Problems

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Solution:

1. 

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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*}

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Solution:

1. \(\,\)
   

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2. 1. Suppose \(y=k\) meets the curve \(C_2\) then \begin{align*} && \tfrac1{16} &= (k^2+(x-1)^2-1)(k^2+(x+1)^2-1) \\ &&&= (k^2+x^2-2x)(k^2+x^2+2x) \\ &&&= k^4+2k^2x^2+x^4-4x^2 \\ &&&= x^4+(2k^2-4)x^2+k^4 \\ \Rightarrow && 0 &= x^4+(2k^2-4)x^2+(k^4-\tfrac1{16}) \\ \\ && \Delta &= 4(k^2-2)^2 - 4 \cdot 1 \cdot (k^4-\tfrac1{16}) \\ &&&= 4(k^4-4k^2+4 - k^4 +\tfrac{1}{16}) \\ &&&= 16(1+\tfrac{1}{64} - k^2) \\ &&&= 16(\tfrac{65}{64} - k^2) \end{align*} Therefore if \(|k| < \frac{\sqrt{65}}{8}\) there are \(4\) intersections. If \(|k| = \frac{\sqrt{65}}{8}\) there are \(2\) intersections, otherwise there are \(0\).
   2. The greatest possible \(y\) value is \( \frac{\sqrt{65}}{8}\) and at this point \(x^2 = \frac{2(2-\frac{65}{64}}{2} =1 - \frac{1}{64} < 1\) so they are close to the \(y\)-axis.
   3. The regions where \(y^2+(x-1)^2-1 < 0\) and \(y^2+(x+1)^2-1 < 0\) is the interior of the two circles from the first part. However, since they don't overlap they can never both be negative. Therefore in our equation both are positive and therefore \(C_2\) is entirely outside \(C_1\)
   4. \(\,\)
      

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Solution:

1. Suppose \((ct, c/t)\) is on \(C_2\) then \begin{align*} && r^2 &= \left ( ct - a \right)^2 + \left ( \frac{c}{t} - b \right)^2 \\ &&&= c^2t^2 - 2cta + a^2 + \frac{c^2}{t^2} - \frac{2cb}{t} + b^2 \\ \Rightarrow && 0 &= c^2t^4 - 2act^3 + (a^2+b^2-r^2)t^2 - 2bct + c^2 \end{align*}
2. Notice that \(\displaystyle \sum t_i = \frac{2a}{c}\) and \(\displaystyle \sum t_it_j = \frac{a^2+b^2-r^2}{c^2}\) so \begin{align*} && \sum t_i^2 &= \left ( \sum t_i \right)^2 - 2 \sum t_it_j \\ &&&= \frac{4a^2}{c^2} - \frac{2a^2+2b^2-2r^2}{c^2} \\ &&&= \frac{2}{c^2} \left (a^2 - b^2 + r^2 \right) \end{align*} Note that \(\frac{1}{t}\) are roots of the \(c^2 - 2act + (a^2+b^2-r^2)t^2 - 2bct^3 + c^2t^4 = 0\) which is the same equation but with \(a \leftrightarrow b\) so \(\displaystyle \sum \frac{1}{t_i^2} = \frac{2}{c^2} (b^2 - a^2 + r^2)\)
3. Therefore \begin{align*} && \sum_{i=1}^4 OP_i^2 &= \sum_{i=1}^4 \left (c^2t_i^2 + \frac{c^2}{t_i^2} \right) \\ &&&= 2(a^2-b^2+r^2) + 2(b^2-a^2+r^2) \\ &&&= 4r^2 \end{align*}
4. If they touch at two distinct points it must be the case that \(t_1 = t_2\) and \(t_3 = t_4\). We must also have \(t_1t_2t_3t_4 = t_1^2t_3^2 = 1\) so \(t_1t_3 = \pm 1\). Therefore our points are \((ct_1, \frac{c}{t_1})\) and \(\pm(\frac{c}{t_1}, ct_1)\) but these are reflections in \(y = \pm x\). But if these two points are reflections of one another the line of reflection is the perpendicular bisector, which must run through the centre of the circle.

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Solution: \begin{align*} && |z-a|^2 &= (z-a)(z-a)^* \\ &&&= (z-a)(z^*-a^*) \\ &&&= zz^*-az^*-a^*z+aa^* \\ &&&= r^2 \end{align*} Therefore the locus of \(P\) is a circle centre \(a\) radius \(r\).

1. \begin{align*} && 0 &= zz^* - az^* - a^*z + aa^* - r^2 \\ &&&= \frac{1}{\omega \omega^{*}} - \frac{a}{\omega^*} - \frac{a^*}{\omega} + aa^*-r^2 \\ \Rightarrow && 0 &= 1-a\omega-a^*\omega^*+(|a|^2-r^2)\omega\omega^* \\ \Rightarrow && 0 &= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)-\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)+ \frac{1}{|a|^2-r^2} \\ &&&= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}-\frac{|a|^2}{(|a|^2-r^2)^2}+ \frac{1}{|a|^2-r^2} \\ &&&=\omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}- \frac{r^2}{(|a|^2-r^2)^2} \end{align*} Therefore \(\displaystyle \left|\omega-\left ( \frac{a^*}{|a|^2-r^2}\right)\right|^2 = \frac{r^2}{(|a|^2-r^2)^2}\) ie \(\omega\) lies on a circle centre \(\frac{a^*}{|a|^2-r^2}\), radius \(\frac{r}{||a|^2-r^2|}\). If these are the same circle then \(r = \frac{r}{||a|^2-r^2|} \Rightarrow (|a|^2-r^2)^2 = 1\) and \(a = \frac{a^*}{|a|^2-r^2} \Rightarrow a = \pm a^*\), ie \(a\) is purely real or imaginary.
2. This is the same story, except we end up with centre \(\frac{a}{|a|^2-r^2}\), so we do not end up with the same conditions

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Solution:

Clearly the distance \(PS\) is \(s - x\), so it remains to determine the heigh \(PQ\). Notice that \(\tan \beta = \frac{PQ}{OP}\) so the height is \(x \tan \beta\) and the area is \(x(s-x)\tan \beta \) Notice that \(R\) has a \(y\)-coordinate of \(x \tan \beta\), but is a distance \(a\) from the origin, so \(s^2 + x^2 \tan^2 \beta = a^2 \Rightarrow s = \sqrt{a^2-x^2 \tan^2 \beta}\) \begin{align*} && \frac{\d A}{\d x} &= (s-x)\tan \beta + x \left (\frac{\d s}{\d x} - 1 \right) \tan \beta \\ &&&= (s-x) \tan \beta + \left (\tfrac12(a^2-x^2\tan^2 \beta)^{-1/2} \cdot (-2x \tan^2 \beta) - 1\right) x \tan \beta \\ &&&= (s-x) \tan \beta + \left ( \frac{-x \tan^2 \beta}{s} -1\right)x \tan \beta \\ &&&= (s-2x) \tan \beta - \frac{x^2}{s}\tan^3\beta \\ \\ \frac{\d A}{\d x} = 0: && 0 &= s(s-2x)-x^2 \tan^2 \beta \\ &&&= s^2-(2x)s-x^2\tan^2 \beta \\ &&&= (s-x)^2-(1+\tan^2\beta)x^2 \\ \Rightarrow && s &= x + x \sec \beta \\ &&&= (1+\sec \beta)x \\ \\ && a^2 &= x^2(1+\sec\beta)^2 + x^2 \tan^2 \beta \\ &&&= x^2(2\sec \beta +2\sec^2 \beta ) \\ &&&= 2x^2 \sec \beta(1+\sec \beta) \\ \\ && A &= x^2\sec \beta \tan \beta \\ &&&= \frac12 a^2 \frac{\sec \beta \tan \beta}{\sec \beta(1+\sec \beta)} \\ &&&= \frac12 a^2 \frac{\tan \beta}{1+\sec \beta} = \frac12 a^2 \tan \frac{\beta}{2}\\ \end{align*} This occurs when \begin{align*} && \frac{RS}{SO} &= \frac{x \tan \beta}{s} \\ &&&= \frac{\tan \beta}{1+\sec \beta} = \tan \frac{\beta}2 \\ \Rightarrow&& \angle ROS &= \frac{\beta}2 \end{align*}

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Solution:

1. \(\,\) \begin{align*} && \frac{\d x}{\d t} &= 2at \\ && \frac{\d y}{\d t} &= 2a \\ \Rightarrow && \frac{\d y}{\d x} &= \frac1t \\ && -p &= \text{grad of normal} \\ &&&= \frac{y-2ap}{x-ap^2} \\ \Rightarrow && y &= -px + ap^3+2ap \\ && 2an &= -pan^2 + ap^3 + 2ap \\ \Rightarrow && 0 &= pan^2+2an-ap(2+p^2) \\ \Rightarrow && n &= p, -\left ( p + \frac2{p}\right) \\ \Rightarrow && n &= -\left ( p + \frac2{p}\right) \end{align*}
2. \(\,\) \begin{align*} && |PN|^2 &= (ap^2-an^2)^2 +(2ap-2an)^2 \\ &&&= a^2(p-n)^2(p+n)^2+4a^2(p-n)^2 \\ &&&= a^2(p-n)^2((p+n)^2+4) \\ &&&= a^2\left(p+p+\frac2p \right)^2 \left ( \left ( -\frac2p\right)^2+4\right)\\ &&&= a^2\left(\frac{2p^2+2}p \right)^2 \left ( \frac{4}{p^2}+4\right)\\ &&&= 16a^2 \frac{(p^2+1)^3}{p^4} \\ \\ && \frac{\d |PN|^2}{\d p^2} &= 16a^2\frac{3(p^2+1)^2p^4-2(p^2+1)^3p^2}{p^8} \\ &&&= 16a^2(p^2+1)^2 \frac{3p^2-2(p^2+1)}{p^6} \\ &&&= 16a^2(p^2+1)^2 \frac{p^2-2}{p^6} \end{align*} Therefore minimized when \(p^2=2\) (clearly a minimum by considering behaviour as \(p^2 \to 0, \infty\))
3. If \(PN\) is the diameter of \(PNQ\) then \(QP\) and \(QN\) are perpendicular, ie \begin{align*} && -1 &= \frac{2ap-2aq}{ap^2-aq^2} \cdot \frac{2aq-2an}{aq^2-an^2} \\ &&&= \frac{2}{p+q} \cdot \frac{2}{q+n} \\ &&&= \frac{2}{p+q} \cdot \frac{2}{q - p -\frac{2}{p}} \\ \Rightarrow && 4 &= (p+q)(p+\frac2{p}-q) \\ &&&= p^2-q^2 + \frac{2q}{p} + 2 \\ \Rightarrow && 2 &= p^2 - q^2 + \frac{2q}{p} \end{align*} Therefore \(q = 0 \Rightarrow p^2 = 2 \Rightarrow |PN|\) is at it's minimum.

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Solution:

1. Let \(\displaystyle y = \frac z {(1+z^2)^{\frac12}}\) then \(\frac{d y}{d x} = \frac{(1+z^2)^{\frac12} - z^2(1+z^2)^{-\frac12}}{1+z^2} = \frac{(1+z^2)-z^2}{(1+z^2)^\frac32} = \frac{1}{(1+z^2)^\frac32}\)
2. \(\kappa = \frac {f''(x)}{\big({1+ (f'(x))^2\big)^{\frac32}}}\) then \begin{align*} && \int \kappa \, dx &= \int \frac{f''(x)}{( 1 + (f'(x))^2)^{\frac32}} \, dx \\ && \kappa x &= \frac{f'(x)}{(1 + (f'(x))^2)^\frac12} + C \\ \Rightarrow && (\kappa x-C)^2 &= \frac{f'(x)^2}{1 + (f'(x))^2} \\ \Rightarrow && f'(x)^2((\kappa x - C)^2 - 1) &= -(\kappa x-C)^2 \\ \Rightarrow && f'(x) &= \frac{\kappa x - C}{\sqrt{1-(\kappa x - C)^2 }} \\ \Rightarrow && f(x) &= \frac{1}{\kappa} \sqrt{1 - (\kappa x - C)^2} \\ \Rightarrow && (\kappa y)^2 + (\kappa x - C)^2 &= 1 \\ \Rightarrow && y^2 + (x - C')^2 &= \frac{1}{\kappa^2} \end{align*} Therefore all the curves are circles and \(\kappa\) is the reciprocal of the radius.

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Solution:

1. \(\,\)
   

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   Note that we have a right angled triangle, with the sides in a ratio of \(m\). So if our target length is \(x\) we have \(x^2 + (mx)^2 = c^2\) and so \(x = c(m^2+1)^{-\frac12}\)
2. The distance from the origin to \(L\) is \(a = c(m^2+1)^{-\frac12}\) so \begin{align*} && a^2(m^2+1) &= c^2 \\ && \frac{c-y(x)}{0-x} &= y' \\ \Rightarrow && c-y &= -xy' \\ \Rightarrow && a^2((y')^2+1) &= (y-xy')^2 \\ \\ && 2a^2y'y'' &= 2(y-xy')(y'-xy''-y') \\ &&&= 2(xy'-y)xy'' \\ \Rightarrow && y'' &= 0 \\ \text{ or } && 2a^2y' &= 2(xy'-y)x \end{align*} If \(y'' = 0\) then \(y = mx + c\) and the result follows immediately. \begin{align*} && 0 &= (a^2-x^2)y' + yx \\ \Rightarrow &&\frac1{y} y' &= -\frac{x}{a^2-x^2} \\ \Rightarrow && \ln y &= \frac12\ln (a^2-x^2) + K \\ \Rightarrow && y^2 &= M(a^2-x^2) \\ \Rightarrow && x^2 + y^2 &= a^2 \end{align*} Where in the last step we know the tangents from an ellipse are not all equidistant to the origin.
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Solution:

1. Suppose we have the shortest distance between the two curves, and the path between the points is not a normal to both curves. Then we could shift the endpoints to reduce the distance. (Assuming we're not at a point of intersection). Therefore, the normal to the curves must be the same (or in other words) the gradients of the curves must be the same. ie we are at a point where \(2y y' = 4a\) we must have \(y' = m\), so \(y = \frac{2a}{m}\) and \(x = \frac{a}{m^2}\) and the distance from this point to the line \(y=mx+c\) is \(\frac{|m \frac{a}{m^2} - \frac{2a}{m}+c|}{\sqrt{m^2+1}} = \frac{|mc-a|}{m\sqrt{m^2+1}} = \frac{mc-a}{m\sqrt{m^2+1}}\). If \(mc \leq a\) then we find \(\frac{a-mc}{m\sqrt{m^2+1}}\) However, we must check that the two curves do not intersect (otherwise the closest distace is \(0\)). ie we need to check if \((mx+c)^2 = 4ax\) has any solutions, this quadratic has discriminant \((2mc-4a)^2 - 4 \cdot m^2 \cdot c^2 = 16a^2-16amc = 16a(a-mc)\) which is clearly greater than \(0\) when \(a \geq mc\). Therefore the shortest distance in this case is \(0\).
2. The distance squared between the point \((p,0)\) and a point of the form \((at^2,2at)\) is \(D^2 = (at^2-p)^2+4a^2t^2 = a^2t^4+(4a^2-2ap)t^2+p^2\) \begin{align*} && \frac{D^2}{a^2} &= t^4 + 2\left(2-\frac{p}{a}\right)t^2 + \frac{p^2}{a^2} \\ &&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 + \frac{p^2}{a^2} - \left (2-\frac{p}{a} \right)^2 \\ &&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 +\frac{4p}{a} -4 \\ \end{align*} Therefore if \(2 \leq \frac{p}{a}\) then we can find a \(t\) such that we attain the minimum for \(D^2/a^2\) of \(\frac{4p}{a}-4\) and so \(D = \sqrt{4pa-4a^2} = 2\sqrt{a(p-a)}\) . If not the smallest value will be when \(t = 0\) and we will have \(|p|\) Now consider all the lines joining points on the parabola to the centre of the circle. The shortest distance from the parabola to the circle will be normal to the circle and therefore will also be a line through the center. Therefore we need only consider the shortest distance from \((p,0)\) to the parabola \(-b\). Case 1: If \(p \geq 2a\) we have \(2\sqrt{a(p-a)} - b\) or \(0\) if \(b \geq 2\sqrt{a(p-a)}\) Case 2: If \(p < 2a\) we have \(p-b\) or \(0\) if \(b \geq p\)

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Solution:

1. \(k = 0\), we have \(x^2 + y^2 + 3x + y = 0\), ie \((x+\tfrac32)^2+(y+\tfrac12)^2 = \frac{10}{4}\).
   

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2. \(3x^2 + 3y^2 +10xy = (3x+y)(x+3y)\) so \(x^2 + y^2 + \tfrac{10}3xy + 3x+y = (3x+y)(\frac{x+3y}{3}+1) = 0\) so we have the line pair \(3x +y =0\), \(x+3y + 3 = 0\)
   

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3. If \(k = 2\) then \((x+y)^2 + (x+y)+2x = 0\). If \(\theta = 45^\circ\) then \( X = \frac1{\sqrt{2}}(x+y), Y = \frac{1}{\sqrt{2}}(y-x)\), ie \(x+y = \sqrt{2}X\) and \(x = \frac{1}{\sqrt2}(X-Y)\), so our equation is: \begin{align*} 0 &= 2X^2 + \sqrt{2}X + \sqrt{2}(X-Y) \\ &= (\sqrt{2}X + 1)^2 - 1 - \sqrt{2} Y \end{align*} which would be a parabola with line of symmetry \(X = -\frac{1}{\sqrt{2}}\). However, we are actually looking at that parabola rotated by \(45^\circ\) anticlockwise.
   

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Solution: \begin{align*} && y &= bx-ba \\ && 1 &= x^2 + y^2 \\ \Rightarrow && 1 &= x^2 + b^2(x-a)^2 \\ \Rightarrow && 0 &= (1+b^2)x^2-2ab^2x+b^2a^2-1 \end{align*} This will have roots which sum to \(\frac{2ab^2}{1+b^2}\), therefore \(M = \left ( \frac{ab^2}{1+b^2}, \frac{ab^3}{1+b^2}-ba \right)=\left ( \frac{ab^2}{1+b^2}, \frac{-ba}{1+b^2} \right)\)

1. Since \(b\) is fixed so is \(\frac{b}{1+b^2} = t\) and all the points are \((bta, -ta)\), ie \(x = -by\). If \(a \in [-1,1]\) we are ranging on the points \((bt, -t)\) to \((-bt, t)\) which is a distance of \begin{align*} && d &= \sqrt{(bt+bt)^2+(-2t)^2} \\ &&&= \sqrt{4(b^2+1)t^2} \\ &&&=2 \sqrt{(b^2+1)\frac{b^2}{(b^2+1)^2}} \\ &&&= \frac{2b}{\sqrt{b^2+1}} \end{align*}
   

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2. • If \(a\) is fixed we have \(\left ( \frac{ab^2}{1+b^2}, -\frac{ba}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= - b \\ \Rightarrow && y &= \frac{a\frac{x}{y}}{1 + \frac{x^2}{y^2}} \\ \Rightarrow && y^2 \left ( 1 + \frac{x^2}{y^2} \right) &= ax \\ \Rightarrow && x^2-ax + y^2 &= 0 \\ \Rightarrow && \left (x - \frac{a}{2} \right)^2 + y^2 &= \frac{a^2}{4} \end{align*} Therefore we will end up with a circle centre \((\tfrac{a}{2}, 0)\) going through the origin.
     

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   • If \(ab = 1\), we have \(\left ( \frac{b}{1+b^2}, -\frac{1}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= -b \\ \Rightarrow && y &= -\frac{1}{1+\frac{x^2}{y^2}} \\ \Rightarrow && y + \frac{x^2}{y} &= - 1 \\ \Rightarrow && y^2 +y+ x^2 &= 0 \\ \Rightarrow && \left ( y + \frac12 \right)^2 + x^2 &= \frac14 \end{align*}
     

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Solution:

\begin{align*} && \mathbf{y}_1 &= \overrightarrow{OG}+\overrightarrow{GY_1} \\ &&&= \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3) -\lambda_1 \left (\mathbf{x}_1 - \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ && 1 &= \mathbf{y}_1 \cdot \mathbf{y}_1 \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \cdot \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac19\left ( (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\mathbf{x}_1 \cdot \mathbf{x}_2+\mathbf{x}_1 \cdot \mathbf{x}_3) + 2(1+\lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right) \\ \Rightarrow && 9 &= (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\beta+\gamma) + 2(1+\lambda_1)^2 \alpha \\ &&&= 3+6\lambda_1^2+2(\beta+\gamma)-2(\beta+\gamma)\lambda_1 - 4\lambda_1^2(\beta+\gamma) + 2\alpha+4\lambda_1\alpha + 2\lambda_1^2 \alpha \\ && 0 &= (-6+2(\alpha+\beta+\gamma))+2(2\alpha-(\beta+\gamma))\lambda_1 + (6+2(\alpha-2(\beta+\gamma)))\lambda_1^2 \\ \Rightarrow && 0 &= ((\alpha+\beta+\gamma)-3)+(2\alpha-(\beta+\gamma))\lambda_1 + (3+\alpha-2(\beta+\gamma))\lambda_1^2 \\ &&&= (\lambda_1+1)((3+\alpha-2(\beta+\gamma))\lambda_1+ ((\alpha+\beta+\gamma)-3)) \\ \Rightarrow && \lambda_1 &= \frac{3-(\alpha+\beta+\gamma)}{3+\alpha-2(\beta+\gamma)} \end{align*} as required. Since \(\dfrac {GX_1}{GY_1} = \frac1{\lambda_1}\) we must have, \begin{align*} && \frac {GX_1}{GY_1} + \frac {GX_2}{GY_2} + \frac {GX_3}{GY_3} &= \frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3} \\ &&&= \frac{(3+\alpha-2\beta-2\gamma)+(3+\beta-2\gamma-2\alpha)+3+\gamma-2\alpha-2\beta)}{3-\alpha-\beta-\gamma} \\ &&&= \frac{9-3(\alpha+\beta+\gamma)}{3-(\alpha+\beta+\gamma)} \\ &&&= 3 \end{align*}

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Solution:

1. • \(\bf (a)\) \((1, \sqrt2)\) is an integer \(2\)-rational point. \((\frac35 + \frac45\sqrt2, \frac45 - \frac{3}{5}\sqrt2)\) is a non-integer \(2\)-rational point.
   • [\bf(b)] First notice that \((\sqrt2)^2 +3^2 = 11\) so then consider \((1 + \tfrac32\sqrt2, 1-\tfrac32\sqrt2)\) will work as \(\pi/4\) degree rotation.
   • [\bf(c)] First notice \(3^2-(\sqrt2)^2 = 2\). Notice that \((k\sec \theta + \sqrt{m} \tan \theta)^2 - (k\tan \theta + \sqrt{m} \sec \theta)^2 = k^2-m\). Taking \(k= 3\) we have \((3 \cdot \frac{13}{5} + \frac{12}{5}\sqrt{2}, 3\cdot\frac{12}5+\frac{13}{5}\sqrt2)\)

Note: we can also find the additional point in the last part by considering lines through \((3, \sqrt2)\), for example \(y = -\frac32x + \sqrt2 + \frac92\) would give the same point.

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Solution: \begin{align*} && 0 &= z^2 + pz + 1\\ &&&= (x+iy)^2 + (x+iy)p + 1 \\ &&& = (x^2-y^2+px+1) + (2xy+py)i \\ \Rightarrow && 0 &= x^2 - y^2 + px + 1 \\ && 0 &= (2x+p)y \\ \Rightarrow && p &= -2x \\ \text{ or } && y &= 0 \\ \Rightarrow && p &= -(x^2+1)/x \end{align*} Therefore as \(p\) varies with either have \(y = 0\) and \(x\) taking any real value except \(0\) ie the real axis minus the origin. Or \(p = -2x\) and \(-y^2-x^2+1 = 0 \Rightarrow x^2 + y^2 = 1\) which is a circle. Suppose \(pz^2 + z + 1 = 0\) \begin{align*} && 0 &= pz^2 + z +1\\ &&&= p(x+iy)^2 + (x+iy) + 1\\ &&&= (px^2-py^2+x+1) + (2xyp + y) i \\ \Rightarrow && 0 &= (2xp+1)y \\ \Rightarrow && y & = 0, p = \frac{-(x+1)}{x^2}, x \neq 0 \\ \text{ or } && p &= -\frac{1}{2x}\\ \Rightarrow && 0 &= -\frac{1}{2}x + \frac{y^2}{2x} + x + 1 \\ &&&= \frac{y^2 - x^2 +2x^2 + 2x}{2x} \\ &&&= \frac{(x+1)^2+y^2-1}{2x} \end{align*} So we either have the real axis (except \(0\)) or a circle radius \(1\) centre \((-1, 0)\) (excluding \(x = 0\)).

Suppose \(pz^2 + p^2 z + 2 = 0\) then \begin{align*} && 0 &= p(x+iy)^2 + p^2(x+iy) + 2 \\ &&&= (p(x^2-y^2) + p^2x + 2) + (2xyp + p^2y)i \\ \Rightarrow && 0 &= py(2x+p) \\ \Rightarrow && y &= 0, \Delta = x^4-8x \\ \Rightarrow && x &\in (-\infty, 0) \cup [2, \infty) \\ \text{ or } && p &= -2x \\ && 0 &= (-2x)(x^2-y^2) + 4x^3+2 \\ &&&= 2x^3+2xy^2+2 \\ \Rightarrow && 0 &= x^3+xy^2+1 \end{align*}