Problems

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Solution:

1. \(\,\)
   

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   Notice that \begin{align*} && (a+b)^2 &= PQ^2 + (a-b)^2 \\ \Rightarrow && PQ^2 &= 4ab \\ && (b+c)^2 &= QR^2 + (c-b)^2 \\ \Rightarrow && QR^2 &= 4bc \\ && (a+c)^2 &= PR^2 + (a-c)^2 \\ \Rightarrow && PR^2 &= 4ac \\ \Rightarrow && 2\sqrt{ac} &= 2\sqrt{ab}+2\sqrt{bc} \\ \Rightarrow && \frac{1}{\sqrt{b}} &= \frac{1}{\sqrt{c}} + \frac1{\sqrt{a}} \\ \end{align*} Let \(x, y, z = \frac{1}{\sqrt{a}}, \frac1{\sqrt{b}}, \frac{1}{\sqrt{z}}\) so we would like to prove that \(2(x^4+y^4+z^4) = (x^2+y^2+z^2)^2\) or \(x^4+y^4+z^4 = 2x^2y^2+2y^2z^2+2z^2x^2\). We also have \begin{align*} && y &= x+z \\ \Rightarrow &&y^2 &= x^2+z^2+2xz \\ \Rightarrow && (y^2-x^2-z^2)^2 &= 4x^2z^2 \\ \Rightarrow && y^4+x^4+z^4 - 2x^2y^2-2y^2z^2+2x^2z^2 &= 4x^2z^2\\ \Rightarrow && y^4+x^4+z^4 &= 2x^2y^2+2y^2z^2+2z^2x^2 \end{align*}
2. Notice that subject to \(y > z > x\) all these steps are reversible, so we must have the equality we desire

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Solution:

Notice that \(\mathbf{x} = m\mathbf{a}\) since \(OX\) is parallel to \(OA\) and \(0 < m < 1\) since \(X\) lies between them. \(\overline{OC} = \overline{OB} + \overline{BC} = \mathbf{b} + k\mathbf{a}\) since \(BC\) is parallel to \(OA\), \(k\) can take any value. The line \(OB\) is \(\lambda \mathbf{b}\), the line \(AC\) is \(\mathbf{a} + \mu (\mathbf{c}-\mathbf{a}) = \mu \mathbf{b} +(1+ \mu(k-1)) \mathbf{a}\) Therefore they meet when \(\mu = \lambda\) and \((1+\mu(k-1)) = 0\), ie \(\mu = \frac{1}{1-k}\) so \(D\) is \(\frac{1}{1-k} \mathbf{b}\) The line \(XD\) is \(m\mathbf{a} + \nu ( \frac{1}{1-k} \mathbf{b} - m \mathbf{a}) \) and \(BC\) is \(\mathbf{b} + \eta \mathbf{a}\) so they meet when \(\nu = 1-k\) and \(\eta = m-(1-k)m = km\). Therefore \(Y = \mathbf{b} + km \mathbf{a}\) Therefore the line \(OY\) is \(\alpha(\mathbf{b} + km \mathbf{a})\) and AB is \(\mathbf{a} + \beta(\mathbf{b} - \mathbf{a})\) so they intersect when \(\alpha = \beta\) and \(\alpha km = (1-\alpha) \Rightarrow \alpha = \frac{1}{1+km}\). Therefore \(Z = \mathbf{a} + \frac{1}{1+km} (\mathbf{b} - \mathbf{a}) = \frac{\mathbf{b}+km\mathbf{a}}{1+km}\) The lines \(DZ\) and \(OA\) are \(\frac{1}{1-k} \mathbf{b} + \gamma \left ( \frac{1}{1-k} \mathbf{b} - \frac{\mathbf{b}+km\mathbf{a}}{1+km} \right)\) and \(\delta \mathbf{a}\). Therefore they intersect when \(\frac{1}{1-k} + \gamma \left (\frac{1}{1-k} - \frac{1}{1+km} \right) = 0 \Rightarrow \gamma = \frac{(1-k)(1+km)}{(k-1)k(m+1)} = -\frac{1+km}{k(m+1)}\) and \(\delta = -\gamma \frac{km}{1+km} = \frac{m}{m+1}\). Therefore \(OT = \frac{m}{m+1} |\mathbf{a}|, OA = |\mathbf{a}|, OX = m|\mathbf{a}|, TA = \frac{1}{m+1}|\mathbf{a}|\), Therefore \(OT \times OA = OX \times TA\). Also \(\frac{1}{OX} + \frac{1}{OA} = \frac{1}{m|\mathbf{a}|} + \frac{1}{|\mathbf{a}|} = \frac{m+1}{m|\mathbf{a}|} = \frac{1}{OT}\)

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Solution:

1. \begin{align*} && 2y \frac{\d y}{\d x} &= 4a \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{2a}{y} \end{align*} Therefore we must have \begin{align*} && \underbrace{-\frac{2aq}{2a}}_{\text{gradient of normal}} &= \underbrace{\frac{2ap-2aq}{ap^2-aq^2}}_{\Delta y / \Delta x} \\ \Rightarrow && -q &= \frac{2}{p+q} \\ && 0 &= 2 + pq+q^2 \end{align*}
2. We must have that \(q,r\) are the two roots of \(x^2+px+2 = 0\) \(QR\) has the equation: \begin{align*} && \frac{y-2aq}{x-aq^2} &= \frac{2ar-2aq}{ar^2-aq^2} \\ \Rightarrow && \frac{y-2aq}{x-aq^2} &= \frac{2}{r+q} \\ \Rightarrow && y &= \frac{2}{q+r}(x-aq^2) +2aq \\ && y &= -\frac{2}{p}x+2a\left(q-\frac{q^2}{q+r} \right) \\ &&y&= -\frac{2}{p}x+2a \frac{qr}{q+r} \\ && y &= -\frac{2}{p}x - 2a \frac{2}{p} \\ && y & = -\frac{2}{p}(x+2a) \end{align*} Therefore the point \((-2a,0)\) lies on all such lines.
3. \(OP\) has equation \(y = \frac{2}{p} x\) \begin{align*} && y &= \frac{2}{p} x \\ && y & = -\frac{2}{p}(x+2a) \\ && 2y &= -\frac{4a}{p} \\ \Rightarrow && y &= -\frac{2a}{p} \\ && x &= -a \end{align*} Therefore \(T\left (-a, -\frac{2a}{p} \right)\) always lies on the line \(x = -a\) The distance to the \(x\)-axis from \(T\) is \(\frac{2a}{|p|}\). We need to show that \(p\) can't be too small. Specifically \(x^2+px+2 = 0\) must have \(2\) real roots, ie \(\Delta = p^2-8 \geq 0 \Rightarrow |p| \geq 2\sqrt{2}\), ie \(\frac{2a}{|p|} \leq \frac{2a}{2\sqrt{2}} = \frac{a}{\sqrt{2}}\) as required.

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Solution: Notice that \(\omega^n = e^{2\pi i} = 1\), so \(\omega\) is a root of \(z^n - 1\), notice also that \((\omega^k)^n =1\) so therefore the \(n\) roots are \(1, \omega, \omega^2, \cdots, \omega^{n-1}\) and so \((z-1)(z-\omega) \cdots (z-\omega^{n-1}) = C(z^n-1)\). By considering the coefficient of \(z^n\) we can see that \(C = 1\).

1. \(P\) lies on the perpendicular bisect of \(1\) and \(\omega\), so \(p = re^{\pi i/n}\), where \(r\) can be positive or negative, but \(|r| = |OP|\). \begin{align*} && |PX_0| \times |PX_1| \times \cdots \times |PX_{n-1}| &= |(p-1)(p-\omega) \cdots (p-\omega^{n-1})| \\ &&&= |p^n - 1| \\ &&&= |r^ne^{\pi i} - 1| \\ &&&= |-|OP|^n - 1| \tag{since \(n\) even} \\ &&&= |OP|^n+1 \end{align*} If \(n\) is odd, depending on the sign of \(r\) we get \(|OP|^n+1\) or \(||OP|^n-1|\).
2. \(\,\) \begin{align*} && (z-\omega) \cdots(z-\omega^{n-1}) &= \frac{z^n-1}{z-1} \\ &&&= 1 + z +\cdots + z^{n-1} \\ && |X_0X_1| \times |X_0X_2| \times \cdots \times |X_0X_{n-1}| &= |(1 - \omega)\cdots(1-\omega^{n-1})| \\ &&&= 1+1+1^2+\cdots + 1^{n-1} \\ &&&= n \end{align*}

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Solution:

The orange guard will always see \(2b+b-a = 3b-a\) The blue guard will see \(b + \frac{b(b-a)}{a} = \frac{b^2}{a}\) if \(b < 3a\) and \(3b + \frac{b(b-3a)}{(b-a)} = \frac{2b(2b-3a)}{b-a}\). Therefore the blue guard always sees more if \(b > 3a\). He sees more in the other case if \begin{align*} && \frac{b^2}{a} &> 3b - a \\ \Leftrightarrow && \frac{b^2}{a^2} &> 3\frac{b}{a} - 1 \\ \Leftrightarrow && x^2 - 3x + 1 &> 0\\ \Leftrightarrow && x > \frac{3 + \sqrt{5}}{2} \text{ or } x < \frac{3-\sqrt{5}}{2} \end{align*} Since \(b > a\) we must have \(b > \frac{3+\sqrt{5}}2 a\) There is an alternative interpretation which is that the orange guard is in the top left corner, ie In this case the green guard will always see \(2b + \frac{2b(b-a)}{b+a} = \frac{4b^2}{b+a}\) Comparing \(\frac{4b^2}{b+a}\) with \(\frac{b^2}{a}\) we can see the former is larger if \(3a > b\). Comparing \(\frac{4b^2}{b+a}\) with $$

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Solution: Let \(O\) be the origin, and let \(\mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}'\) be the four points. The conditions give us \begin{align*} && \mathbf{a} \cdot \mathbf{b} &= 0 \\ && |\mathbf{a}| &= |\mathbf{b}| \\ && \mathbf{a}' \cdot \mathbf{b}' &= 0 \\ && |\mathbf{a}'| &= |\mathbf{b}'| \\ \end{align*} So \begin{align*} \text{midpoint }AB \text{ to midpoint } BA' &= (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}'))\cdot (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}')) \\ &= \frac12(\mathbf{a}-\mathbf{a}')\cdot \frac12(\mathbf{a} - \mathbf{a}') \\ \text{midpoint }BA' \text{ to midpoint } A'B' &= (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}')) \cdot (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}'))\\ &= \frac12(\mathbf{b}-\mathbf{b}')\cdot \frac12(\mathbf{b} - \mathbf{b}') \\ &= \frac14 (|\mathbf{b}|^2 + |\mathbf{b}'|^2 - 2\mathbf{b}\cdot\mathbf{b}')\\ &= \frac14(|\mathbf{a}|^2 + |\mathbf{a}'|^2 - 2\mathbf{b}\cdot\mathbf{b}') \\ \text{midpoint }A'B' \text{ to midpoint } B'A &= (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a})) \cdot (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a}))\\ &= \frac12(\mathbf{a}'-\mathbf{a})\cdot \frac12(\mathbf{a}' - \mathbf{a}) \\ \text{midpoint }B'A \text{ to midpoint } AB &= (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b})) \cdot (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b}))\\ &= \frac12(\mathbf{b}'-\mathbf{b})\cdot \frac12(\mathbf{b}' - \mathbf{b}) \\ \end{align*} So it's sufficient to prove \(\mathbf{a}\cdot \mathbf{a}' = \mathbf{b}\cdot \mathbf{b}'\) but this is clear from looking at a diagram for 1 second. Given the length of the square is what it is, we want to minimise \(\mathbf{b}\cdot \mathbf{b}'\) which is when they are vertically opposite each other, ie \(\angle BOA' = 90^\circ\)

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Solution:

1. Notice that \(P\) lies on \(C_1C_2\), and that the triangles formed from \(C_iPT_i\) where \(T_i\) are the tangent points are similar, with ratios \(\frac{r_1}{r_2}\). Therefore \(\frac{C_1P}{r_1} = \frac{C_2P}{r_2}\), and hence \(\frac{C_1P}{C_1C_2} = \frac{C_1P}{C_1P-C_2P} = \frac{1}{1-\frac{r_2}{r_1}} = \frac{r_1}{r_1-r_2}\) So we have \(\mathbf{p} = \mathbf{x_1} + (\mathbf{x}_2 - \mathbf{x}_1)\cdot\frac{r_1}{r_1-r_2} = \frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1-r_2}\)
2. Suppose \(\mathbf{x}_3 = \binom{\alpha}{\beta}\) in the basis of \(\{ \mathbf{x}_1, \mathbf{x}_2 \}\), then we can see that \begin{align*} && \mathbf{p} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} \\ && \mathbf{q} &= \frac{r_1(\alpha \mathbf{x}_1 +\beta \mathbf{x}_2) - r_3\mathbf{x}_1}{r_1-r_3} \\ &&&= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ && \mathbf{r} &=\frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ && \mathbf{p}-\mathbf{q} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} - \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ &&&= \frac{1}{(r_1-r_2)(r_1-r_3)} \binom{(r_1-r_3)(-r_2)-(r_1-r_2)(r_1\alpha-r_3)}{(r_1-r_3)r_1 - (r_1-r_2)r_1\beta} \\ &&&= \frac{r_1}{(r_1-r_2)(r_1-r_3)} \binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \\ && \mathbf{q} - \mathbf{r} &= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} - \frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(r_2-r_3)(r_1\alpha-r_3) - (r_1-r_3)r_2\alpha)}{(r_2-r_3)r_1\beta - (r_1-r_3)(r_2\beta - r_3)} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(-r_2r_3+r_3^2) - \alpha(r_1r_3-r_3r_2)}{r_3(r_1-r_3)-\beta(r_1-r_2)} \\ &&&= \frac{r_3}{(r_1-r_3)(r_2-r_3)}\binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \end{align*} Therefore they are clearly parallel, and hence lie on a line.
3. \(Q\) is halfway between \(P\) and \(R\) if \begin{align*} && \frac{r_1}{(r_1-r_2)(r_1-r_3)} &= \frac{r_3}{(r_1-r_3)(r_2-r_3)} \\ \Leftrightarrow && r_1(r_2-r_3) &= r_3(r_1-r_2) \\ \Leftrightarrow && r_1r_2 - r_1r_3 &= r_1r_3 - r_2r_3 \\ \Leftrightarrow && r_2 &= \frac{2r_1r_3}{r_1+r_3} \end{align*}

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Solution:

Notice that \(\mathbf{d} = \frac{1}{r+1} \mathbf{b}\) and \(\mathbf{e} = \frac{s}{s+1}\mathbf{a}\). We must also have that the line \(AD\) is \(\mathbf{a} + \lambda \left (\mathbf{a} - \frac{1}{r+1} \mathbf{b}\right)\) and \(BE\) is \(\mathbf{b} + \mu \left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right)\) at their point of intersection we must have \begin{align*} && \mathbf{a} + \lambda \left (\mathbf{a} - \frac{1}{r+1} \mathbf{b}\right) &= \mathbf{b} + \mu \left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right) \\ [\mathbf{a}]: && 1 + \lambda &= -\frac{\mu s}{s+1} \\ [\mathbf{b}]: && -\frac{\lambda}{r+1} &= 1 + \mu \\ \Rightarrow && \lambda &= -\frac{1+s+\mu s}{s+1} \\ \Rightarrow && \mu &= \frac{1+s+\mu s}{(1+r)(1+s)} - 1 \\ \Rightarrow && (1+r+rs)\mu &= 1+s - 1 - r - s - rs \\ \Rightarrow && \mu &= -\frac{r+rs}{1+r+rs} \\ \Rightarrow && \mathbf{g} &= \mathbf{b} -\frac{r+rs}{1+r+rs}\left (\mathbf{b} - \frac{s}{s+1} \mathbf{a}\right) \\ &&&= \frac{rs}{1+r+rs} \, \mathbf{a} + \frac 1 {1+r+rs} \, \mathbf{b} \end{align*} \item The line \(OG\) is \(\lambda \mathbf{g}\). The line \(AB\) is \(\mathbf{a} + \mu(\mathbf{b}-\mathbf{a})\), so we need \begin{align*} && \lambda \mathbf{g} &= \mathbf{a} + \mu(\mathbf{b}-\mathbf{a}) \\ [\mathbf{a}]: && \lambda \frac{rs}{1+r+rs} &= 1-\mu \\ [\mathbf{b}]: && \lambda \frac{1}{1+r+rs} &= \mu \\ \Rightarrow && \lambda \frac{1+rs}{1+r+rs} &= 1 \\ \Rightarrow && \lambda &= \frac{1+r+rs}{1+rs} \\ \Rightarrow && \mu &= \frac{1}{1+rs} \end{align*} Therefore the line is divided in the ratio \(rs : 1\), and therefore we have proven Ceva's Theorem.

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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.
3. 

   + - ↺

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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. Suppose \(X*Y = Y*X\), then \begin{align*} && X * Y &= \lambda \mathbf{x} + (1-\lambda) \mathbf{y} \\ && Y * X &= \lambda \mathbf{y} + (1-\lambda) \mathbf{x}\\ \Rightarrow && 0 &= (2\lambda - 1)(\mathbf{x} -\mathbf{y}) \end{align*} Therefore, either \(\mathbf{x} = \mathbf{y}\) or \(\lambda = \frac12\). Since we assumed \(X,Y\) were distinct, \(\mathbf{x} \neq \mathbf{y}\) and so \(X*Y\) and \(Y*X\) are distinct unless \(\lambda = \frac12\)
2. Suppose \((X*Y)*Z = X*(Y*Z)\) \begin{align*} &&(X*Y)*Z &= (\lambda \mathbf{x} + (1-\lambda) \mathbf{y}) * \mathbf{z} \\ &&&= (\lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)\mathbf{z}\\ &&X*(Y*Z) &=\mathbf{x}* (\lambda \mathbf{y} + (1-\lambda) \mathbf{z}) \\ &&&= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z}\\ \Rightarrow && 0 &= (\lambda^2 - \lambda)\mathbf{x} + ((1-\lambda) - (1-\lambda)^2)\mathbf{z} \\ &&&=(1-\lambda)(-\lambda \mathbf{x} +\lambda \mathbf{z}) \\ &&&= \lambda(1-\lambda)(\mathbf{z}-\mathbf{x}) \end{align*} Therefore they are distinct unless \(\lambda = 1, 0\) or \(\mathbf{x} = \mathbf{z}\).
3. Claim: \((X*Y)*Z = (X*Z)*(Y*Z)\) Proof: \begin{align*} && (X*Y)*Z &= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ && (X*Z)*(Y*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) * (\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) + (1-\lambda)(\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda) \mathbf{z} \end{align*} Claim: \(X*(Y*Z) = (X*Y)*(X*Z)\) Proof: \begin{align*} X*(Y*Z) &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ (X*Y)*(X*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y})*(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda (\lambda \mathbf{x} + (1-\lambda)\mathbf{y}) + (1-\lambda)(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \end{align*}
4. \(P_1 = X*Y\) divides the line segment into the ratio \(\lambda:(1-\lambda)\). \(P_n\) divides the line segment \(P_{n-1}Y\) into the ratio \(\lambda:(1-\lambda)\), therefore it divides the line segment \(XY\) in the ratio \(\lambda^n : 1- \lambda^n\) Alternatively, \begin{align*} P_1 &= \lambda \mathbf{x} + (1-\lambda)\mathbf{y} \\ P_2 &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y} )*\mathbf{y} \\ &= \lambda^2 \mathbf{x} + (1-\lambda^2) \mathbf{y} \end{align*} Suppose \(P_k = \lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}\) then \begin{align*}P_{k+1} &= (\lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}) * \mathbf{y} \\ &= \lambda^{k+1}\mathbf{x} + \lambda(1-\lambda^k)\mathbf{y} + (1-\lambda)\mathbf{y}\\ & = \lambda^{k+1}\mathbf{x} + (1-\lambda^{k+1})\mathbf{y}\end{align*}

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Solution: \begin{questionparts} \item The tangent to \(y = \ln x\) is \begin{align*} && \frac{y - \ln x_1}{x - x_1} &= \frac{1}{x_1} \\ \Rightarrow && \frac{x_1y -x_1 \ln x_1}{ x- x_1} &= 1 \\ \Rightarrow && x_1 y - x_1 \ln x_1 &= x - x_1 \end{align*} So to run through the origin, we need \(\ln x_1 = 1 \Rightarrow x_1 = e\) so the line will be \(y = \frac1{e} x\) If \(ma = \ln a \Rightarrow m = \frac{\ln a}{a} = \frac{\ln b}{b} \Rightarrow b \ln a = a \ln b \Rightarrow a^b = b^a\). \item

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Solution:

1. First, not that the point must be ony the curve:
   

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   Since otherwise it's pretty clear we could make the area of the rectangle larger by moving the point onto the curve. Therefore \(A = x(f(x)-f(t))\). To maximise this we need \(xf'(x) + f(x)-f(t) = 0\), ie \(x_0f'(x_0) + f(x_0) = f(t)\)
2. Suppose \(\displaystyle \g(t) =\frac 1t \int_0^t \f(x) \d x\) then \begin{align*} && \g(t) &=\frac 1t \int_0^t \f(x) \d x\\ \Rightarrow && tg(t) &= \int_0^t \f(x) \d x \\ \Rightarrow && tg'(t) +g(t) &= f(t) \\ \Rightarrow && tg'(t) &= f(t) - g(t) \end{align*}
   

   + - ↺
   Clearly the blue area + green area is larger than the green area. So \(\displaystyle \int_0^t (f(x) - f(t))\d x > A_0(t)\). Notice that \(f(t) = \frac1{t} \int_0^t f(t) \d x \) so \(-t^2g'(t) = \int_0^t f(x) \d x > A_0(t)\)
3. Not that if \(f(x) = \dfrac{1}{1+x}\), the \(f'(x) = -\frac{1}{(1+x)^2}\) and so \begin{align*} && -\frac{x_0}{(1+x_0)^2} + \frac{1}{1+x_0} &= \frac{1}{1+t} \\ && \frac{1}{(1+x_0)^2} &= \frac{1}{1+t} \\ \Rightarrow && x_0 &= \sqrt{1+t} - 1 \\ && A_0(t) &= (\sqrt{1+t} - 1) \left ( \frac{1}{\sqrt{1+t}} - \frac{1}{t+1} \right) \\ &&&= 1 - \frac{1}{\sqrt{1+t}} - \frac{1}{\sqrt{1+t}} + \frac{1}{1+t} \\ &&&= \frac{2+t}{1+t} - \frac{2}{\sqrt{1+t}} \\ && g(t) &= \frac{1}{t} \int_0^t \frac{1}{1+x} \d x \\ &&&= \frac{\ln(1+t)}{t} \\ \Rightarrow && g'(t) &= \frac{\frac{t}{1+t} - \ln(1+t)}{t^2} \\ \Rightarrow && -t^2g(t) &= \ln(1+t) - \frac{t}{1+t} \\ \Rightarrow && \ln(1+t) - \frac{t}{1+t} &> \frac{2+t}{1+t} - \frac{2}{\sqrt{1+t}} \\ \Rightarrow && \ln \sqrt{1+t} & > 1 - \frac{1}{\sqrt{1+t}} \end{align*}

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Solution: Note that \({\bf p}_i \cdot {\bf p}_i = 1\) and \({\bf p}_i \cdot {\bf p}_j\) are all equal when \(i \neq j\) by symmetry and commutativity. \begin{align*} && 0 &= {\bf p}_i \cdot \left ( {\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4 \right) \\ &&&= 1 + \sum_{j \neq i} {\bf p}_i \cdot {\bf p}_j \\ &&&= 1 + 3 {\bf p}_i \cdot {\bf p}_j \\ \Rightarrow && {\bf p}_i \cdot {\bf p}_j &= -\frac13 \end{align*}

1. \(\,\) \begin{align*} && (XP_i)^2 &= ({\bf p}_i- {\bf x)}\,.\,({\bf p}_i- {\bf x}) \\ &&&= {\bf p}_i \cdot {\bf p}_i - 2 {\bf p}_i \cdot {\bf x} + {\bf x} \cdot {\bf x} \\ &&&= 2 - 2 {\bf p}_i \cdot {\bf x} \\ \Rightarrow && \sum_i (XP_i)^2 &= \sum_i \left (2 - 2 {\bf p}_i \cdot {\bf x} \right) \\ &&&= 8 - 2 \sum_i {\bf p}_i \cdot {\bf x} \\ &&&= 8 - 2 \left ( {\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4 \right) \cdot {\bf x} \\ &&&= 8 \end{align*}
2. Notice we have \(1 = \left \|\begin{pmatrix} a \\0 \\b \end{pmatrix} \right \|= a^2 + b^2\) and \(-\frac13 = \begin{pmatrix} a \\0 \\b \end{pmatrix} \cdot \begin{pmatrix} 0 \\0 \\ 1 \end{pmatrix} = b\). So \(b = -1/3\) and \(a = \sqrt{1-b^2} = 2\sqrt{2}/3\). Suppose another of the vertices has coordinates \((u,v,w)\) we must have \begin{align*} && 1 &= u^2+v^2+w^2 \\ && -\frac13&=w \\ && -\frac13 &= \frac{2\sqrt{2}}3 u +\frac19 \\ \Rightarrow && u &= -\frac{\sqrt2}3 \\ \Rightarrow && 1 &= \frac19 + \frac29 + v^2 \\ \Rightarrow && v &= \pm \sqrt{\frac{2}{3}} \end{align*} So \(P_3, P_4 = (-\frac{\sqrt2}3, \pm \frac{\sqrt{6}}3, -\frac13)\)
3. \(\,\) \begin{align*} && \sum_{i=1}^4 (XP_i)^4 &= \sum_i \left (2 - 2 {\bf p}_i \cdot {\bf x} \right)^2 \\ &&&= 4 \sum_i \left (1 - {\bf p}_i \cdot {\bf x} \right)^2 \\ &&&= 4 \sum_i (1 - 2{\bf p}_i \cdot {\bf x} + ({\bf p}_i \cdot {\bf x})^2) \\ &&&= 16 + 4\sum_i ({\bf p}_i \cdot {\bf x})^2 \\ &&&=16+ 4\left ( z^2+\left (\frac{2\sqrt{2}}3x-\frac13z \right)^2 +\left (-\frac{\sqrt{2}}3x-\frac{\sqrt{6}}3y-\frac13z \right)^2 +\left (-\frac{\sqrt{2}}3x+\frac{\sqrt{6}}3y-\frac13z \right)^2 \right) \\ &&&= 16+4 \left ( \frac43z^2 + \left (\frac89 + \frac29+\frac29 \right)x^2+\left (\frac69 + \frac69 \right)y^2 + 0xz + 0yz + 0zx \right) \\ &&&= 16+ 4\cdot\frac43(x^2+y^2+z^2) \\ &&&=16+\frac{16}{3}=\frac{64}{3} \end{align*}

Note: It may be better to view the last part of this question in terms of linear transformations. There are two possible approaches. One is to show \(T:{\bf x} \mapsto \sum_i ({\bf p}_i \cdot x) {\bf p}_i\) is \(\frac43I\) (easy since it has three eigenvectors with the same eigenvalue which span \(\mathbb{R}^3\) and we are interested in the value \({\bf x} \cdot T\mathbf{x} = \frac43 \lVert {\bf x} \rVert^2\). The second is to consider \(\sum_I ({\bf p}_i \cdot {\bf x})^2 = {\bf x}^TM{\bf x}\) where \(M = \sum_i {\bf p}_i{\bf p}_i^T\) and note that this matrix is invariant under rotations.