A {\em proper factor} of an integer \(N\) is a positive integer, not \(1\) or \(N\), that divides \(N\).
Solution:
A curve has the equation \[ y^3 = x^3 +a^3+b^3\,, \] where \(a\) and \(b\) are positive constants. Show that the tangent to the curve at the point \((-a,b)\) is \[ b^2y-a^2x = a^3+b^3\,. \] In the case \(a=1\) and \(b=2\), show that the \(x\)-coordinates of the points where the tangent meets the curve satisfy \[ 7x^3 -3x^2 -27x-17 =0\,. \] Hence find positive integers \(p\), \(q\), \(r\) and \(s\) such that \[ p^3 = q^3 +r^3 +s^3\,. \]
Solution: \begin{align*} && y^3 &= x^3 + a^3 + b^3 \\ \Rightarrow && 3y^2 \frac{\d y}{\d x} &= 3x^2 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{x^2}{y^2} \end{align*} Therefore the tangent at the point \((-a,b)\) has gradient \(\frac{a^2}{b^2}\), ie \begin{align*} && \frac{y-b}{x+a} &= \frac{a^2}{b^2} \\ \Rightarrow && b^2y - b^3 &= a^2 x + a^3 \\ \Rightarrow && b^2 y-a^2 x &= a^3 + b^3 \end{align*} Notice that tangent will be, \(4y-x = 9\) so substituting this we obtain: \begin{align*} && \left (\frac{9+x}{4} \right)^3 &= x^3 + 9 \\ \Rightarrow && 9^3 + 3 \cdot 9^2 x + 3 \cdot 9x^2 + x^3 &= 64x^3 + 64 \cdot 9 \\ \Rightarrow && 9 \cdot (9^2 - 8^2) + 9 \cdot (3 \cdot 9) + 9 \cdot 3x^2 -9 \cdot 7x^3 &= 0 \\ \Rightarrow && 7x^3-3x^2-27x-17 &= 0 \\ \Rightarrow && (x+1)^2(7x-17) &= 0 \tag{repeated root since tangent} \end{align*} So we have another point on the curve \(y^3 = x^3 + 2^3 + 1^3\), namely \((\frac{17}7, \frac{17+9 \cdot 7}{28}) = (\frac{17}7, \frac{20}{7})\), so \begin{align*} 20^3 &= 17^3 + 14^3 + 7^3 \end{align*}
The sides of a triangle have lengths \(p-q\), \(p\) and \(p+q\), where \(p>q> 0\,\). The largest and smallest angles of the triangle are \(\alpha\) and \(\beta\), respectively. Show by means of the cosine rule that \[ 4(1-\cos\alpha)(1-\cos\beta) = \cos\alpha + \cos\beta \,. \] In the case \(\alpha = 2\beta\), show that \(\cos\beta=\frac34\) and hence find the ratio of the lengths of the sides of the triangle.
Solution: The largest angle will be opposite the side with length \(p+q\). Similarly the smallest angle will be opposite the side with length \(p-q\). The cosine rule tells us that: \begin{align*} && (p+q)^2 &= p^2 + (p-q)^2 - 2p(p-q) \cos \alpha \\ && 0 &= p(p-4q-2(p-q)\cos \alpha)\\ && 0 &= p(1-2\cos \alpha) + q(2\cos \alpha - 4)\\ \Rightarrow && \frac{p}{q} & = \frac{4-2 \cos \alpha}{1-2 \cos \alpha} \\ && (p-q)^2 &= p^2 + (p+q)^2 - 2p(p+q) \cos \beta \\ && 0 &= p(p+4q-2(p+q) \cos \beta) \\ && 0 &= p(1-2\cos \beta)+q(4-2\cos \beta) \\ \Rightarrow && \frac{p}{q} &= \frac{2\cos \beta - 4}{1-2\cos \beta} \\ \Rightarrow && \frac{4-2 \cos \alpha}{1-2 \cos \alpha} &= \frac{2\cos \beta - 4}{1-2\cos \beta} \\ \Rightarrow && (2-\cos \alpha)(1-2\cos \beta) &= (\cos \beta - 2)(1 - 2 \cos \alpha) \\ \Rightarrow && 2 - \cos \alpha -4\cos \beta+2\cos \alpha \cos \beta &= \cos \beta - 2-2\cos \alpha \cos \beta + 4 \cos \alpha \\ \Rightarrow && 4-4\cos \alpha - 4\cos \beta+4\cos \alpha\cos \beta &= \cos \alpha + \cos \beta \\ \Rightarrow && 4(1-\cos \alpha)(1-\cos \beta) &= \cos \alpha + \cos \beta \end{align*} If \(\alpha = 2 \beta\), and let \(c = \cos \beta\) \begin{align*} && 4 (1- \cos 2 \beta)(1-\cos \beta) &= \cos 2 \beta + \cos \beta \\ \Rightarrow && 4(1-(2c^2-1))(1-c) &= 2c^2-1+c\\ \Rightarrow && 8(1+c)(1-c)^2 &= (2c-1)(c+1) \\ \Rightarrow && 0 &= (c+1)(8(1-c)^2-(2c-1)) \\ &&&= (c+1)(8c^2-18c+9) \\ &&&= (c+1)(4c-3)(2c-3) \\ \end{align*} Therefore \(c = -1, \frac32, \frac34\). Clearly \(\cos \beta \neq -1, \frac32\), since they are not valid angles in a triangle (or valid values of \(\cos \beta\)). \(\frac{p}{q} = \frac{2 \cdot \frac34-4 }{1 - 2\cdot \frac34} = \frac{3-8}{2-3} = 5\) so \(4:5:6\)
A right circular cone has base radius \(r\), height \(h\) and slant length \(\ell\). Its volume \(V\), and the area~\(A\) of its curved surface, are given by \[ V= \tfrac13 \pi r^2 h \,, \ \ \ \ \ \ \ A = \pi r\ell\,. \] \vspace*{-1cm}
Show that, for any integer \(m\), \[ \int_0^{2\pi} \e^x \cos mx \, \d x = \frac {1}{m^2+1}\big(\e^{2\pi}-1\big)\,. \]
Solution: \begin{align*} && I &= \int_0^{2 \pi} e^{x} \cos m x \d x \\ &&&= \left [e^x \cos m x \right]_0^{2 \pi}-\int_0^{2 \pi} e^x m (-\sin mx) \d x\\ &&&= e^{2\pi}-1 + m\int_0^{2\pi}e^x \sin m x \d x \\ &&&= e^{2\pi}-1 + m\left [e^x \sin m x \right]_0^{2\pi} - m \int_0^{2\pi} e^x m \cos x \d x \\ &&&= e^{2\pi}-1+0 - m^2 I\\ \Rightarrow && (m^2+1)I &= e^{2\pi}-1 \\ \Rightarrow && I &= \frac{1}{m^2+1} (e^{2\pi}-1) \end{align*}
Two particles \(P\) and \(Q\) are projected simultaneously from points \(O\) and \(D\), respectively, where~\(D\) is a distance \(d\) directly above \(O\). The initial speed of \(P\) is \(V\) and its angle of projection {\em above} the horizontal is \(\alpha\). The initial speed of \(Q\) is \(kV\), where \(k>1\), and its angle of projection {\em below} the horizontal is \(\beta\). The particles collide at time \(T\) after projection. Show that \(\cos\alpha = k\cos\beta\) and that \(T\) satisfies the equation \[ (k^2-1)V^2T^2 +2dVT\sin\alpha -d^2 =0\,. \] Given that the particles collide when \(P\) reaches its maximum height, find an expression for~\(\sin^2\alpha\) in terms of \(g\), \(d\), \(k\) and \(V\), and deduce that \[ gd\le (1+k)V^2\,. \]
A triangular wedge is fixed to a horizontal surface. The base angles of the wedge are \(\alpha\) and \(\frac\pi 2-\alpha\). Two particles, of masses \(M\) and \(m\), lie on different faces of the wedge, and are connected by a light inextensible string which passes over a smooth pulley at the apex of the wedge, as shown in the diagram. The contacts between the particles and the wedge are smooth.