Problems

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Solution: \begin{align*} && (x+y+z)^2 &= x^2 + y^2 + z^2 + 2(xy+yz+zx) \\ \Rightarrow && 1^2 &= 2 + 2(xy+yz+zx) \\ \Rightarrow && xy+yz+zx &= -\frac12 \end{align*} \begin{align*} && 1 \cdot 2 &= (x+y+z)(x^2+y^2+z^2) \\ &&&= x^3 + y^3 + z^3 + x^2y+x^2z+y^2z+y^2x+z^2x+z^2y \\ &&&= 3 + x^2y+x^2z+y^2z+y^2x+z^2x+z^2y\\ \Rightarrow && -1 &= x^2y+x^2z+y^2z+y^2x+z^2x+z^2y \end{align*} \begin{align*} && (x+y+z)^3 &= x^3 + y^3 + z^3 + \\ &&&\quad \quad 3xy^2 + 3xz^2 + \cdots + 3zx^2 + 3zy^2 + \\ &&&\quad \quad \quad 6xyz \\ \Rightarrow && 1 &= 3 + 3(-1) + 6xyz \\ \Rightarrow && xyz &= \frac16 \end{align*} Since we have \(f(t) = (t-x)(t-y)(t-z) = t^3-t^2-\frac12 t - \frac16\) is zero for \(x,y,z\) we can notice that: \(t^{n+1} = t^n +\frac12 t^{n-1} + \frac16 t^{n-2}\) is also true for \(x,y,z\) (by multiplying by \(t^{n-2}\). Therefore: \(S_{n+1} = S_n + \frac12 S_{n-1} + \frac16 S_{n-2}\)

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Solution: \begin{align*} && \underbrace{(c-a)}_{\geq 0}\underbrace{(d-b)}_{\geq 0} & \geq 0 \\ \Leftrightarrow && cd -bc -ad + ab &\geq 0 \\ \Leftrightarrow && ab +cd &\geq bc+ad \\ \end{align*}

1. Applying \((*)\) with \(c=d=x\) and \(a=b=y\) we obtain: \(x^2 + y^2 \geq xy + xy = 2xy\) Similarly, applying \((*)\) with \(c=x, d=y, a=b=z\) we obtain: \(z^2 + xy \geq zx+zy\) so \(x^2+y^2+z^2 \geq 2xy + z^2 \geq xy + zx+zy\) There was nothing special about our choice of ordering \(x,y,z\) so it is true for all \(x,y,z\)
2. \begin{align*} \frac st +\frac tr +\frac rs +\frac ts +\frac rt +\frac sr &=\left ( \frac st+\frac ts \right)+\left ( \frac tr +\frac rt \right)+\left ( \frac rs +\frac sr \right) \\ & \geq 2 \sqrt{\frac st \frac ts} + 2 \sqrt{\frac tr \frac rt} + 2 \sqrt{\frac rs \frac sr} \\ & = 2 + 2 + 2 \\ &= 6 \end{align*}

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Solution: This is a second order recurrence relation, so we need to find \(m\) and \(n\) such that; \begin{align*} &&\frac15 &= m\frac13 + n \\ &&\frac17 &= m \frac15 + n\frac13 \\ \Rightarrow && m,n &= \frac67, - \frac{3}{35} \end{align*} So we now need to solve the characteristic equation: \(\lambda^2 - \frac67 \lambda + \frac{3}{35} = 0\) So \(x,y = \frac{15 \pm 2 \sqrt{30}}{35}\). We need, \begin{align*} && 1 &= a+ b \\ && \frac13 &= a \frac{15 + 2 \sqrt{30}}{35} + b \frac{15 - 2 \sqrt{30}}{35} \\ && \frac13 &= \frac{15}{35} + \frac{2 \sqrt{30}}{35}(a-b) \\ \Rightarrow && -\frac{\sqrt{30}}{18} &= a-b \\ \Rightarrow && a &= \frac{18-\sqrt{30}}{36} \\ && b &= \frac{18+\sqrt{30}}{38} \end{align*} So our two answers are: \[ (a,b,x,y) = \left (\frac{18\pm\sqrt{30}}{36} ,\frac{18\mp\sqrt{30}}{36},\frac{15 \pm 2 \sqrt{30}}{35},\frac{15 \mp 2 \sqrt{30}}{35}, \right)\]

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Solution: \begin{align*} && x^3 - 3xbc+b^3 + c^3 &= (x+b+c)(x^2-x(b+c)+b^2+c^2-bc) \\ &&&= \tfrac12(x+b+c)((x-b)^2+(x-c)^2+(b-c)^2) \\ \end{align*} We must have: \begin{align*} && 0 &= ak^2 + bk+c \tag{1}\\ &&0 &= bk^2+ck+a \tag{2}\\ b*(1)&& 0 &= abk^2 + b^2k+cb \\ a*(2)&& 0 &= abk^2 + ack + a^2 \\ \Rightarrow && 0 &= k(ac-b^2)+a^2-bc \\ \Rightarrow && (ac-b^2)k &= bc-a^2 \\ \\ c*(1) && 0 &= ack^2+bck+c^2 \\ b*(2) && 0 &= b^2k^2+bck+ab \\ \Rightarrow && 0 &= (ac-b^2)k^2 +c^2-ab \\ \Rightarrow && (ac-b^2)k^2 &= ab-c^2 \\ \\ \Rightarrow && \frac{ab-c^2}{ac-b^2} &= k^2 = \left (\frac{bc-a^2}{ac-b^2} \right)^2 \\ \Rightarrow && (ab-c^2)(ac-b^2) &= (bc-a^2)^2 \\ \Rightarrow && a^2bc - ab^3-ac^3+b^2c^2 &= b^2c^2-2a^2bc+a^4 \\ \Rightarrow && 0 &= a^4+ab^3+ac^3-3a^2bc \\ &&&= a(a^3+b^3+c^3-3abc) \\ \underbrace{\Rightarrow}_{a \neq 0} && 0 &= a^3+b^3+c^3-3abc \\ &&&= (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \end{align*} Therefore \(a+b+c = 0\). (Since otherwise \(a=b=c\) but \(ac \neq b^2\)). This means \(1\) is a root of our equations. Therefore, either \(k = 1\) or they have both roots in common, ie they are the same equation up to a scalar factor. ie \(b = la, c = lb, a= lc \Rightarrow l^3 = 1 \Rightarrow l = 1\). Therefore, they are the same equation.

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Solution: \begin{align*} \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) &= \frac{\tan(\theta_1 + \theta_2) + \tan(\theta_3 + \theta_4)}{1 - \tan(\theta_1 +\theta_2)\tan(\theta_3+\theta_4)} \\ &= \frac{\frac{t_1+t_2}{1-t_1t_2}+\frac{t_3+t_4}{1-t_3t_4}}{1-\frac{t_1+t_2}{1-t_1t_2}\frac{t_3+t_4}{1-t_3t_4}} \\ &= \frac{(t_1+t_2)(1-t_3t_4)+(t_3+t_4)(1-t_1t_2)}{(1-t_1t_2)(1-t_3t_4)-(t_1+t_2)(t_3+t_4)} \\ &= \frac{t_1 +t_2+t_3+t_4 - (t_1t_2t_3+t_1t_2t_4+t_1t_3t_4+t_2t_3t_4)}{1-t_1t_2-t_1t_3-t_1t_4-t_2t_3-t_2t_4-t_3t_4} \end{align*} If \(t_1, t_2, t_3, t_4\) are the roots of \(at^4+bt^3+ct^2+dt+e = 0\), then \(t_1+t_2+t_3+t_4 = -\frac{b}{a}, t_1t_2+t_1t_3+t_1t_4+t_2t_3+t_2t_4+t_3t_4 = \frac{c}{a}, t_1t_2t_3+t_1t_2t_4+t_1t_3t_4+t_2t_3t_4 = -\frac{d}{a}\), therefore the expression is: \begin{align*} \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) &= \frac{-\frac{b}{a}+\frac{d}{a}}{1 - \frac{c}{a}} \\ &= \frac{d-b}{a-c} \end{align*} \begin{align*} &&0 &= p \cos 2\theta + \cos (\theta - \alpha) + p \\ &&&= p (2\cos^2 \theta -1) + \cos \theta \cos \alpha - \sin \theta \sin \alpha + p \\ &&&= 2p \cos^2 \theta + \cos \theta \cos \alpha - \sin \theta \sin \alpha\\ \Rightarrow && 0 &=2p \cos \theta + \cos \alpha - \tan \theta \sin \alpha \\ \Rightarrow && -2p \cos \theta&= \cos \alpha - \tan \theta \sin \alpha \\ \Rightarrow && 4p^2 \cos^2 \theta &= \cos^2 \alpha - 2 \sin \alpha \cos \alpha \tan \theta + \sin^2 \alpha \tan^2 \theta \\ && 4p^2 \frac{1}{1 + \tan^2 \theta} &= \cos^2 \alpha - \sin 2\alpha \tan \theta + \sin^2 \alpha \tan^2 \theta \\ \Rightarrow && 4p^2 &= \cos^2 \alpha - \sin 2\alpha t+t^2-\sin2\alpha t^3+\sin^2 \alpha t^4 \\ \Rightarrow && \tan (\theta_1+\theta_2 + \theta_3+ \theta_4) &= \frac{0}{\sin^2 \alpha - 1} \\ &&&= 0 \\ \Rightarrow && \theta_1 + \theta_2 + \theta_3 + \theta_4 &= n\pi \end{align*}

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Solution: The complex numbers represent an equilateral triangle iff \(\gamma\) is a \(\pm 60^\circ\) rotation of \(\beta\) around \(\alpha\), ie \begin{align*} && \gamma - \alpha &= \omega(\beta - \alpha) \\ \Leftrightarrow && \omega &= \frac{\gamma - \alpha}{\beta - \alpha} \\ \Leftrightarrow && -1 &= \left (\frac{\gamma - \alpha}{\beta - \alpha} \right)^3 \\ \Leftrightarrow && -(\beta - \alpha)^3 &=(\gamma - \alpha)^3 \\ \Leftrightarrow && 0 &= (\gamma-\alpha)^3+(\beta-\alpha)^3 \\ &&&= \gamma^3-3\gamma^2\alpha +3\gamma\alpha^2-\alpha^3 +\beta^3-3\beta^2\alpha+3\beta\alpha^2-\alpha^3 \\ &&&= (\beta + \gamma - 2\alpha)(\alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta) \\ \Leftrightarrow && 0 &= \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta \end{align*} The roots of the equation \(z^3+az^2+bz+c = 0\) represents the vertices of an equilateral triangle iff \(a^2-3b = (\alpha+\beta+\gamma^2) - 3(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta = 0\) as erquired. Suppose \(a^2 = 3b\), then consider \(z = pw +q\), we must have \begin{align*} && 0 &= (pw+q)^3+a(pw+q)^2 + b(pw+q)+c \\ &&&= p^3w^3 +(3p^2q+ap^2)w^2+(3pq^2+2apq+bp)w+(q^3+aq^2+bq+c) \\ &&&= p^3w^3+p^2(3q+a)w^2+p(3q^2+2aq+b)w+(q^3+aq^2+bq+c) \\ \end{align*} We need to check if \(\left(\frac{3q+a}{p} \right)^2 = 3 \left (\frac{3q^2+2qa+b}{p^2} \right)\). Clearly the denominators match, so consider the numerators \begin{align*} && (3q+a)^2 &= 9q^2+6aq+a^2 \\ &&&= 9q^2+6aq+3b \\ &&&= 3(3q^2+2qa+b) \end{align*} as required

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Solution: \begin{align*} && f(0) &= 0^n + a_1\cdot 0^{n-1} + \cdots + a_n \\ &&&= a_n \\ && f(0) &= (0+k_1)(0+k_2) \cdots (0+k_n) \\ &&&= k_1 k_2 \cdots k_n \\ \Rightarrow && a_n &= k_1 k_2 \cdots k_n \\ \\ &&f(1) &= 1^n + a_1 \cdot 1^{n-1} + \cdots + a_n \\ &&&= 1 + a_1 + a_2 + \cdots + a_n \\ && f(1) &= (1 + k_1) (1 + k_2) \cdots (1+k_n) \\ \Rightarrow && (k_1+1)\cdots (k_n+1) &= 1 + a_1 + \cdots + a_n \\ \\ && f(-1) &= (-1)^{n} + a_1 \cdot (-1)^{n-1} + \cdots + a_n \\ &&&= a_n - a_{n-1} + \cdots + (-1)^{n-1} a_1 + (-1)^{n} \\ && f(-1) &= (-1+k_1)(-1+k_2) \cdots (-1+k_n) \\ &&&= (k_1-1)(k_2-1)\cdots(k_n-1) \\ \Rightarrow && (k_1-1)\cdots(k_n-1) &= a_n - a_{n-1} + \cdots + (-1)^{n-1} a_1 + (-1)^{n} \end{align*} \(576 = 2^6 \cdot 3^2\). Notice that \(1 - 22 + 172 -552 + 576 = 175 = 5^2 \cdot 7\) and \(1+22 + 172+552+576 = 1323 = 3^3 \cdot 7^2\). \(k_i = 2, 6, 6, 8\) therefore the roots are \(-2, -6, -6, -8\)

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Solution:

Since there are two turning points the derivative (a quadratic) has two distinct real roots. \begin{align*} && f'(x) &= 3x^2-2(p+q+r)x+(pq+qr+rp) \\ && 0 &< \Delta = 4(p+q+r)^2 - 4\cdot 3(pq+qr+rp) \\ \Rightarrow && (p+q+r)^2 &> 3(pq+qr+rp) \end{align*} If \(g^2 > 4h\) then \(p(x) = (x^2+gx+h)(x-k)\) has at least 2 real roots (possibly one repeated, and in particular it has two turning point, ie \begin{align*} && p'(x) &= (2x+g)(x-k)+(x^2+gx+h) \\ &&&= 3x^2+(2g-2k)x + (h-kg) \\ && 0 &< \Delta = 4(g-k)^2 - 4\cdot 3 (h-gk) \\ \Rightarrow && (g-k)^2 &> 3(h-gk) \end{align*} Pick \(g = h = 1\) and \(k = 1000\) then \((-999)^2 > 0 > 3(1-1000)\) so it is sufficient but not necessary.

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Solution: \begin{align*} && ( x^2 - a x + b ) ( x^2 + a x + c ) &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\ \Rightarrow && x^4 + p x^2 + q x + r &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\ \Rightarrow && p &= b+c-a^2 \tag{1}\\ && q &= a(b-c) \tag{2}\\ && r &= bc \tag{3} \end{align*} \begin{align*} (1): && p+a^2 &= b+ c \\ (2): && \frac{q}{a} &= b - c \\ \Rightarrow && b &= \frac12 (p+a^2 + \frac{q}{a}) \\ && c &= \frac12 (p+a^2 - \frac{q}{a}) \\ (3): && r &= \frac12 (p+a^2 + \frac{q}{a}) \frac12 (p+a^2 - \frac{q}{a}) \\ \Rightarrow && 4ra^2 &= (pa + a^3 + q)(pa+a^3-q) \\ &&&= (pa+a^3)^2 - q^2 \\ &&&= a^2(p+a^2)^2 -q^2 \\ &&&= a^2(p^2 + 2pa^2 + a^4) - q^2 \\ &&&= pa^2 + 2pa^4 + a^6 - q^2 \\ \end{align*} Therefore \(a^2\) is a root of \(u^3 + 2pu^2 + pu - q^2 = 4ru\), ie the given equation. When \(u = 0\), this equation is \(-q^2\), therefore the cubic is negative. But as \(u \to \infty\) the cubic tends to \(\infty\), therefore it must cross the \(x\)-axis and have a positive root. If \(p=-1, q = -6, r = 15\) then the cubic is: \(u^3 - 2u^2 + (1-60)u -36\) and so when \(u = 9\) we have \begin{align*} 9^3 - 2\cdot 9^2 -59 \cdot 9 -36 &= 9(9^2-2\cdot 9 - 29 -4) \\ &= 9(81 -18-59-4) \\ &= 0 \end{align*} so \(u = 9\) is a root Let \(y=z + 2\) \begin{align*} &&y^4 - 8 y^3 + 23 y^2 - 34 y + 39 &= (z+2)^4-8(z+2)^3 + 23(z+2)^2 - 34(z+2) + 39 \\ &&&= z^4+8z^3+24z^2+32z+16 - \\ &&&\quad -8z^3-48z^2-96z-64 \\ &&&\quad\quad +23z^2+92z+92 \\ &&&\quad\quad -34z-68 + 39 \\ &&&= z^4-z^2-6z+15 \end{align*} So conveniently this is \(p = -1, q = -6, r = 15\), so we know that \(a = 3\) is a sensible thing to true. \(b = \frac12(-1 + 9 + \frac{-6}{3}) = 3\) \(c = \frac12(-1+9-\frac{-6}{3}) = 5\) so \begin{align*} && z^4-z^2-6z+15 &= (z^2-3z+3)(z^2+3z+5) \\ &&y^4 - 8 y^3 + 23 y^2 - 34 y + 39 &= ((y-2)^2-3(y-2)+3)((y-2)^2+3(y-2)+5) \\ &&&= (y^2-4y+4-3y+6+3)(y^2-4y+4+3y-6+5) \\ &&&= (y^2-7y+13)(y^2-y+3) \end{align*}

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Solution:

1. If the roots are \(ak^{-1}, a, ak\) then we must have that \(p = a(k^{-1}+1+k)\), \(q = a^2(k^{-1}+k+1)\) and \(r = a^3\), therefore \(a = \frac{q}{p}\) (ie one of the roots is \(\frac q p\) and \(r = \left ( \frac{q}{p} \right)^3 \Rightarrow q^3 =rp^3 \Rightarrow q^3-rp^3 = 0\)
2. Suppose \(r = q^3/p^3\) then \(\left (\frac{q}{p} \right)^3 - p\left (\frac{q}{p} \right)^2+q\left (\frac{q}{p} \right) - r = \frac{--pq^2+pq^2}{p^2} =0 \), therefore \(q/p\) is a root by the factor theorem. We must also have the product of the three roots is \(q^3/p^3\) but one of the roots is \(q/p\) therefore the product of the other two roots is \(q^2/p^2\), but the condition \(ac = b^2\) is precisely the condition that \(a,b,c\) is a geometric progression.
3. If the three roots are \(a-d, a, a+d\) then \(p = 3a\), \(q = a^2-da+a^2+da+a^2-d^2 = 3a^2-d^2\), \(r = a(a^2-q^2)\), therefore \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\) Similarly, suppose \(\frac{p}{3}\) is a root, then the other two roots must sum to twice this and therefore they are in arithmetic progression. The condition \(\frac{p}{3}\) is a root is equivalent to: \(\frac{p^3}{27} - \frac{p^3}{9} + \frac{qp}{3} - r = 0\), ie exactly \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\), therefore this condition is both necessary and sufficient.

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Solution:

1. Suppose \(pqr = qrs\), since the roots are positive, we can divide by \(qr\) to obtain \(p=s\) (a contradiction. Therefore all those terms are distinct.
2. \(4c^3 = pqr+qrs+rsp+spq\), \(d^4 = pqrs\). Applying AM-GM, we obtain: \begin{align*} && c^3 = \frac{ pqr+qrs+rsp+spq}{4} & > \sqrt[4]{p^3q^3r^3s^3} = d^{3} \\ \Rightarrow && c &> d \end{align*}
3. There must be a turning point between each root (since there are no repeated roots).
4. \(f'(x) = 4x^3-12ax^2+12b^2-4c^3 = 4(x^3-3ax^2+3b^2-c^3)\). Letting the roots of this polynomial be \(\alpha, \beta, \gamma\) and again applying AM-GM, we must have: \begin{align*} && b^2 = \frac{\alpha\beta + \beta \gamma+\gamma \alpha}{3} &> \sqrt[3]{\alpha^2\beta^2\gamma^2} = c^2 \\ \Rightarrow && b &> c \end{align*}
5. Again, since there are turning points between the roots of \(f'(x)\) we must have distinct roots for \(f''(x)\), ie: \(f''(x) = 3x^2-6ax+6b^2 = 3(x^2-2ax+b^2)\) has distinct real roots. But for this to occur we must have that \((2a)^2-4b^2 = 4(a^2-b^2) > 0\), ie \(a>b\)

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Solution: Consider the linear \(3\times 3\) system in \(yz, zx, xy\), then \begin{align*} \left(\begin{array}{ccc|c} 2 & 1 & -5 & 2 \\ 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ 2 & 1 & -5 & 2 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 3 & -9 & 0 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 0 & 3 & 6 \\ \end{array}\right) \\ \end{align*} Therefore \(yz = 2, zx = 6, xy = 3 \Rightarrow (xyz)^2 = 36 \Rightarrow xyz = \pm 6\). If \(xyz = 6, x = 3, y = 1, z = 2\), if \(xyz = -6, x = -3, y = -1, z = -2\)