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. If \(\vec{PQ} = \vec{SR}\) we have a parallelogram. \(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\) then we have a rhombus.
2. If the four points lie in a plane then \((\vec{RS} \times \vec{RP}) \cdot \vec{RQ} =0\), so \begin{align*} && 0 &=\left ( \begin{pmatrix}-r\\ s-1 \\ 0 \end{pmatrix} \times \begin{pmatrix}p-r\\ -1 \\ -1 \end{pmatrix}\right) \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ && &= \begin{pmatrix}1-s \\ -r \\r+(p-r)(1-s) \end{pmatrix} \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ &&&= (1-s)(1-r)-r(q-1)-r-(p-r)(1-s) \\ &&&=(1-s)(1-r-p+r)-rq \\ \Rightarrow && rq &= (1-s)(1-p) \end{align*}
   1. \(\,\) \begin{align*} && \vec{PQ} &= \vec{SR} \\ \Leftrightarrow && \begin{pmatrix}1-p\\q \\ 0 \end{pmatrix} &= \begin{pmatrix}r\\1-s \\ 0 \end{pmatrix} \\ \Leftrightarrow && 1-p = r & \quad ; \quad q = 1-s\\ \Leftrightarrow && 1= r+p & \quad ; \quad 1 = q+s\\ \end{align*} The centroid is \(\frac14 (p+1+r, q+s+1, 2)\) which is clearly \(\frac12(1,1,1)\) iff those equations are true.
   2. \(\,\) \begin{align*} && |\vec{PQ}| &= |\vec{PS}| \\ \Leftrightarrow && (1-p)^2+q^2+ 0^2 &= p^2+s^2+1)\\ \Leftrightarrow && 1-2p+p^2+q^2 &= p^2 + s^2 + 1 \\ \Leftrightarrow && -2p+q^2 &= s^2 \end{align*} From the previous equations we have \(r = 1-p\), and \(-2p+(1-s)^2 = s^2 \Rightarrow -2p + 1 -2s = 0 \Rightarrow s = \frac12 - p\) and \(q = \frac12 + p\) \begin{align*} && \cos PQR &= \frac{\vec{QP}\cdot \vec{QR}}{|\vec{QP}||\vec{QR}|} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -q \\ 0 \end{pmatrix} \cdot \begin{pmatrix}r-1\\ 1-q \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+q^2}\sqrt{(r-1)^2+(1-q)^2+1^2}} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -\frac12-p \\ 0 \end{pmatrix} \cdot \begin{pmatrix}-p\\ \frac12-p \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+(-\frac12-p)^2}\sqrt{p^2+(\frac12-p)^2+1^2}} \\ &&&= \frac{ p-p^2-\frac14+p^2}{\sqrt{p^2-2p+1+\frac14+p+p^2}\sqrt{p^2+\frac14-p+p^2+1}} \\ &&&= \frac{4p-1}{\sqrt{8p^2-4p+5}\sqrt{8p^2-4p+5}}\\ &&&= \frac{4p-1}{8p^2-4p+5}\\ \end{align*} For \(PQRS\) to be a square \(\cos PQR = 0\), ie \(p = \frac14\) and so \((p,q,r,s) = (\frac14, \frac34, \frac34, \frac14)\) and \(|PQ| = \sqrt{(1-p)^2+q^2} = \sqrt{\left ( \frac34 \right)^2 + \left ( \frac34 \right)^2 } = \frac{3\sqrt{2}}4\), notice that \(\left ( \frac{21}{20} \right)^2 = \frac{441}{400} < \frac{9}{8}\) (\(441 < 450\)) therefore the sides are at least as long as \(\frac{21}{20}\)

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Solution:

Notice the total area is \(xf(x)\) and it is made up of the sum of the two integrals.

1. Suppose \((g(t))^3 + g(t) = t\). Notice that \((g(0))^3 + g(0) =0 \Rightarrow g(0)((g(0))^2 + 1) = 0 \Rightarrow g(0) = 0\). \begin{align*} && t &= (g(t))^3 + g(t) \\ \Rightarrow && 1 &= 3(g(t))^2 g'(t) + g'(t) \\ \Rightarrow && g'(t) &= \frac{1}{1 + 3(g(t))^2} > 0 \end{align*}
   

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   From our sketch, we can see we are interested in: \begin{align*} && \int_0^2 g(t) \d t &= 2 - \int_0^1 (x^3 + x) \d x \\ &&&= 2 - \frac14 - \frac12 = \frac54 \end{align*}
2. \(\,\)
   

   + - ↺
   From our second sketch, we can see that: \begin{align*} && \int_0^8 h(t) \d t &= 16 - \int_1^2 (x^3+x-2) \d x \\ &&&= 16 - \left ( \frac{8}{4} + \frac{2^2}{2} - 2 \cdot 2 \right)+ \left ( \frac{1}{4} + \frac{1}{2} - 2 \right) \\ &&&= \frac{59}{4} \end{align*}

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Solution:

1. Given \(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\), we can form the following results: \begin{align*} && \begin{cases} \mathbf{a} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \mathbf{b} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \mathbf{c} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) &= 0 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= 0 \\ \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= 0 \\ \mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} &= 0 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot\mathbf{b} + \mathbf{a} \cdot\mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -1 \\ \mathbf{a} \cdot \mathbf{b} +\mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} &= -\frac12 \\ \end{cases} \\ \Rightarrow && \begin{cases} \mathbf{a} \cdot \mathbf{b} = -\frac12 \\ \mathbf{a} \cdot \mathbf{c} = -\frac12 \\ \mathbf{b} \cdot \mathbf{c} = -\frac12 \\ \end{cases} \end{align*} The triangle must be equilateral since the angles between each vertex are the same.
2. We have \(\displaystyle \sum_{i=1}^{4} \mathbf{a}_i = \mathbf{0}\) so \(\displaystyle \mathbf{a}_i \cdot \sum_{i=1}^{4} \mathbf{a}_i = 0\) or for each \(i\), \(\displaystyle \sum_{j \neq i} \mathbf{a}_i \cdot \mathbf{a}_j = -1\). \begin{align*} && \begin{cases} \mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_1 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_2 \cdot \mathbf{a}_4 + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \end{cases} \\ && \text{adding the first two, subtracting the last two} \\ \Rightarrow && \begin{cases} \mathbf{a}_1 \cdot \mathbf{a}_2 +\cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} = -1 \\ \mathbf{a}_1 \cdot \mathbf{a}_2 + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} = -1 \\ \cancel{\mathbf{a}_1 \cdot \mathbf{a}_3} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_3} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \cancel{\mathbf{a}_1 \cdot \mathbf{a}_4} + \cancel{\mathbf{a}_2 \cdot \mathbf{a}_4} + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1 \\ \end{cases} \\ \Rightarrow && 2 (\mathbf{a}_1 \cdot \mathbf{a}_2) - 2(\mathbf{a}_3 \cdot \mathbf{a}_4) = 0 \end{align*} Rather than adding the first two and last two, we could have done any pair, resulting in the relations: \begin{align*} \mathbf{a}_1 \cdot \mathbf{a}_2 &= \mathbf{a}_3 \cdot \mathbf{a}_4 \\ \mathbf{a}_1 \cdot \mathbf{a}_3 &= \mathbf{a}_2 \cdot \mathbf{a}_4 \\ \mathbf{a}_1 \cdot \mathbf{a}_4 &= \mathbf{a}_2 \cdot \mathbf{a}_3 \end{align*}
   1. The shape must be a parallelogram (from the angle requirement, but also cyclic quadrilateral (since all vectors are unit length), therefore it must be a rectangle
   2. Given it's a regular tetrahedron, \(\mathbf{a}_i \cdot \mathbf{a}_j\) must be the same for all \(i \neq j\), ie \(-\frac13\). We are interested in \(|\mathbf{a}_i - \mathbf{a}_j|\) so consider, \begin{align*} |\mathbf{a}_i - \mathbf{a}_j|^2 &= (\mathbf{a}_i - \mathbf{a}_j) \cdot (\mathbf{a}_i - \mathbf{a}_j) \\ &= \mathbf{a}_i \cdot \mathbf{a}_i - 2 \mathbf{a}_i \cdot \mathbf{a}_j + \mathbf{a}_j \cdot \mathbf{a}_j \\ &= 1 - \frac23 + 1 \\ &= \frac43 \end{align*} Therefore the unit side lengths are \(\frac{2}{\sqrt{3}}\)

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Solution:

1. Suppose \((x,y)\) is on the line of invariant points, then \begin{align*} &&\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &&&= \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} \\ \Rightarrow && \begin{cases} (a-1)x + by = 0 \\ (cx + (d-1)y = 0 \end{cases} \tag{*} \end{align*} Therefore either \(x = 0, y = 0\) or \((a-1)(d-1)-bc = 0\) \(\Rightarrow ((a-1)(d-1)-bc)xy = 0\). We also know this is true for all values \(x,y\) on the line of invariant points. If there is one where both \(x \neq 0, y \neq 0\) we are done, otherwise the line of invariant points must be one of the axes. ie but then one of \(\begin{pmatrix} a \\ c \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) or \(\begin{pmatrix} b \\ d \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) is true and we'd also be done. If the line doesn't go through the origin then there are points on every line, not equal to the origin which are fixed. But then every point on those lines is fixed (since \(\mathbf{A}\) is a linear operator) and so every point is fixed. ie \(\mathbf{A} = \mathbf{I}\).
2. Suppose \((a-1)(d-1) -bc = 0\) and \(b \neq 0\) then I claim that \(y = \frac{1-a}{b}x\) is a line of invariant points. It's clear that the first equation will be satisfied in \((*)\) so it suffices to check the second, but the first condition is equivalent to the equations being linearly dependent, ie both equations are satisfied. If \(b = 0\) then \((a-1)(d-1) = 0\), so our matrix must look like \(\begin{pmatrix} 1 & 0 \\ c & d\end{pmatrix}\) (if \(d \neq 1\))or \(\begin{pmatrix} * & 0 \\ * & 1\end{pmatrix}\). In the first case, the line \(y = \frac{c}{1-d}x\) and in the second \(x = 0\) is an invariant line.
3. Suppose the invariant line is \(y = mx+k\) then we must have that \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ mx + k \end{pmatrix} &= \begin{pmatrix} (a + mb)x + bk \\ (c+dm)x + dk \end{pmatrix} \end{align*} and \((c+dm)x + dk = m((a + mb)x + bk) +k \Rightarrow k(d-mb-1) = x(-c+(a-d)m+m^2b)\) Since this equation must be true for all values of \(x\), and \(k \neq 0\) we can say that \(mb = d-1\) and \(-c+(a-d)m+m^2b = 0\), ie \(-c + (a-d)m + m(d-1) = 0 \Rightarrow (a-1)m-c = 0\) if \(m \neq 0\) then \((a-1)\frac{(d-1)}{b} - c = 0\) ie our desired relation is true. If \(m = 0\) then we must have that \(y = k\) is an invariant line, ie \(d-1=0\) and \(c=0\) which also satisfies our relation.

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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:

1. Let \(A = (0,0,0)\) and then \(B = 22b \mathbf{i}, D = 2d\mathbf{j}, C = 2b\mathbf{i}+2d\mathbf{j}\) and \(V = b \mathbf{i} + d\mathbf{j} + h\mathbf{k}\) We also have \begin{align*} && \tan \alpha &= \frac{h}{d}\\ && \tan \beta &= \frac{d}{b} \\ && \vec{AV} \times \vec{VB} &= \begin{pmatrix} b \\ d \\ h \end{pmatrix} \times \begin{pmatrix} -b \\ d \\ h \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ -2bh \\ 2db \end{pmatrix} \\ &&&= 2b \begin{pmatrix} 0 \\ -d \tan \alpha \\ d \end{pmatrix} \\ &&&= k \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \end{align*} similarly for the vector perpendicular to the other face it must be \(\begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix}\) Looking at the angle between these perpendicular (to find the angles between the faces we see: \begin{align*} \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \cdot \begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix} &= \cos \alpha \cos \beta \end{align*} But this is also \(\pi -\) the angle between the planes, ie \(\cos \theta = \cos \alpha \cos \beta\)
2. \(\,\) \begin{align*} && \cot^2 \phi &= \frac{b^2+d^2}{h^2} \\ && \cot^2 \alpha &= \frac{d^2}{h^2} \\ && \cot^2 \beta &= \frac{b^2}{h^2} \\ \Rightarrow && cot^2 \phi &= \cot^2 \beta+\cot^2 \alpha \end{align*} \begin{align*} && \cos^2 \phi &= \frac{b^2+d^2}{b^2+d^2+h^2} \\ && \cos^2 \alpha &= \frac{d^2}{d^2+h^2} \\ && \cos^2 \beta &= \frac{b^2}{b^2+h^2} \\ && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &= \frac{\frac{d^2}{d^2+h^2}+\frac{b^2}{b^2+h^2}-2\cdot \frac{d^2}{d^2+h^2} \cdot \frac{b^2}{b^2+h^2}}{1 - \frac{d^2}{d^2+h^2} \cdot\frac{b^2}{b^2+h^2}} \\ &&&= \frac{d^2(b^2+h^2)+b^2(d^2+h^2)-2d^2b^2}{(d^2+h^2)(b^2+h^2)-d^2b^2} \\ &&&= \frac{h^2(b^2+d^2)}{h^2(b^2+d^2+h^2)} \\ &&&= \frac{b^2+d^2}{b^2+d^2+h^2} \\ &&&= \cos^2\phi \end{align*} Also notice that \begin{align*} && \cos^2 \alpha + \cos^2 \beta &\underbrace{\geq}_{AM-GM} 2 \cos \alpha \cos \beta \\ &&&= 2 \cos \theta \\ \Rightarrow && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &\geq \frac{2 \cos \theta - 2\cos^2 \theta}{1-\cos^2 \theta} \\ &&&= \frac{2\cos \theta}{1+\cos \theta} = \cos \theta \frac{2}{1+\cos \theta} \\ &&&> \cos^2 \theta \\ \Rightarrow && \phi &< \theta \end{align*}

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Solution:

The line \(AN\) is \(\mathbf{a} + \alpha (\mathbf{n}-\mathbf{a})\) \\ The line \(BM\) is \(\mathbf{b} + \beta (\mathbf{m} - \mathbf{b})\) The point \(OQ = OB + BQ = \mathbf{b} + \frac{1}{\mu+1} (\mathbf{m}-\mathbf{b})\) It is also \(OQ = OA + AQ = \mathbf{a} + \frac{1}{\nu+1} ( \mathbf{n} - \mathbf{a})\) \begin{align*} && \mathbf{q} &= \mathbf{a} + \frac{1}{\nu+1} ( n\mathbf{b} - \mathbf{a}) \\ && \mathbf{q} &= \mathbf{b} + \frac{1}{\mu+1} ( m\mathbf{a} - \mathbf{b}) \\ \Rightarrow && \frac{\nu}{\nu+1} &= \frac{m}{\mu+1} \\ \Rightarrow && m &= \frac{(1+\mu)\nu}{1+\nu} \\ \Rightarrow && \mathbf{m} &= \frac{(1+\mu)\nu}{1+\nu} \mathbf{a} \end{align*} By similar triangles (\(\triangle OAN \sim \triangle OML\), we can observe that \(\lambda = \mu \nu\). The significance of \(\mu \nu < 1\) \(L\) lies on the side \(OB\) and both \(M\) and \(N\) lie between \(O\) and \(A\) and \(B\) respectively.

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Solution:

1. \(A\), \(Q\), \(C\) lie on a straight line if \(q = \lambda a + (1-\lambda)c\) for some \(\lambda \in \mathbb{R}\), \begin{align*} && q &= \lambda a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)(a-c) \\ \Leftrightarrow && \frac{q - a}{c-a} &= 1-\lambda \\ \end{align*} therefore \(\frac{q-a}{c-a} \in \mathbb{R}\) \begin{align*} && \frac{q-a}{c-a} & \in \mathbb{R} \\ \Leftrightarrow && \left (\frac{q-a}{c-a} \right)^* &= \frac{q-a}{c-a} \\ \Leftrightarrow && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \end{align*} Given \(aa^* = cc^* = 1\), \begin{align*} && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \Leftrightarrow && q^*(c-a) - \frac{c}{a}+1 &= q \frac{a-c}{ca} - \frac{a}{c}+1 \\ \Leftrightarrow && (c-a)\l q^* +\frac{q}{ca}\r &= \frac{c}{a} - \frac{a}{c} \\ &&&= \frac{c^2-a^2}{ac} \\ \Leftrightarrow && q^* +\frac{q}{ca} &= \frac{c+a}{ac} \\ \Leftrightarrow && q^*ac +q &= a+c \end{align*}
2. Since \(Q\) lies on \(AC\) and \(BD\) we must have \begin{align*} &&& \begin{cases} q^*ac +q &= a+c \\ q^*bd +q &= b+d \\ \end{cases} \\ \Rightarrow && q^*(ac-bd) &= (a+c)-(b+d) \\ \Rightarrow && q(ac-bd) &= (b+d)ac-(a+c)bd \\ \Rightarrow && (q+q^*)(ac-bd) &= (a+c)(1-bd)+(b+d)(ac-1) \\ &&&=a-abd+c-bcd+abc-b+acd-d \\ &&&= a(1+cd)-b(1+cd)+c(1+ab)-d(1+ab) \\ &&&= (a-b)(1+cd)+(c-d)(1+ab) \end{align*}
3. If \(AB\) and \(CD\) meet at \(p\) we must have \(p^*ab + p = a+b\), ie \(p(1+ab) = a+b\) amd \(p(1+cd) = c+d\), so \begin{align*} && (q+q^*)(ac-bd) &= (a-b) \frac{c+d}{p} + (c-d) \frac{a+b}{p} \\ \Leftrightarrow && p(q+q^*)(ac-bd) &= (a-b)(c+d)+(c-d)(a+b) \\ &&&= ac+ad-bc-bd+ac+bc-ad-bd \\ &&&= 2(ac-bd) \\ \Leftrightarrow && p(q+q^*) &= 2 \end{align*}

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Solution: \begin{align*} && 2yy' &= 4a \\ \Rightarrow && y' &= \frac{2a}{y} = \frac{2a}{2ap} = \frac1p \\ \Rightarrow && \frac{y-2ap}{x-ap^2} &= \frac1p \\ \Rightarrow && y &= \frac1p x +ap \end{align*} The other tangent will be \(y = \frac1qx+aq\) \begin{align*} &&& \begin{cases} py-x &= ap^2 \\ qy - x &= aq^2 \end{cases} \\ \Rightarrow && y(p-q) &= a(p^2-q^2) \\ \Rightarrow && y &= a(p+q) \\ && x &= apq \end{align*} Therefore \(R(apq, a(p+q)), S(0, ap), T(0, aq)\).

The line \(PQ\) has equation \begin{align*} && \frac{y - 2ap}{x-ap^2} &= \frac{2aq-2ap}{aq^2-ap^2} \\ &&&= \frac{2}{p+q} \\ y= 0: && x - ap^2 &= -(p+q)ap \\ \Rightarrow && x&= -apq \end{align*} So set \(X(-apq, 0)\) \begin{align*} && [RST] &= \frac12 \cdot a(p-q) \cdot (-apq) = \frac12 a^2 |qp(p-q)| \\ \\ && [OPQ] &= [OPX] + [OQX] \\ &&&= \frac12 \cdot (-apq) \cdot 2ap + \frac12 \cdot (-apq) \cdot (-2aq) \\ &&&= -\frac12a^2pq \left (2p-2q \right) = a^2|pq(p-q)| = 2[RST] \end{align*} as required

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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: The line through \(A\) perpendicular to \(BC\) is \(\mathbf{a} + \lambda\mathbf{u}\). The line through \(B\) perpendicular to \(CA\) is \(\mathbf{b} + \mu \mathbf{v}\). They intersect when \(\mathbf{a} + \lambda\mathbf{u} = \mathbf{b} + \mu \mathbf{v}\). Since \(\mathbf{v}\) is perpendicular to \(CA\), we must have \begin{align*} && \mathbf{a} + \lambda\mathbf{u} &= \mathbf{b} + \mu \mathbf{v} \\ \Rightarrow && \mathbf{a}\cdot(\mathbf{c}-\mathbf{a}) + \lambda\mathbf{u}\cdot(\mathbf{c}-\mathbf{a}) &= \mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) + \mu \mathbf{v}\cdot(\mathbf{c}-\mathbf{a}) \\ \\ \Rightarrow && \lambda &= \frac{\mathbf{b}\cdot(\mathbf{c}-\mathbf{a}) -\mathbf{a}\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u}\cdot(\mathbf{c}-\mathbf{a})} \\ &&&= \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \end{align*} Therefore the point is \(\mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u}\). The line \(CP\) is \(\mathbf{c} + \nu \left (\mathbf{p} - \mathbf{c} \right)\), to check this is perpendicular with \(AB\) we should dot \(\mathbf{p}-\mathbf{c}\) with \(\mathbf{a}-\mathbf{b}\), ie \begin{align*} && (\mathbf{p}-\mathbf{c}) \cdot (\mathbf{a}-\mathbf{b}) &= \left ( \mathbf{a} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} - \mathbf{c}\right) \cdot ( \mathbf{a}-\mathbf{b}) \\ &&&= \left ( \mathbf{a}- \mathbf{c} + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})} \mathbf{u} \right) \cdot ( \mathbf{a}-\mathbf{c}+(\mathbf{c}-\mathbf{b})) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) + \frac{(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})}{\mathbf{u} \cdot(\mathbf{c}-\mathbf{a})}\mathbf{u} \cdot (\mathbf{a}-\mathbf{c}) + \\ &&&\quad (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) + \lambda \underbrace{\mathbf{u} \cdot (\mathbf{c}-\mathbf{b})}_{=0} \\ &&&=(\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}) -(\mathbf{b}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{a})+ (\mathbf{a}-\mathbf{c})\cdot(\mathbf{c}-\mathbf{b}) \\ &&&= (\mathbf{a}-\mathbf{c})\cdot(\mathbf{a}-\mathbf{c}+\mathbf{b}-\mathbf{a}+\mathbf{c}-\mathbf{b}) \\ &&&= 0 \end{align*} as required.

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Solution:

1. We can map \(a \mapsto 0\), rotate the whole plane, then shift the plane back to \(a\), so \(z \mapsto (z-a) \mapsto e^{i \theta}(z-a) \mapsto a + e^{i \theta}(z-a) = e^{i \theta}z + a(1 - e^{i\theta})\)
2. \(z \mapsto e^{i \theta}z + a(1 - e^{i\theta}) \mapsto e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi})\) \begin{align*} e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi}) &= e^{i(\phi + \theta)}z + ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \\ \end{align*} Therefore this is rotation by angle \(\phi + \theta\) and about \begin{align*} \frac{ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi})}{1 - e^{i(\phi + \theta)}} &= \frac{e^{-i\frac{(\phi + \theta)}{2}} \l ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{\l ae^{i\frac{\phi-\theta}{2}} - ae^{i \frac{(\theta + \phi)}{2}} + b(e^{-i\frac{(\phi + \theta)}{2}} -e^{i\frac{(\phi - \theta)}{2}}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{ae^{i\frac{\phi}{2}} 2i\sin(\frac{\theta}{2}) + be^{-i\frac{\theta}{2}}2i\sin\frac{\phi}{2} }{2i \sin(\frac{\phi + \theta}2)} \\ \end{align*} as required. If \(\phi + \theta = 2\pi\), then \(z \mapsto z + (b-a)(1 - e^{i\phi})\) which is a translation.
3. If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \begin{align*} && a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi &= b\,\e^{ {\mathrm i}\theta/2}\sin \tfrac12\phi + a\,\e^{-{\mathrm i} \phi/2}\sin \tfrac12 \theta \\ && a\,(\e^{ {\mathrm i}\phi/2}-\e^{-{\mathrm i}\phi/2})\sin \tfrac12\theta + b\,(\e^{-{\mathrm i} \theta/2}-\e^{+{\mathrm i} \theta/2})\sin \tfrac12 \phi &= 0 \\ && a \sin \frac{\phi}{2} \sin \frac{\theta}{2}-b \sin \frac{\theta}{2} \sin \frac{\phi}{2} &= 0 \\ \Leftrightarrow && a = b \text{ or } \sin \frac{\theta}{2} = 0 \text{ or } \sin \frac{\phi}{2} = 0 \\ \Leftrightarrow && a = b \text{ or } \theta = 0 \text{ or } \phi = 0 \\ \end{align*} If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \(b = a\) or \(e^{i\phi} = e^{i \theta}\) ie rotation through the same angle.