Problems

Solution: There are many ways to do the counting in each question, possibly the clearest way is to always consider the order in which discs are taken, although all methods should work equally well. For some examples Bayes rule also offers a fast solution.

1. There are we are choose \(4\) objects in order from \(5\) (ie \({^5\P_4}\)) to obtain valid draws, this is out of a total of picking \(4\) objects from \(11\) (\({^{11}\P_4}\)). Ie the probability is: \(\displaystyle \frac{{^5\P_4}}{{^{11}\P_4}} = \frac{5! \cdot 7!}{11!} = \frac{1}{66}\) Alternatively, there are \(\binom{5}{4}\) ways to choose four numbered discs, out of \(\binom{11}{4}\) ways to choose four discs. ie \(\displaystyle \binom{5}{4} \Big / \binom{11}{4} = \frac{5 \cdot 4! \cdot 7!}{11!} = \frac{5 \cdot 4 \cdot 3 \cdot 2}{11 \cdot 10 \cdot 9 \cdot 8} = \frac1{66}\)
2. \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{^4\P_3 \big / {^{11}\P_4}}{1/11} \\ &= 11\cdot \frac{4!}{1!} \Bigg / \frac{11!}{7!} \\ &= \frac{4! \cdot 7! \cdot 11}{11!} \\ &= \frac{4\cdot 3 \cdot 2}{10 \cdot 9 \cdot 8} \\ &= \frac{1}{30} \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{3} \Bigg / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac{1}{30} \end{align*} Where we are calculating this as "choose one number", then "choose 3 more", which can happen ending up with 3, number, number, number in \(\binom{4}{3}\) ways, and there are \(11 \cdot \binom{10}{3}\) was overall. Another alternative using Bayes rule: \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \mathbb{P}( \text{first disc is 3} | \text{all four discs are numbered}) \frac{ \mathbb{P}( \text{all four discs are numbered} }{ \mathbb{P}( \text{first disc is 3} )} \\ &= \frac{\frac{1}{5} \cdot \frac{1}{66}}{\frac{1}{11}} \\ &= \frac1{30} \end{align*}
3. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{3 \cdot {^4\P_1} \cdot {^{6}\P_2} \big / {^{11}\P_4}}{\frac1{11} } \\ &= \frac12 \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac12 \end{align*}
4. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and 3 taken})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / \binom{11}{4}}{\frac{4}{11}} \\ &= \frac{\frac{2}{11}}{\frac4{11}} \\ &= \frac12 \end{align*} Using Bayes rule: \(\mathbb{P}( \text{3 taken}) = \frac{1}{11} + \frac{10}{11}\frac{1}{10} + \frac{10}{11}\frac{9}{10}\frac{1}{9} + \frac{10}{11}\frac{9}{10}\frac89\frac18 = \frac{4}{11}\) \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{3 taken | exactly two discs are numbered})\mathbb{P}(\text{exactly two discs are numbered})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\frac{4}{10} \cdot \binom{5}{2} \binom{6}{2} \Big / \binom{11}{4}}{4/11} \\ &= \frac{\frac4{10}{5 / 11}}{4/11} \\ &= \frac{1}{2} \end{align*}
5. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc first}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc first})}{\mathbb{P}( \text{numbered disc first})} \\ &= \frac{3 \cdot {^5\P_1}\cdot{^4\P_1}\cdot{^6\P_2} \Big / {^{11}\P_4}}{\frac{5}{11}} \\ &= \frac{1}{2} \end{align*}
6. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc taken})}{\mathbb{P}(\text{numbered disc taken})} \\ &= \frac{\mathbb{P}(\text{exactly two discs are numbered})}{1 - \mathbb{P}(\text{not numbered discs taken})} \\ &= \frac{\binom{5}{2}\binom{6}{2} \Big / \binom{11}{4}}{1 - \binom{6}{4} \Big / \binom{11}{4}} \\ &= \frac{\frac{5}{11}}{\frac{21}{22}} \\ &= \frac{10}{21} \neq \frac12 \end{align*}

Solution: Let \(X \sim Po(\lambda)\), then

1. \(\,\)
   

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   Suppose \(\mathbb{P}(X \geq 2) = 1-p\) then \(\mathbb{P}(X=0) + \mathbb{P}(X=1) = p\), ie \(e^{-\lambda} +\lambda e^{-\lambda} = p\) If \(f(x) = (1+x)e^{-x}\) we have see it is strictly decreasing on \(x \geq 0\) and takes all values from \(1\) to \(0\), therefore we can find a unique value such that \(f(\lambda) = p\) which is our desired \(\lambda\).
2. Note that \(\mathbb{P}(X = 1) = \lambda e^{-\lambda}\)
   

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   Sketching \(y = xe^{-x}\) and finding it's turning point we can see that there is a unique value of \(\lambda = 1\) when \(q = \frac{1}{e}\), otherwise there is either no solution (\(p > \frac1{e}\)) or two solutions (\(0 < q > \frac1{e}\)).
3. Suppose \(\mathbb{P}(X = 1 | X \leq 2) = r\), ie \begin{align*} && r &= \frac{\lambda e^{-\lambda}}{e^{-\lambda} + \lambda e^{-\lambda} + \frac12 \lambda^2 e^{-\lambda}} \\ &&&= \frac{2\lambda}{2+2\lambda+\lambda^2} \\ \Rightarrow && 0 &= r\lambda^2 + 2(r-1) \lambda + 2r\\ \Rightarrow && \Delta &= 4(r-1)^2 - 4\cdot r \cdot 2 r \\ &&&= 4((r-1)^2-2r^2) \\ &&&= 4(r-1-\sqrt{2}r)(r-1+\sqrt{2}r) \\ &&&= -4((\sqrt{2}-1)r + 1)((1+\sqrt{2})r-1) \end{align*} Therefore our quadratic in \(r\) has a unique solution if \(r = \frac{1}{1+\sqrt{2}}\). If it has a positive solution then note since \(2r > 0\) both solutions are positive, so \(\lambda\) is not unique by excluding other solutions.

Solution:

1. Using the generalise binomial theorem \begin{align*} \left( 1+\frac k {100} \right)^{\frac12} &= 1 + \frac12 \frac{k}{100} + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)}{2!} \left (\frac{k}{100} \right)^2 + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)\cdot \left ( -\tfrac32\right)}{3!} \left (\frac{k}{100} \right)^3 + \cdots \\ &= 1 + \frac{1}{200}k - \frac{1}{80\,000}k^2 + \frac{1}{16\,000\,000}k^3 + \cdots \end{align*}
   1. If \(k = 8\), \begin{align*} && \left( 1+\frac 8 {100} \right)^{\frac12} &= 1 + \frac{1}{200}8 - \frac{1}{80\,000}8^2 + \frac{1}{16\,000\,000}8^3 + \cdots \\ \Rightarrow && \frac{6\sqrt{3}}{10} &\approx 1 + 0.04 - 0.0008 + 0.000032 \\ &&&= 1.039232\\ \Rightarrow && \sqrt{3} &\approx 1.73205 \, (5\, \text{d.p.}) \end{align*}
   2. If \(k = -4\), \begin{align*} && \left( 1-\frac 4 {100} \right)^{\frac12} &= 1 - \frac{1}{200}4 - \frac{1}{80\,000}4^2 - \frac{1}{16\,000\,000}4^3 + \cdots \\ \Rightarrow && \frac{4\sqrt{6}}{10} &\approx 1 -0.02-0.0002 -0.000004 \\ &&&= 0.979796\\ \Rightarrow && \sqrt{6} &\approx 2.44949\, (5\, \text{d.p.}) \end{align*}
2. \(\,\) \begin{align*} && \left( 1+\frac k {1000} \right)^{\!\frac13} &= 1 + \frac13 \frac{k}{1000} + \cdots \\ &&&= 1 + \frac{k}{3\,000} + \cdots \\ && 3 \times 7^3 &= 1029 \\ \Rightarrow && \left( 1+\frac {29} {1000} \right)^{\!\frac13} &\approx 1 + \frac{29}{3\,000} \\ \Rightarrow && \frac{7\sqrt[3]{3}}{10} &\approx \frac{3\,029}{3000} \\ \Rightarrow && \sqrt[3]{3} &= \frac{3\,029}{2\,100} \end{align*}

Solution:

1. \(\,\) \begin{align*} && y &= 2x^3-bx^2+cx \\ \Rightarrow && y' &= 6x^2-2bx+c \end{align*} We must have \(p, q\) are the roots of this equation, ie \(\frac13b = p+q, \frac16c = pq\)
2. The point of inflection will be at \(\frac{b}6\) The graph will have rotational symmetry of \(180^{\circ}\) about the point of inflection.
   

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3. \begin{align*} && m-n &= 2(p^3-q^3)-b(p^2-q^2)+c(p-q) \\ &&&= (p-q)(2(p^2+qp+q^2)-b(p+q)+c) \\ &&&= (p-q)(2(p^2+qp+q^2)-3(p+q)^2+6pq) \\ &&&= (p-q)(-p^2-q^2+2pq) \\ &&&= (q-p)^3 \end{align*}
4. The area of \(R\) is \begin{align*} A &= \frac12 bh \\ &= \frac12 (q-p)(m-n) = \frac12(q-p)^4 \end{align*} as required.

Solution: \begin{align*} && I &= \int \frac{1}{a^2+x^2} \d x\\ x = a \tan \theta, \d x =a \sec^2 \theta \d \theta &&&= \int \frac{1}{a^2+a^2\tan^2 x} a \sec^2 \theta \d \theta \\ &&&=\int \frac{\sec^2 \theta}{a \sec^2 \theta} \d \theta \\ &&&= \frac1a \theta + C \\ &&&= \frac1a \arctan \frac{x}{a} + C \end{align*}

1. 1. \(\,\) \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac {\cos x}{1+\sin^2 x} \d x \\ &&&= \left [ \arctan (\sin x) \right]_0^{\pi/2} \\ &&&= \arctan(1) - \arctan(0) = \frac{\pi}{4} \end{align*}
   2. \(\,\) \begin{align*} && t &= \tan \frac{x}{2} \\ \Rightarrow && \sin x &= \frac{2t}{1+t^2} \\ && \cos x &= \frac{1-t^2}{1+t^2} \\ && \d x &= \frac{2}{1+t^2} \d t \\ \Rightarrow && I &= \int_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x } \d x \\ &&&= \int_{t=0}^{t = 1} \frac{\frac{1-t^2}{1+t^2}}{1 + \left (\frac{2t}{1+t^2} \right)^2} \frac{2}{1+t^2} \d t \\ &&&= 2 \int_0^1 \frac{1-t^2}{(1+t^2)^2+(2t)^2} \d t\\ &&&= 2 \int_0^1 \frac{1-t^2}{1+6t^2+t^4} \d t\\ \end{align*} From which the conclusion follows
2. \(\,\) \begin{align*} && J &= \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{1+14t^2+t^4}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{(t^2+1)^2+3(2t)^2}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \frac12\int_{x=0}^{x=\pi/2} \frac {\cos x}{1+3 \sin^2 x} \d x \\ &&&= \frac{1}{6}\left[ \sqrt{3} \arctan(\sin \sqrt{3}x)\right]_0^{\pi/2} \\ &&&= \frac16 \sqrt{3} \frac{\pi}{3} \\ &&&= \frac{\sqrt{3}\pi}{18} \end{align*}

Solution: \begin{align*} && \alpha \sin(A-B) + \beta \cos (A + B) &= \gamma \sin(A+B) \\ \Leftrightarrow && \alpha \sin A \cos B - \alpha \cos A \sin B + \beta \cos A \cos B - \beta \sin A \sin B &= \gamma \sin A \cos B + \gamma \cos A \sin B \\ \Leftrightarrow && \alpha \tan A - \alpha \tan B + \beta - \beta \tan A \tan B &= \gamma \tan A + \gamma \tan B \\ \Leftrightarrow && \beta \tan A \tan B +(\gamma-\alpha) \tan A + (\gamma +\alpha)\tan B&=\beta \\ \Leftrightarrow && \tan A \tan B +\left (\frac{\gamma-\alpha}{\beta} \right) \tan A + \left (\frac{\gamma +\alpha}{\beta} \right)\tan B&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) - \frac{\gamma^2 - \alpha^2}{\beta^2}&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) &= \frac{\beta^2+\gamma^2-\alpha^2}{\beta^2}\\ \end{align*} Which has the desired form iff \(\beta^2+\gamma^2 = \alpha^2\).

1. \(\,\) \begin{align*} && 2\sin(x-\tfrac14\pi) + \sqrt 3 \cos(x+\tfrac14\pi) &=\sin(x+\tfrac14\pi) \\ && 3 + 1 &= 4 \\ \Rightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{4} + \frac{1-2}{\sqrt3}\right) &= 0\\ \Rightarrow && \tan x &= -\sqrt3 \\ \Rightarrow && x &= \tfrac23\pi, \tfrac53\pi \end{align*}
2. \(\,\) \begin{align*} && 2\sin(x-\frac16\pi) + \sqrt 3 \cos(x+\frac16\pi) &=\sin(x+\frac16\pi) \\ \Leftrightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{3} + \frac{1-2}{\sqrt3}\right) &= 0\\ && x &\in [0, 2\pi) \end{align*}
3. \(\,\) \begin{align*} && 2\sin(x+\frac13\pi) + \sqrt 3 \cos(3x) = \sin(3x) \\ && A-B =x + \tfrac13\pi, A+B &= 3x \\ \Rightarrow && A = 2x + \tfrac\pi6, B &= x-\tfrac{\pi}{6} \\ \Rightarrow && \tan (2x+\tfrac\pi6)&=-\sqrt3 \\ && 2x + \tfrac{\pi}{6} &= \tfrac23\pi, \tfrac53\pi, \tfrac83 \pi, \tfrac{11}3\pi \\ && x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} \\ && \tan(-x-\tfrac{\pi}{6}) &= \frac1{\sqrt{3}} \\ \Rightarrow && x-\tfrac{\pi}{6} &= \ldots, \tfrac{\pi}{6}, \tfrac{7\pi}{6}, \ldots \\ \Rightarrow && x &= \tfrac{\pi}3, \tfrac{4\pi}{3} \\ \\ \Rightarrow && x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} , \tfrac{\pi}3, \tfrac{4\pi}{3} \end{align*}

Solution:

1. \(\,\) \begin{align*} && f(x) &= \frac{x+\sqrt3}{1-\sqrt3x} \\ \Rightarrow && f(f(x)) &= \frac{f(x)+\sqrt3}{1-\sqrt3f(x)} \\ &&&= \frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{x+\sqrt{3}+\sqrt3(1-\sqrt3x)}{1-\sqrt3x-\sqrt3(x+\sqrt3)} \\ &&&= \frac{-2x+2\sqrt3}{-2-2\sqrt3x} \\ &&&= \frac{x-\sqrt3}{1+\sqrt3 x} \\ \\ && f^3(x) &= f^2(f(x)) \\ &&&= \frac{f(x)-\sqrt3}{1+\sqrt3 f(x)} \\ &&&=\frac{\frac{x+\sqrt3}{1-\sqrt3x}-\sqrt3}{1+\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{(x+\sqrt3)-\sqrt3(1-\sqrt3 x)}{(1-\sqrt3x)+\sqrt3 (x+\sqrt3)} \\ &&&= \frac{-2x}{-2} = x \\ \\ && f^{2007}(x) &= x \end{align*}
2. If \(x = \tan \theta\) then \(f(x) = \frac{\tan \theta + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{3} \tan \theta} = \tan (\theta + \frac{\pi}{3})\) and hence \(f^n(x) = \tan (\theta + \frac{n \pi}{3})\)
3. Note that if \(t = \sin \theta\) then \(g(t) = \cos \frac{\pi}{6} t\sin \theta + \frac12 \cos \theta = \sin(\theta + \frac{\pi}6)\) therefore \(g^n(t) = \sin(\sin^{-1}(t) + \frac{n\pi}{6})\)

Solution:

1. \(\,\) \begin{align*} && y &= \ln (x + \sqrt{3+x^2}) \\ \Rightarrow && y' &= \frac{1}{x + \sqrt{3+x^2}} \cdot \left (1 + \frac{x}{\sqrt{3+x^2}} \right) \\ &&&= \frac{1}{\sqrt{3+x^2}} \\ \\ && y &= x\sqrt{3+x^2} \\ && y' &= \sqrt{3+x^2} + \frac{x^2}{\sqrt{3+x^2}} \\ &&&= 2\sqrt{3+x^2} - \frac{3}{\sqrt{3+x^2}} \\ \\ \Rightarrow && \sqrt{3+x^2} &= \frac12(x \sqrt{3+x^2})' + \frac32(\ln(x+\sqrt{3+x^2})' \\ \Rightarrow && \int \sqrt{3+x^2}\, \d x &= \frac12x\sqrt{3+x^2} + \frac32 \ln (x+\sqrt{3+x^2}) + C \end{align*}
2. \(\,\) \begin{align*} && 3 \left ( \frac{\d y}{\d x} \right)^2 + 2x \frac{\d y}{\d x} &= 1 \\ && \frac{\d y}{\d x} &= \frac{-x \pm \sqrt{x^2+3} }3 \\ \Rightarrow && y &= -\frac{x^2}{6} \pm \frac16x\sqrt{3+x^2} \pm \frac12 \ln (x+\sqrt{3+x^2}) + C \\ y = 0, x = 1: && 0 &= -\frac16 \pm \frac13 \pm \frac12 \ln 3 \\ \Rightarrow && y &= -\frac{x^2}{6} \pm \frac12x\sqrt{3+x^2} \pm \frac32 \ln (x+\sqrt{3+x^2}) + \frac16 \mp \frac13 \mp \frac12 \ln 3 \end{align*}

Solution: \begin{align*} && f(x) &= \sin x \\ \Rightarrow && f''(x) &= -\sin x \end{align*} which is clearly negative on \((0,\pi)\) since \(\sin\) is positive on this interval. \begin{align*} && f(x) &= \ln x \\ \Rightarrow && f''(x) &= -1/x^2 \end{align*} which is clearly negative for \(x > 0\)

1. Since \(A,B,C\) are angles in a triangle, we must have \(0 < A,B,C< \pi\) and so we can apply Jensen with \(f = \sin\) to obtain: \begin{align*} &&\frac13( \sin A + \sin B + \sin C) &\leq \sin \left ( \frac{A+B+C}{3}\right) \\ &&&= \sin \frac{\pi}{3} = \frac{\sqrt{3}}2 \\ \Rightarrow && \sin A + \sin B + \sin C &\leq\frac{3\sqrt{3}}2 \end{align*}
2. Suppose \(f(x) = \ln x\), then applying Jensen on the positive numbers \(t_1, \ldots, t_n\) we obtain \begin{align*} && \frac1n \left ( \sum_{i=1}^n \ln t_n \right) &\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \frac1n \ln\left (\prod_{i=1} t_n\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \ln\left (\left (\prod_{i=1} t_n\right)^{1/n}\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \left (\prod_{i=1} t_n\right)^{1/n}&\leq\frac1n\sum_{i=1}^n t_n \\ \Rightarrow && \sqrt[n]{t_1t_2 \cdots t_n}&\leq\frac1n(t_1 + t_2 + \cdots + t_n) \tag{AM-GM}\\ \end{align*}
   1. Applying AM-GM with \(t_1 = x^4, t_2 = y^4, t_3 = z^4, t_4 = 2^4\) we have \begin{align*} && \frac{x^4+y^4+z^4+16}{4} & \geq \sqrt[4]{x^4y^4z^42^4} \\ \Rightarrow && x^4+y^4+z^4+16 &\geq 8xyz \end{align*}
   2. Applying AM-GM with \(t_1 = x^5, t_2 = y^5, t_3 = z^5, t_4 = 1^5, t_5 = 1^5\) we have \begin{align*} && \frac{x^5+y^5+z^5+1+1}{5} & \geq \sqrt[5]{x^5y^5z^5} \\ \Rightarrow && x^5+y^5+z^5+2 &\geq 5xyz \\ \Rightarrow && x^5+y^5+z^5 - 5xyz &\geq -2 \end{align*} Therefore the minimum is \(-2\)

Solution:

1. \(X\) lies on the line including \(B\) and \(C\).
2. points on the line \(AP\) have the form \(\lambda(\beta \mathbf{b}+\gamma\mathbf{c})\), and the point \(D\) will be the point where \(\lambda\beta + \lambda \gamma = 1\). \begin{align*} && \frac{|BD|}{|DC|} &= \frac{|\mathbf{b} -\lambda(\beta \mathbf{b}+\gamma\mathbf{c})| }{|\lambda(\beta \mathbf{b}+\gamma\mathbf{c})- \mathbf{c}|} \\ &&&= \frac{|(1-\lambda \beta)\mathbf{b} - \lambda \gamma \mathbf{c}|}{|\lambda \beta \mathbf{b}+(\lambda \gamma -1)\mathbf{c}|}\\ &&&= \frac{|\lambda \gamma\mathbf{b} - \lambda \gamma \mathbf{c}|}{|\lambda \beta \mathbf{b}-(\lambda \beta)\mathbf{c}|} \\ &&&= \frac{\gamma}{\beta} \end{align*}
3. The line \(BP\) is \(\mathbf{b} + \mu(\beta \mathbf{b}+\gamma\mathbf{c})\) and will meet \(CA\) when \(1+\mu\beta = 0\), ie \(\mu = -\frac{1}{\beta}\), therefore \(E\) is \(-\frac{\gamma}{\beta}\mathbf{c}\), and so \(\frac{|CE|}{|EA|} = \frac{1+\gamma/\beta}{\gamma/\beta} = \frac{\beta+\gamma}{\gamma}\). Similarly, \(F\) is \(-\frac{\beta}{\gamma}\mathbf{b}\) and \(\frac{|AF|}{|FB|} = \frac{\beta/\gamma}{1+\frac{\beta}{\gamma}} = \frac{\beta}{\gamma+\beta}\), and so \[\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = \frac{\beta}{\gamma+\beta} \frac{\gamma}{\beta} \frac{\beta+\gamma}{\gamma} = 1 \]