Problems

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Solution: \begin{align*} x^4 - 2x^3+x^2 &= x^2(x^2-2x+1) \\ &= x^2(x-1)^2 > 0 \end{align*} Since \(x\)and \(x-1\) can't both be zero, and square cannot be negative. \begin{align*} x^2 - 8x+17 &= (x-4)^2 +1 \geq 1 > 0 \end{align*} If \(x \leq 2\) then \(x^4 - 2x^3+x^2 > 0\) and \(17-8x \geq 1\) so \(x^4-2x^3+x^2-8x+17 > 0\) If \(x > 2\) then \(x^4-2x^3 = x^3(x-2) \geq 0\) and \(x^2-8x+17 > 0\) so \(x^4-2x^3+x^2-8x+17 > 0\), so for all real numbers our polynomial is positive and therefore cannot have any roots. Note that: \(x^4-x^3+x^2 = x^2(x^2-x+1) > 0\) and \(x^2-4x+4 =(x-2)^2 \geq 0\) If \(x < 1\) then \(x^4-x^3+x^2 > 0\) and \(4(1-x) > 0\) so \(x^4-x^3+x^2-4x+4 > 0\). If \(x > 1\) then \(x^4-x^3 = x^3(x-1) > 0\) and \(x^2-4x+4 \geq 0\) therefore \(x^4-x^3+x^2-4x+4 > 0\). Therefore \(x^4-x^3+x^2-4x+4 > 0\) for all real \(x\) and hence there are no real roots.

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Solution: \begin{align*} \sin(A+B)\sin(A+D) - \sin B \sin D &= \sin (A+B)\sin(\pi - B-C) - \sin B \sin (\pi - A - B - C) \\ &= \sin (A+B)\sin(B+C) - \sin B \sin(A+B+C) \\ &= \sin(A+B)\sin (B+C) - \sin B (\sin (A+B)\cos C +\cos(A+B) \sin C) \\ &= \sin(A+B)\cos B \sin C + \cos(A+B)\sin B \sin C \\ &= \sin A \sin C \cos^2 B + \cos A \sin B \cos B \sin C - \cos A \cos B \sin B \sin C + \sin A \sin^2 B \sin C \\ &= \sin A \sin C (\cos^2 B + \sin^2 B) \\ &= \sin A \sin C \end{align*}

Using the extended form of the sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\) where \(R\) is the circumradius, we have \begin{align*} PR \times QS &= 2R \sin (A+D) \times 2R \sin (A+B) \\ &= 4R^2 \l \sin A \sin C + \sin B \sin D \r \\ &= 2R \sin A \times 2R \sin C + 2R \sin B 2R \sin D \\ &= PS \times QR + PQ \times RS \end{align*} as required.

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Solution:

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\((i)\) \((v)\) \((vii)\)

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Solution: With \(n=1\) line, the plane is divided in half. With \(n=2\) lines the plane is divided into four pieces. (Each of the previous pieces are split in half) With \(n=3\) lines the plane is divided into up to \(7\) pieces. (The new line crosses two lines in two places dividing \(3\) regions into \(2\), thus increasing the number of regions by \(3\)). With \(n=4\) lines the plane is divided into \(11\) pieces. (The new line crosses three lines in three places doubling the number of regions of \(4\) places). Claim: With \(n\) lines the plane is divided into \(\frac12(n^2+n+2)\) regions. Proof: (By induction) (Base case) When \(n=1\) clearly the line is divided into \(2\) regions, and \(\frac12 (1^2 + 1^2 + 2) = 2\) so the base case is true. (Inductive step) Suppose our formula is true for \(n=k\), so we have placed \(k\) lines in general position and divided the plane into \(\frac12(k^2+k+2)\) regions. When we place a new line it will meet those \(k\) lines in \(k\) places (since no lines are parallel) and there will be k+1 regions the line will run through (since no three lines meet at a point). Each of those \(k+1\) regios is now split in half, so there are \(k+1\) "new regions". Therefore there are now \(\frac12(k^2+k+2)+(k+1) = \frac12(k^2+k+1+2k+2) = \frac12 ((k+1)^2+(k+1)+1)\) regions, ie our hypothesis is true for \(n=k+1\). (Conclusion) Therefore since our statement is true for \(n=1\) and since if it is true for some \(n=k\) it is true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 1\) Proof: (Alternative). There are \(\binom{n}{2}\) places where the lines meet. Each intersection is the most extreme point (say lowest) for one region (if two are tied, rotate by a very small amount) so this is a unique mapping. There will be \(n+1\) regions which are infinite and don't have a most extreme point, hence \(\binom{n}{2} + n+1 = \frac12(n^2-n)+n+1 = \frac12(n^2+n+2)\)

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Solution: The perpendicular bisector of the chords runs through the origin, therefore \(LM\) is perpendicular to \(PQ\) if and only if \(\frac{l+m}{2}\) is perpendicular to \(\frac{p+q}{2}\), ie \begin{align*} && (l+m) &= it (p+q) \\ \Leftrightarrow && \frac{l+m}{p+q} & \in i \mathbb{R} \\ \Leftrightarrow && 0 &= \frac{l+m}{p+q} + \frac{l^*+m^*}{p^*+q^*} \\ &&&= \frac{l+m}{p+q} + \frac{\frac{r^2}{l}+\frac{r^2}{m}}{\frac{r^2}{p}+\frac{r^2}{q}} \\ &&&=\frac{l+m}{p+q} + \frac{l+m}{p+q} \frac{pq}{lm} \\ &&&= \frac{l+m}{p+q} \left ( \frac{lm+pq}{lm} \right) \end{align*} Therefore as long as \(l+m, p+q \neq 0\) \(lm+pq = 0\) is equivalent to the chords being perpendicular. In the case where (say) \(l,m\) is a diameter, then the condition for the chords to be perpendicular is that \(p,q\) is also a diameter and at right angles, but clearly this is also equivalent to our condition. Suppose \(A_1, A_2, A_3\) are distinct points on \(S\), and \(A_1'\) is given and suppose \(a_i, a_i'\) are the corresponding complex numbers, then the conditions are: \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_1'' \perp A_3A_1: && 0 &= a_3'a_1'' + a_3a_1 \\ \\ \Rightarrow && a_2' &= -\frac{a_1a_2}{a_1'} \\ && a_3' &= -\frac{a_2a_3}{a_2'} \\ &&&= \frac{a_1'a_2a_3}{a_1a_2} \\ &&&= \frac{a_1'a_3}{a_1} \\ && a_1'' &= - \frac{a_3a_1}{a_3'} \\ &&&= \frac{a_3a_1a_1}{a_1'a_3} \\ &&&= \frac{a_1^2}{a_1'} \\ \Rightarrow && a_1'a_1'' &= a_1^2 \end{align*} Therefore \(a_1' = a_1''\) if \(a_1' = \pm a_1\) Suppose we have \(4\) points, then \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_4' \perp A_3A_4: && 0 &= a_3'a_4' + a_3a_4 \\ A_4'A_1'' \perp A_4A_1: && 0 &= a_4'a_1'' + a_4a_1 \\ \\ \Rightarrow && a_4' &= -\frac{a_3a_4}{a_3'} \\ &&&= -\frac{a_1a_3a_4}{a_1'a_3} \\ &&&= -\frac{a_1a_4}{a_1'} \\ \Rightarrow && a_1'' &= -\frac{a_4a_1}{a_4'} \\ &&&= \frac{a_4a_1a_1'}{a_1a_4} \\ &&&= a_1' \end{align*} So they coincide. For \(n\) points if there are an even number of points they coincide, an odd number and there are two points when they coincide.

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Solution:

1. If \(b > a\) and \(c > 1\) then \(c \geq 2 \underbrace{\Rightarrow}_{\times c} bc \geq 2c = c+c \underbrace{\geq}_{c \geq 2} 2 + c\)
2. If \(a < b\) and \(d < c\) then \(b \geq a+1\) and \(c \geq d+1\) so \begin{align*} bc - ad &\underbrace{\geq}_{b \geq a+1} (a+1)c - ad \\ &\underbrace{\geq}_{d \leq c-1} (a+1)c - a(c-1) \\ &= a+c \end{align*}
3. If \(\displaystyle \frac{a}{b} < p < \frac{c}{d}\) then \(a < pb\) and \(pd < c\) so by the previous part \((pb)c - a(pd) \geq a + c \Leftrightarrow (bc-ad)p \geq a+c\).
4. If \(\displaystyle \frac{a}{b} < \frac{p}{q} < \frac{c}{d}\) then \(\displaystyle \frac{qa}{b} < p < \frac{qc}{d}\) and so by the previous part we must have \((bc-ad)qp \geq q(a+c) \Rightarrow p \geq \frac{a+c}{bc-ad}\). Similarly we have \(\frac{d}{c} < \frac{q}{p} < \frac{b}{a}\) and so \(q \geq \frac{b+d}{bc-ad}\)

Suppose \(\frac{p}{q}\) is a fraction such that \(q \leq 20\) and \(\frac89 < \frac{p}q < \frac9{10}\) then: \begin{align*} q & \leq 20 \\ p & \geq \frac{17}{81-80} = 17 \\ q & \geq \frac{19}{81-80} = 19 \end{align*} Therefore the only fraction is \(\frac{17}{19}\) since \(\frac{18}{19} > \frac{18}{20} = \frac{9}{10}\)

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Solution: Suppose \(f(t) = Ae^{pt}\) is a solution, then \begin{align*} && 0 &= Ape^{pt} + Ae^{pt} - Ake^{p(t-1)} \\ \Leftrightarrow && 0 &= p +1 - ke^{-p} \\ \Leftrightarrow && p+1 &= ke^{-p} \end{align*}

If we sketch \(y = x+1\) and \(y = ke^{-x}\) we can see that when \(k \geq 0\) there will be exactly one solution. If \(k < 0\) there can be no solutions (if \(k\) is large and negative) and two solutions if \(k\) is small and negative. There will be exactly one real root if \(y = x+1\) is tangent to \(y = ke^{-x}\). The gradient of \(y = ke^{-x}\) is \(-ke^{-x}\) so to have a gradient \(1\) we must have \(x+1 = -1 \Rightarrow x = -2\), but also \(x = -\ln(-k) \Rightarrow k = -e^{-2}\). Therefore there will be so solution as long as \(k \geq -e^{-2}\). The solutions will tend to \(0\) as \(t \to + \infty\) as long as the intersection point is less than zero. For \(k \geq 0\) they intersect at \(0\) when \(k =1\), so we would want \(k < 1\). All negative values of k will work since the intersection has to happen for negative \(p\). Therefore the range is \(-e^{-2} \leq k < 1\)

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Solution: Consider \(\frac12 x(t)^2\) then differentiating we obtain \(x(t)x'(t) = x(t)y(t)\). Also note that \(x(0) = \int_0^0 y(u) \d u = 0\) Therefore, \begin{align*} \int_0^1 x(t)y(t) \d t &= \left [ \frac12 x(t)^2 \right]_0^1 \\ &= \frac12[x(1)]^2 \end{align*} \begin{align*} && 0 &= y'' + (y^2-1)y' + y \\ \Rightarrow && 0 &= \int_0^1 \l y'' + (y^2-1)y' + y \r \d t \\ &&&= \left [y'(t) + \frac13y^3-y+x \right]_0^1 \\ &&&= x(1) \end{align*} Therefore \(x(1) = 0\). \begin{align*} && 0 &= xy'' + (y^2-1)y' x+ yx \\ \Rightarrow && 0 &= \int_0^1 \l xy'' + (y^2-1)y'x + xy \r \d t \\ &&&= \left [ x y' +(\frac13 y^3-y)x \right]_0^1 - \int_0^1 yy'+\frac13y^4-y^2 \d t \\ &&&= 0 - \frac13 \int_0^1 [y(t)]^4 \d t - \int_0^1 [y(t)]^2 \d t \\ \Rightarrow && \int_0^1 [y(t)]^2 \d t &= \frac13 \int_0^1 [y(t)]^4 \d t \end{align*} \begin{align*} && 0 &= yy'' + (y^2-1)y' y+ y^2 \\ \Rightarrow && 0 &= \int_0^1 \l yy'' + (y^2-1)y'y + y^2 \r \d t \\ &&&= \left [ y y' +(\frac14 y^4-\frac12y^2) \right]_0^1 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ &&&= 0 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ \Rightarrow && \int_0^1 [y'(t)]^2 \d t &= \int_0^1 [y(t)]^2 \d t \end{align*}

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Solution:

The curves will intersect when: \begin{align*} && \frac{1}{1+\cos \theta} &= 4 (1 + \cos \theta) \\ \Rightarrow && 1 + \cos \theta &= \pm \frac{1}{2} \\ \Rightarrow && \cos \theta &= -\frac12 \\ \Rightarrow && \theta &= \pm \frac{2\pi}{3}, \end{align*} Therefore we can measure the two sides of the boundaries. For the cardioid it will be: \begin{align*} s &= \int_{-2\pi/3}^{2 \pi /3} \sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &= \int_{-2\pi/3}^{2 \pi /3}\sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &= \int_{-2\pi/3}^{2 \pi /3}\sqrt{16(1 + \cos \theta)^2 + 16 \sin^2 \theta} \d \theta \\ &= 4\int_{-2\pi/3}^{2 \pi /3}\sqrt{2 + 2 \cos \theta} \d \theta \\ &= 8\int_{-2\pi/3}^{2 \pi /3}\sqrt{\cos^2 \frac{\theta}{2}} \d \theta \\ &= 8\int_{-2\pi/3}^{2 \pi /3}|\cos \frac{\theta}{2}| \d \theta \\ &= 16\int_{\pi}^{2\pi/3}(-\cos \frac{\theta}{2}) \d \theta + 8\int_{-\pi}^{\pi}\cos \frac{\theta}{2} \d \theta \\ &= 16 \cdot \left [ 2\sin \frac{\theta}{2}\right]_{\pi}^{2\pi/3}+ 8 \cdot 4 \\ &= 16 \cdot (\sqrt{3}-2) + 8 \cdot 4 \\ &= 16\sqrt{3} \end{align*} For the parabola we have that \(\sqrt{x^2+y^2} + x = 1 \Rightarrow x^2 + y^2 = 1 - 2x + x^2 \Rightarrow y^2 = 1-2x\). So we can parameterise our parabola as \(y = t, x = \frac{1-t^2}2\). And we are interested in the points \(t = -\sqrt{3}\) and \(t =\sqrt3\) \begin{align*} &&s &= \int_{-\sqrt3}^\sqrt3 \sqrt{\left ( \frac{\d x}{\d t} \right)^2 + \left ( \frac{\d y}{\d t} \right)^2 } \d t \\ &&&= \int_{-\sqrt3}^\sqrt3 \sqrt{t^2+1^2} \d t \\ \sinh u = t, \frac{\d t}{\d u} = \cosh u&&&= \int_{-\sinh^{-1} \sqrt3}^{\sinh^{-1}\sqrt3} \cosh^2 u \d u \\ &&&= \left [\frac12 u + \frac14 \sinh(2u) \right ]_{-\sinh^{-1} \sqrt3}^{\sinh^{-1}\sqrt3} \\ &&&= \sinh^{-1} \sqrt{3} + 2\sqrt{3} \\ &&&= \ln(2 + \sqrt{3}) + 2\sqrt{3} \end{align*} Therefore the total distance is as required.

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Solution: Since \(0 = \det \mathbf{0} = \det \mathbf{AB} = \det \mathbf{A} \det\mathbf{B}\) at least one of \(\det \mathbf{A}\) or \(\det \mathbf{B}\) is zero. If \(\det \mathbf{B} \neq 0\) then \(\mathbf{B}\) is invertible, and multiplying on the right by \(\mathbf{B}^{-1}\) gives us \(\mathbf{A} = \mathbf{0}\). If \(\mathbf{A} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 1 & 0 \\1 & 0 \end{pmatrix}\), then \(\det \mathbf{A} = \det \mathbf{B} = 0\), but \(\mathbf{AB} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \neq \mathbf{0}\) Since \(\mathbf{M}\) commutes with itself and the identity matrix, this is equivalent to factorising the polynomial over the reals. Therefore $$\mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=(\mathbf{M}-2\mathbf{I})(\mathbf{M}+\mathbf{I})(\mathbf{M}+3\mathbf{I}),$$ Since we now know at least one of \(\det (\mathbf{M}-2\mathbf{I})\), \(\det (\mathbf{M}+\mathbf{I})\), \(\det (\mathbf{M}+3\mathbf{I})\), we should look at cases: Since at least one of those must be non-zero, we must have the following cases: \((a,b) = (2,-1), (-1,2), (-1,-3), (-3,-1), (2,-3), (-3,2)\) In each of those cases, we will have: \(\begin{pmatrix} 0 & c \\ 0 & b+k \end{pmatrix}\begin{pmatrix} a+l & c \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}\) and so all of those solutions are valid. So \(c\) can be anything as long as \((a,b)\) are in that set of solutions