Problems

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Solution: Note that \begin{align*} && \frac{\d a \sec \theta}{\d \theta} &= a \sec \theta \tan \theta \\ && \frac{\d b \tan \theta}{\d \theta} &= b \sec^2 \theta \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{b \sec^2 \theta}{a \sec \theta \tan \theta} \\ &&&= \frac{b}{a} \frac{1}{\sin \theta} \\ \Rightarrow && \frac{y-b \tan \theta}{x - a \sec \theta} &= \frac{b}{a} \frac{1}{\sin \theta} \\ \Rightarrow && a \sin \theta y - ab \tan \theta \sin \theta &= bx -ab \sec \theta \\ \Rightarrow && bx-ay\sin \theta &= ab \sec x (1 - \sin ^2 \theta) \\ &&&= ab \cos \theta \end{align*}

1. \begin{align*} S: &&& \begin{cases} bx-ay &= 0 \\ bx-ay \sin \theta &= ab \cos \theta \end{cases} \\ \Rightarrow && ay(1-\sin \theta) &= ab \cos \theta \\ \Rightarrow && y &= \frac{b \cos \theta}{1-\sin \theta} \\ &&x &=\frac{a\cos \theta}{1-\sin \theta} \\ T: &&& \begin{cases} bx+ay &= 0 \\ bx-ay \sin \theta &= ab \cos \theta \end{cases} \\ \Rightarrow && ay(1+\sin \theta) &= -ab \cos \theta \\ \Rightarrow && y &= \frac{-b \cos \theta}{1+\sin \theta} \\ &&x &=\frac{a\cos \theta}{1+\sin \theta} \\ M: && x &= \frac{a \cos \theta}{2} \frac{2}{1-\sin^2 \theta} \\ &&&= a \sec \theta \\ && y &= \frac{b \cos \theta}{2} \frac{2 \sin \theta}{1-\sin^2 \theta} \\ &&&= b \tan \theta \end{align*} The midpoint of \(ST\) is the same as \(P\).
2. The tangents are perpendicular, therefore \(\frac{b}{a} \cosec \theta = - \frac{a}{b} \sin \phi\), ie \(b^2 = -a^2 \sin \phi \sin \theta\) The will intersect at: \begin{align*} &&& \begin{cases} bx - ay \sin \theta &= ab \cos \theta \\ bx - ay \sin \phi &= ab \cos \phi \end{cases} \\ \Rightarrow && ay ( \sin \theta - \sin \phi) &= ab(\cos \phi - \cos \theta) \\ \Rightarrow && y &= \frac{b(\cos \phi - \cos \theta)}{(\sin \theta - \sin \phi)} \\ && y^2 &= \frac{-a^2 \sin \phi \sin \theta (\cos\phi - \cos \theta)^2}{(\sin \theta - \sin \phi)^2} \\ \Rightarrow && bx(\sin \phi - \sin \theta) &= ab(\cos \theta \sin \phi - \cos \phi \sin \theta) \\ \Rightarrow && x &= \frac{a(\cos \theta \sin \phi - \cos \phi \sin \theta)}{\sin \phi - \sin \theta} \\ &&&= \frac{a^2(\cos \theta \sin \phi - \cos \phi \sin \theta)^2}{(\sin \phi - \sin \theta)^2} \end{align*} Therefore \begin{align*} && x^2+y^2 &= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\cos \theta \sin \phi- \cos \phi \sin \theta)^2 - \sin \phi \sin \theta (\cos\phi - \cos \theta)^2 \r \\ &&&= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\sin \phi - \sin \theta)(\cos^2 \theta \sin \phi - \sin \theta \cos^2 \phi) \r \\ &&&=a^2-b^2 \end{align*}

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Solution:

Note that the sine rule gives us \begin{align*} && \frac{x}{\sin \alpha} &= \frac{AB}{\sin (180-3\alpha)} \\ \Rightarrow && AB &= x \frac{\sin 3\alpha}{\sin \alpha} \\ &&&= x \frac{\sin \alpha \cos 2\alpha + \cos \alpha \sin 2\alpha}{\sin \alpha} \\ &&&= x \left ( \frac{\sin \alpha (1-2\sin^2\alpha) + 2(1-\sin^2 \alpha)\sin \alpha}{\sin \alpha} \right) \\ &&&= x (3 - 4\sin^2 \alpha) \end{align*} Note that \(AD = (\tfrac32 - 2 \sin^2\alpha)x\) and \(AE = (3-4\sin^2\alpha-\cos2\alpha)x\) so \begin{align*} DE &= (3-4\sin^2\alpha-\cos2\alpha)x - (\tfrac32 - 2 \sin^2\alpha)x \\ &= (\tfrac32 - 2\sin^2 \alpha - (1-2\sin^2 \alpha))x \\ &= \tfrac12x \end{align*} Since \(DE = \frac12x\) if we drop the perpendicular from \(F\) to \(EC\) we have a \(30-60-90\) triangle. Therefore \(\angle FCE = 30^\circ\). Notice that \(\angle CEB = 90^{\circ} - 2\alpha\) and \(\angle ACB = 180^\circ - 3\alpha\), therefore \begin{align*} \angle ACF &= \angle ACB - \angle FCE - \angle ECB \\ &= (180^\circ - 3\alpha) - 30^\circ - (90^{\circ} - 2\alpha) \\ &= 60^\circ - \alpha \\ &= \frac13 \angle ACB \end{align*}

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Solution: The vector representing the side \(AB\) is \(b - a\) and the vector representing the side \(DC\) is \(c - d\). \(ABCD\) is a parallelogram if and only if these opposite sides are parallel and equal in length, which is given by \(b - a = c - d\), or equivalently \(a + c = b + d\). Similarly, if \(a + c = b + d\), then \(c - b = d - a\), so the side \(BC\) is parallel and equal in length to the side \(AD\). Thus, \(a + c = b + d\) is the necessary and sufficient condition for \(ABCD\) to be a parallelogram. In a parallelogram, the shape is a square if and only if the diagonals are equal in length and perpendicular to each other. The diagonals are represented by the vectors \(c - a\) and \(d - b\). For these to be equal in length and perpendicular, one must be a \(90^\circ\) rotation of the other. Since \(A, B, C, D\) are labeled anticlockwise, a \(90^\circ\) anticlockwise rotation of the vector \(\vec{AC}\) (which is \(c-a\)) would point in the direction of \(\vec{DB}\) (which is \(b-d\) if we consider the relative orientation). Specifically: \(i(c - a) = d - b \implies -i(a - c) = d - b \implies i(a - c) = b - d\). Thus, \(ABCD\) is a square if and only if \(i(a - c) = b - d\).

1. The midpoint of the side \(PQ\) is \(\frac{1}{2}(p + q)\). To find the centre \(X\) of the square built externally on \(PQ\), we start at the midpoint and move a distance equal to half the side length in a direction perpendicular to \(PQ\). Since \(P, Q, R, S\) are anticlockwise, the outward direction is a \(90^\circ\) clockwise rotation of the vector \(\vec{PQ}\). A clockwise rotation of \(90^\circ\) corresponds to multiplication by \(-i\). \[ x = \frac{p+q}{2} + (-i)\left(\frac{q-p}{2}\right) = \frac{p + q - iq + ip}{2} = \frac{1}{2} \big( p(1+i) + q(1-i) \big) \]
2. From part (i), we have the representations for the centres: \begin{align*} x &= \tfrac{1}{2}(p(1+i) + q(1-i)) \\ y &= \tfrac{1}{2}(q(1+i) + r(1-i)) \\ z &= \tfrac{1}{2}(r(1+i) + s(1-i)) \\ t &= \tfrac{1}{2}(s(1+i) + p(1-i)) \end{align*} As shown in the first part of the problem, \(XYZT\) is a square if and only if: (1) \(x+z = y+t\) (it is a parallelogram) (2) \(i(x-z) = y-t\) (it is a square) First, examine condition (1): \begin{align*} x+z - (y+t) &= \tfrac{1}{2} \big[ (p+r)(1+i) + (q+s)(1-i) - (q+s)(1+i) - (r+p)(1-i) \big] \\ &= \tfrac{1}{2} \big[ (p+r)(1+i - (1-i)) - (q+s)(1+i - (1-i)) \big] \\ &= \tfrac{1}{2} \big[ (p+r)(2i) - (q+s)(2i) \big] \\ &= i(p+r - (q+s)) \end{align*} Thus, \(x+z = y+t\) if and only if \(p+r = q+s\), which is the condition that \(PQRS\) is a parallelogram. Next, examine condition (2): \begin{align*} i(x-z) &= \tfrac{1}{2} i \big[ p(1+i) + q(1-i) - r(1+i) - s(1-i) \big] \\ &= \tfrac{1}{2} \big[ p(i-1) + q(i+1) - r(i-1) - s(i+1) \big] \\ y-t &= \tfrac{1}{2} \big[ q(1+i) + r(1-i) - s(1+i) - p(1-i) \big] \\ \text{So, } i(x-z) - (y-t) &= \tfrac{1}{2} \big[ p(i-1 + 1-i) + q(i+1 - 1-i) + r(-i+1 - 1+i) + s(-i-1 + 1+i) \big] \\ &= 0 \end{align*} Since \(i(x-z) = y-t\) is an identity (always true for any \(PQRS\)), \(XYZT\) is a square if and only if it is a parallelogram. As established above, this occurs if and only if \(PQRS\) is a parallelogram.

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Solution:

1. The line \(PO\) has gradient \(\frac{p^2}{p} = p\) and teh line \(QO\) has gradient \(q\), therefore we must have that \(pq = -1\). Therefore, \(R\) is the point \begin{align*} && R &= \left ( \frac{p-\frac{1}{p}}{2}, \frac{p^2+\frac{1}{p^2}}{2} \right) \\ &&&= \left ( \frac12\left ( p - \frac{1}{p} \right),2\left (\frac12 \left(p-\frac{1}{p}\right) \right)^2+1 \right) \\ &&&= \left ( t, 2t^2+1\right) \end{align*} So we are looking at another parabola.
   

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2. The tangents are \(y = 2px+c\), ie \(p^2 = 2p^2+c\), ie \(y = 2px -p^2\) so we have \begin{align*} && y - 2px &= -p^2 \\ && y - 2qx &= -q^2 \\ \Rightarrow && (2p-2q)x &= p^2-q^2 \\ \Rightarrow && x &= \frac12 (p+q)\\ && y &= p(p+q)-p^2 \\ && y &= pq = -1 \end{align*} Therefore \(x = \frac12(p - \frac1p), y= -1\), so we have the line \(y = -1\) (the directrix)
   

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Solution: Since \(B\) and \(C\) are reflections of each other in the plane \(y+z=2\) (since that's what it means to be symmetrical), we must have that \(N\) also lies on the plane \(y+z=2\). Since \(\overrightarrow{AY}=3\overrightarrow{YN}.\) we must have \(\overrightarrow{AN}=\overrightarrow{AY}+\overrightarrow{YN} = \frac43\overrightarrow{AY} = \frac43\begin{pmatrix} 3\\-6\\6\end{pmatrix} = \begin{pmatrix} 4\\-8\\8\end{pmatrix}\) and \(N\) is the point \((4,-4,6)\) (which fortunately is on our plane as expected). \(Y\) is the point \((3,-2,4)\) \(|\overrightarrow{AY}| = \sqrt{3^2+(-6)^2+6^2} = 3\sqrt{1+4+4} = 9\)

Notice that \(BN^2 + 3^2 = 9^2 \Rightarrow BN^2 = 3\sqrt{3^2-1} = 6\sqrt2\). Therefore \(\overrightarrow{NB} = \pm 6\sqrt{2} \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1\end{pmatrix}\) and \(\{ B, C\} =\{ (4, 2, 12), (4, -10, 0)\}\). Suppose \(D = (x,y,z)\) then \begin{align*} && \begin{pmatrix} -1 \\ 2 \\ -2\end{pmatrix} \cdot \begin{pmatrix} x- 3 \\ y+2 \\ z-4\end{pmatrix} &= 3 \cdot 9 \cdot \frac13 = 9\\ \Rightarrow && 9 &= 3-x+2(y+2)-2(z-4) \\ &&&= -x+2y-2z+15 \\ \Rightarrow && 6 &= x-2y+2z \\ && 2 &= x -4y \\ \\ \Rightarrow && 81 &= (4y+2-3)^2+(y+2)^2+(2-y-4)^2 \\ &&&= (4y-1)^2+2(y+2)^2 \\ &&&= 16y^2-8y+1+2y^2+8y+8 \\ &&&= 18y^2+9 \\ \Rightarrow && y^2 &= 2 \\ \Rightarrow && y &= \pm 2 \end{align*} Therefore \(\displaystyle D \in \left \{ (10, 2, 0), (-6, -2, 4) \right \}\)

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Solution: Let \(\mathbf{a}\) denote the vector position of \(A\) and similarly for \(B, C, D\). Then we know that \((\mathbf{b}-\mathbf{a})\cdot(\mathbf{b}-\mathbf{a})=(\mathbf{c}-\mathbf{d})\cdot(\mathbf{c}-\mathbf{d})\) and \((\mathbf{b}-\mathbf{c})\cdot(\mathbf{b}-\mathbf{c})=(\mathbf{a}-\mathbf{d})\cdot(\mathbf{a}-\mathbf{d})\). Subtracting these two equations we see that \(|\mathbf{a}|^2 -2\mathbf{a}\cdot\mathbf{b}+2\mathbf{c}\cdot\mathbf{b} - |\mathbf{c}|^2 = |\mathbf{c}|^2-2\mathbf{c}\cdot\mathbf{d}+2\mathbf{a}\cdot\mathbf{d}-|\mathbf{a}|^2\) or \(2|\mathbf{a}|^2 -2\mathbf{a}\cdot\mathbf{b}+2\mathbf{c}\cdot\mathbf{b} - 2|\mathbf{c}|^2 +2\mathbf{c}\cdot\mathbf{d}-2\mathbf{a}\cdot\mathbf{d}=0\) The midpoints of the diagonals \(AC\) and \(BD\) are \(\frac{\mathbf{a}+\mathbf{c}}{2}\) and \(\frac{\mathbf{b}+\mathbf{d}}{2}\), so the line is parallel to: \(\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}\). The diagonals are parallel to \(\mathbf{a}-\mathbf{c}\) and \(\mathbf{b}-\mathbf{d}\). So it suffices to prove that \((\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}) \cdot (\mathbf{a}-\mathbf{c}) = 0\) (since the other will follow by symmetry, \begin{align*} (\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}) \cdot (\mathbf{a}-\mathbf{c}) &= |\mathbf{a}|^2-\mathbf{a}\cdot\mathbf{b}-\mathbf{a}\cdot \mathbf{d}+\mathbf{b}\cdot \mathbf{c}-|\mathbf{c}|^2+\mathbf{c}\cdot \mathbf{d} \\ \end{align*} But this is exactly half the equation we determined earlier, so we are done.