \(1 \leq p \leq n-1\), ie \(S_p = q^p\). Note that \begin{align*} S_n &= \sum \alpha_i^n \\ &= q\sum \alpha_i^{n-1} - \sum r \\ &= qS_{n-1} - nr \\ &= q^n - nr \\ \\ S_{n+1} &= \sum \alpha_i^{n+1} \\ &= q \sum \alpha_i^{n} - r \sum \alpha_i \\ &= q^{n+1} - rq \\ \\ S_{n+2} &= \sum \alpha_i^{n+2} \\ &= q \sum \alpha_i^{n+1} - r \sum \alpha_i^2 \\ &= q^{n+2} - rq^2 \\ \end{align*} Claim: \(S_{n+m} = q^{n+m} - rq^{m}\) Proof: The obvious

Solution: Writing \(c = \cos \theta, s = \sin \theta\) then de Moivre states that: \begin{align*} && \cos 9 \theta + i \sin 9 \theta &= (c +i s)^9 \\ &&&= c^9 + 9ic^8s - 36c^7s^2-84ic^6s^3+126c^5s^4 + 126ic^4s^5 -84c^3s^6 -36ic^2s^7+9cs^8+is^9 \\ &&&= (c^9-36c^7s^2+126c^5s^3-84c^3s^6+8cs^8)+i(9c^8s-75c^6s^3+126c^4s^5-36c^2s^7+s^9) \\ \Rightarrow && \tan 9\theta &= \frac{(9c^8s-75c^6s^3+126c^4s^5-36s^2c^7+s^9)}{(c^9-36c^7s^2+126c^5s^4-84c^3s^6+8cs^8)} \\ &&&= \frac{9t-75t^3+126s^5-36t^7+t^9}{1-36t^2+126t^4-84t^6+8t^8} \\ &&&= \frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)} \end{align*} If we consider \(\tan 9\theta = 0\) it will have the roots \(\theta = \frac{n \pi}{9}, n \in \mathbb{Z}\), in particular, the numerator of our fraction for \(\tan 9 \theta\) will be zero for \(t = 0, \tan \frac{\pi}{9}, \tan \frac{2\pi}{9}, \tan \frac{3\pi}{9}, \tan \frac{4 \pi}{9}, \tan \frac{5\pi}{9}, \tan \frac{6 \pi}{9}, \tan \frac{7 \pi}{9}, \tan \frac{8\pi}{9}\). It's worth noting all other values of \(\theta\) will repeat these values. Also note that \(0,\tan \frac{\pi}{3}, \tan \frac{2\pi}{3}\) are the roots of \(t\) and \(t^2-3\) respectively. Therefore the other values are the roots of our sextic. However, also note that \(\tan \frac{8\pi}{9} = - \tan \frac{\pi}{9}\) and similar, therefore we can notice that all the roots in pairs can be mapped to \(\tan \frac{\pi}{9}, \tan \frac{2 \pi}{9}\) and \(\tan \frac{4 \pi}{9}\) and all those values are squared, so the roots of: \(x^3 - 33x^2+27x-3\) will be \(\tan^2 \frac{\pi}{9}, \tan^2 \frac{2 \pi}{9}\) and \(\tan^2 \frac{4 \pi}{9}\). The product of the roots will be \(3\), so \begin{align*} && \tan^2 \frac{\pi}{9} \tan^2 \frac{2 \pi}{9} \tan^2 \frac{4 \pi}{9} &= 3 \\ \Rightarrow && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \pm \sqrt{3} \\ \underbrace{\Rightarrow}_{\text{all positive}} && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \sqrt{3} \\ \end{align*} Notice that \(x^3 + y^3 +z^3 - 3xyz = (x+y+z)((x+y+z)^2-3(xy+yz+zx))\) Therefore \begin{align*} \tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right) &= 33(33^2-3\cdot27) + 3 \cdot 3 \\ &= 33\,273 \end{align*}

Solution: \begin{align*} && \begin{cases} 2a-3y=\frac{\left(z-x\right)^{2}}{y} \\ 2a-3z=\frac{\left(x-y\right)^{2}}{z} \end{cases} \\ \Rightarrow && \begin{cases} 2ay-3y^2=\left(z-x\right)^{2} \\ 2az-3z^2=\left(x-y\right)^{2} \end{cases} \\ \Rightarrow && 2a(y-z)-3(y+z)(y-z) &= (z-x+x-y)(z-x-x+y) \\ \Rightarrow && (y-z)(2a-3y-3z) &= (z-y)(z-2x+y) \\ \Rightarrow && 2a-3y-3z &= 2x-y-z \tag{\(y \neq z\)} \\ \Rightarrow && a &= x+y+z \\ \end{align*} This is is our first result. \begin{align*} && 2a-3y-3z &= 2x-y-z \\ \Rightarrow && 2a-3y-3x &= 3z-y-x \\ \Rightarrow && (y-x)2a-3(y-x)(y+x) &= (y-x)(2z-x-y) \\ \Rightarrow && 2a(y-x)-3(y^2-x^2) &= (z-y)^2-(x-z)^2 \\ \Rightarrow && 2ax - 3x^2 &= (y-z)^2 \\ \Rightarrow && 2a - 3x &= \frac{(y-z)^2}{x} \end{align*} Suppose \(x = \frac23 a, y = z = \frac16 a\) then all equations are satisfied, but \(y = z\).

Solution: Suppose \(k\) is a root of our quadratic. There are two possibilities, if \(k^{-1} = k\) then we must have \(k^2 = 1\) so either \(\pm 1\) is a root or we must have \((x-k)(x-k^{-1}) = x^2+bx+c\). In the first case we can have: \begin{align*} x^2+bx +c = (x-1)^2 &\Rightarrow b = -2, c = 1 \\ x^2+bx +c = (x+1)^2 &\Rightarrow b = 2, c = 1 \\ x^2+bx +c = (x-1)(x+1) &\Rightarrow b = 0, c = 1 \\ \end{align*} In the other cases, \(c = 1\) and \(b = k^{-1}+k\). Therefore we must have \(c = 1\) and \(b\) can take any values. The statement "if \(k\) is a root then \(1-k\) is a root" implies these two roots are different, so we must have \(1-k = k^{-1} \Rightarrow k-k^2 = 1 \Rightarrow k^2-k+1 = 0\) so \(b = -1, c = 1\). Suppose \(x^3+px^2+qx+r = 0\) has the first property, then for any root \(k\) we must have: \(k^3 + pk^2 + qk + r = 0\) and \(1 + pk^{-1} + qk^{-2} + rk^{-3} = 0\) therefore \(x^3+px^2+qx+r\) and \(rx^3+qx^2+px+1 = 0\) must have identical roots (since \(x = \pm\) either wont work here since they imply having the roots \(1, 0\) or \(-1, 2, \frac12\) which is exactly our equation. Therefore \(r = 1, p = q\). Suppose \(x^3 + px^2 + px+1 = 0\) has the property that if \(k\) is a root \(1-k\) is a root, therefore: \begin{align*} 0 &= (1-k)^3+p(1-k)^2 + p(1-k) + 1 \\ &= 1 -3k+3k^2-k^3+p-2kp+pk^2+p-pk+1 \\ &= -k^3+(3+p)k^2+(-3-3p)k+(2+2p) \end{align*} Since these roots must be the same as the original roots, we must have \(3+p = -p, -3-3p = -p, 2+2p = -1 \Rightarrow p = -\frac32\)

Solution: The point \(ze^{i\theta}\) is obtained by rotating the point \(z\) about \(0\) by an angle \(\theta\) anticlockwise. The complex numbers \(\alpha, \beta\) and \(\gamma\) will form an equilateral triangle iff the angles between each side are \(\frac{\pi}{3}\), ie \begin{align*} \begin{cases}{\gamma - \beta} &= e^{i \frac{\pi}{3}}({\beta - \alpha}) \\ {\alpha- \gamma} &= e^{i \frac{\pi}{3}}({\gamma- \beta}) \\ {\beta- \alpha} &= e^{i \frac{\pi}{3}}({\alpha- \gamma})\end{cases} \end{align*} We don't need all these equations, since the first two are equivalent to the third. Combining the first two equations, we have \begin{align*} && \frac{\gamma - \beta}{\beta-\alpha} &= \frac{\alpha-\gamma}{\gamma - \beta} \\ \Leftrightarrow && (\gamma - \beta)^2 &= (\alpha-\gamma)(\beta-\alpha) \\ \Leftrightarrow && \gamma^2 +\beta^2 - 2\gamma \beta &= \alpha\beta-\alpha^2-\gamma\beta+\gamma\alpha \\ \Leftrightarrow && \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta&=0 \end{align*} as required. If the roots of \(z^{3}+az^{2}+bz+c=0\) are \(\alpha, \beta, \gamma\) then \(\alpha+\beta+\gamma = -a\) and \(\beta\gamma+\gamma\alpha+\alpha\beta = b\). We also have that \(a^2 - 2b = \alpha^2+\beta^2+\gamma^2\). Therefore there roots will lie at the vertices of an equilateral triangle iff \(a^2-3b = 0\)