Problems

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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*} && \frac{\d y}{\d x} &= -\frac1{2x^2} \\ \Rightarrow && \frac{y - \frac{1}{2p}}{x - p} &= - \frac{1}{2p^2} \\ \Rightarrow && y - \frac1{2p} &= -\frac{1}{2p^2}x +\frac1{2p} \\ \Rightarrow && y +\frac{1}{2p^2}x &= \frac1p \\ \Rightarrow && 2p^2 y + x &= 2p\\ \Rightarrow && 2q^2 y + x &= 2q \\ \Rightarrow && (p^2-q^2)y &= p-q \\ \Rightarrow && y &= \frac{1}{p+q} \\ && x &= \frac{2pq}{p+q} \end{align*} \begin{align*} \text{normal}: && \frac{y-\frac1{2p}}{x-p} &= 2p^2 \\ \Rightarrow && y - \frac1{2p} &= 2p^2x - 2p^3 \\ \Rightarrow && 2py -4p^3x &= 1-4p^4 \\ \Rightarrow && 2qy -4q^3x &= 1-4q^4 \\ pq = \tfrac12: && y - 2p^2 x &= q-2p^3 \\ && y - 2q^2 x &= p-2q^3 \\ \Rightarrow && (2q^2-2p^2)x &= q-p +2q^3-2p^3 \\ &&&= (q-p)(q+p+2q^2+1+2p^2) \\ \Rightarrow && x &= \frac{1+2(p^2+q^2)+1}{2(p+q)} \\ &&&= \frac{1+2(p^2+q^2+2pq-1)+1}{2(p+q)} \\ &&&= p+q\\ && y &= 2p^2 \left ( p+q \right) + q - 2p^3 \\ &&&= p+q \end{align*} So \(N(p+q, p+q)\) and \(T\left (\frac{1}{p+q}, \frac{1}{p+q} \right)\), so both points lie on \(y = x\). \[ OT \cdot ON = \frac{\sqrt{2}}{p+q} \cdot (p+q)\sqrt{2} = 2 \] which is clearly constant.

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Solution: The curve \(x^4 + y^4 = u\) has lines of symmetry:

• \(y = 0\)
• \(x = 0\)
• \(y = x\)
• \(y = -x\)

The curve \(xy = v\) has lines of symmetry:

• \(y = x\)
• \(y = -x\)

The points are \(A = (\alpha, \beta), B = (\beta, \alpha), C = (-\alpha, -\beta), D = (-\beta, -\alpha)\) \(AD\) has gradient \(\frac{\beta+\alpha}{\alpha+\beta} = 1\), \(BC\) has the same gradient. \(AB\) has gradient \(\frac{\alpha-\beta}{\beta-\alpha} = -1\), as does \(CD\). Therefore it has two sets of perpendicular and parallel sides, hence a rectangle. The area is \(|AD||AB| = \sqrt{2(\alpha+\beta)^2}\sqrt{2(\alpha-\beta)^2} = 2(\alpha^2-\beta^2)\) The squared area is \(4(\alpha^4+\beta^4 - 2 \alpha^2\beta^2) = 4(u - 2v^2)\) ie the area is \(2\sqrt{u-2v^2}\) When \(u = 81, v = 4\) we have the area is \(2 \sqrt{81 - 2 \cdot 16} = 14\) as required.

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Solution:

\(Q = (-a,-b)\) \begin{align*} && \frac{\d y}{\d x} &= -\frac{1}{x^2} \\ \Rightarrow && \frac{y-b}{x-a} &= -\frac{1}{a^2} \\ \Rightarrow && 0 &= a^2y+x-a^2b-a \\ &&&= a^2y+x - 2a\\ \\ && 2x - 2y \frac{\d y}{\d x} &= 0 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{x}{y} \\ \Rightarrow && \frac{y+b}{x+a} &= \frac{a}{b} \\ \Rightarrow && 0 &= by-ax+b^2 - a^2 \\ &&&= by - ax -2 \end{align*} Notice that \((-b,a)\) is on both lines, therefore it is their point of intersection. The coordinates of \(N\) will be \((a,-b)\). We can see this is a square by noting each point is a rotation (centre the origin) of \(90^\circ\) of each other.

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Solution: \begin{align*} \mathbf{PAP} &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}-1 & 8\\ 8 & 11 \end{pmatrix}\begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix} \\ &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}15 & -10\\ 30 & 5 \end{pmatrix} \\ &= \begin{pmatrix}75 & 0\\ 0 & -25 \end{pmatrix} \end{align*} Which is diagonal as required. Letting \(\mathbf{x}=\mathbf{P}\mathbf{x}_{1}+\mathbf{a}\) \begin{align*} && \mathbf{x}^{\mathrm{T}}\mathbf{Ax}+\mathbf{b}^{\mathrm{T}}\mathbf{x}-11&=0 \\ \Leftrightarrow && (\mathbf{P}\mathbf{x}_{1}+\mathbf{a})^T\mathbf{A}(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 + \mathbf{x}_{1}^T\mathbf{PAa} + \mathbf{a}^T\mathbf{AP}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \end{align*} It would be nice if we picked \(\mathbf{a}\) such that \(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T = 0\), if \(\mathbf{a} = \begin{pmatrix} a_1 \\a_2 \end{pmatrix}\) then this equation becomes: \begin{align*} && 2\begin{pmatrix}-a_1 + 8a_2 & 8a_1+11a_2 \end{pmatrix} + \begin{pmatrix}18 & 6 \end{pmatrix} &= 0 \\ \Rightarrow && a_1 = 1, a_2 = -1 \end{align*} So our equation is now \begin{align*} && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1-6 +12 - 11 &= 0 \\ \Leftrightarrow && 25(3x_1^2 - y_1^2) &= 5 \\ \Leftrightarrow && 3x_1^2 - y_1^2 &= \frac{1}{5} \end{align*}

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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 \(\displaystyle \frac{(2+z)z}{2(z+1)} = \left (1 - \frac{z}{2z+2} \right)z\)