STEP3 2013 -- Pure Mathematics

STEP3 2013 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2013  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions are pure mathematics. Solutions and examiner commentary are included below.

Overview

Q	Topic	Difficulty	Key Techniques
1	积分 Integration	Challenging	万能代换,递推关系,部分分式分解
2	级数 Series	Hard	莱布尼茨法则,麦克劳林展开,微分方程递推
3	向量与几何 Vectors and Geometry	Challenging	向量点积,坐标几何,对称性
4	复数 Complex Numbers	Challenging	单位根因式分解,复数代入,三角恒等式
5	数论 Number Theory	Hard	互质论证,素因数分析,反证法
6	复数 Complex Numbers	Challenging	复数模的展开,共轭乘积化简,三角不等式,不等式推导
7	微分方程 Differential Equations	Hard	能量函数构造,首次积分,微分验证,不等式估计
8	复数与几何 Complex Numbers and Geometry	Hard	几何级数求和,圆锥曲线焦点准线性质,三角恒等式,复数方法

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Question 1

Topic: 积分 Integration  |  Difficulty: Challenging  |  Marks: 20

Problem

1 Given that $t = \tan \frac{1}{2}x$, show that $\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{2}(1 + t^2)$ and $\sin x = \frac{2t}{1 + t^2}$.

Hence show that

$\int_{0}^{\frac{1}{2}\pi} \frac{1}{1 + a \sin x} \, \mathrm{d}x = \frac{2}{\sqrt{1 - a^2}} \arctan \frac{\sqrt{1 - a}}{\sqrt{1 + a}} \qquad (0 < a < 1).$

Let

$I_n = \int_{0}^{\frac{1}{2}\pi} \frac{\sin^n x}{2 + \sin x} \, \mathrm{d}x \qquad (n \geqslant 0).$

By considering $I_{n+1} + 2I_n$, or otherwise, evaluate $I_3$.

Hint

1. Thus if $\sqrt[n]{N} = \frac{p}{q}$ where $p$ and $q$ are coprime, it is rational and can be written in lowest terms, then $q^n N = p^n$ and so $q = 1$ and thus $\sqrt[n]{N}$ is an integer. Otherwise, $\sqrt[n]{N}$ cannot be written as $\frac{p}{q}$, that is, it is irrational.

For (ii), using the same logic as in part (i), as $b^a$ divides $a^a d^b$, $b^a$ divides $d^b$, so $d^b = k b^a$, for some $k$. Likewise, $a^a = k' c^b$, for some integer $k'$, and thus $k k' = 1$, so $k = k' = 1$, and $d^b = b^a$. If $p$ is a prime factor of $d$, then $p$ divides $d^b$, and so $b^a$ too. Writing $b^a = b b^{a-1}$, using the logic of the very first part of the question, if $p$ does not divide $b$, $p$ divides $b^{a-1}$, and repetition of this argument leads to a contradiction. So $p$ is a prime factor of $b$. $p^{mb}$ and $p^{na}$ is the highest power of $p$ that divides $d^b = b^a$. So $mb = na$, and $b = \frac{na}{m}$. So $p^n$ divides $na$, but as $a$ and $b$ are coprime, $p^n$ divides $n$ and thus $p^n \le n$. By the given result, this means $p = 1$, and as $b$ is only divisible by 1, $b = 1$. If $r$ is a positive rational $\frac{a}{b}$, such that $r^r = \frac{c}{d}$ is rational, then $a^a d^b = b^a c^b$ so $b = 1$ and $r$ is a positive integer.

Model Solution

Showing that $\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{2}(1 + t^2)$:

We have $t = \tan \frac{1}{2}x$, so by the chain rule:

$\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{2}\sec^2\!\tfrac{1}{2}x = \frac{1}{2}\!\left(1 + \tan^2\!\tfrac{1}{2}x\right) = \frac{1}{2}(1 + t^2).$

Showing that $\sin x = \frac{2t}{1 + t^2}$:

Using the double angle formula $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$:

$\sin x = 2\sin\tfrac{x}{2}\cos\tfrac{x}{2} = \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\!\frac{x}{2} + \sin^2\!\frac{x}{2}} \cdot \frac{\sec^2\!\frac{x}{2}}{\sec^2\!\frac{x}{2}} = \frac{2\tan\frac{x}{2}}{1 + \tan^2\!\frac{x}{2}} = \frac{2t}{1 + t^2}.$

Evaluating the integral using the substitution:

Let $I = \int_{0}^{\frac{1}{2}\pi} \frac{1}{1 + a\sin x}\,\mathrm{d}x$.

With $t = \tan\frac{x}{2}$: when $x = 0$, $t = 0$; when $x = \frac{\pi}{2}$, $t = \tan\frac{\pi}{4} = 1$. We have $\mathrm{d}x = \frac{2}{1+t^2}\,\mathrm{d}t$ and $\sin x = \frac{2t}{1+t^2}$, so:

$I = \int_{0}^{1} \frac{1}{1 + \frac{2at}{1+t^2}} \cdot \frac{2}{1+t^2}\,\mathrm{d}t = \int_{0}^{1} \frac{2}{(1+t^2) + 2at}\,\mathrm{d}t = \int_{0}^{1} \frac{2}{t^2 + 2at + 1}\,\mathrm{d}t.$

Completing the square in the denominator: $t^2 + 2at + 1 = (t+a)^2 + (1-a^2)$.

$I = \int_{0}^{1} \frac{2}{(t+a)^2 + (1-a^2)}\,\mathrm{d}t.$

Let $u = t + a$, so $\mathrm{d}u = \mathrm{d}t$:

$I = \int_{a}^{1+a} \frac{2}{u^2 + (1-a^2)}\,\mathrm{d}u = \frac{2}{\sqrt{1-a^2}}\left[\arctan\frac{u}{\sqrt{1-a^2}}\right]_{a}^{1+a}$

$= \frac{2}{\sqrt{1-a^2}}\left(\arctan\frac{1+a}{\sqrt{1-a^2}} - \arctan\frac{a}{\sqrt{1-a^2}}\right).$

Since $0 < a < 1$, we have $\sqrt{1-a^2} = \sqrt{(1-a)(1+a)}$, so:

$\frac{1+a}{\sqrt{1-a^2}} = \frac{1+a}{\sqrt{(1-a)(1+a)}} = \sqrt{\frac{1+a}{1-a}}, \qquad \frac{a}{\sqrt{1-a^2}} = \frac{a}{\sqrt{(1-a)(1+a)}}.$

Using the subtraction formula $\arctan A - \arctan B = \arctan\frac{A-B}{1+AB}$:

$\frac{\sqrt{\frac{1+a}{1-a}} - \frac{a}{\sqrt{(1-a)(1+a)}}}{1 + \sqrt{\frac{1+a}{1-a}} \cdot \frac{a}{\sqrt{(1-a)(1+a)}}} = \frac{\frac{1+a}{\sqrt{(1-a)(1+a)}} - \frac{a}{\sqrt{(1-a)(1+a)}}}{1 + \frac{a}{1-a}}.$

The numerator is $\frac{1}{\sqrt{(1-a)(1+a)}} = \frac{1}{\sqrt{1-a^2}}$ and the denominator is $\frac{1}{1-a}$, so the ratio is:

$\frac{1-a}{\sqrt{1-a^2}} = \frac{1-a}{\sqrt{(1-a)(1+a)}} = \sqrt{\frac{1-a}{1+a}}.$

Therefore:

$I = \frac{2}{\sqrt{1-a^2}}\arctan\sqrt{\frac{1-a}{1+a}},$

as required.

Evaluating $I_3$:

We have $I_n = \int_{0}^{\frac{1}{2}\pi} \frac{\sin^n x}{2+\sin x}\,\mathrm{d}x$. Note that $\sin^n x = \sin^{n-1}x \cdot \sin x = \sin^{n-1}x\!\left[(2+\sin x) - 2\right]$, so:

$\sin^n x = (2+\sin x)\sin^{n-1}x - 2\sin^{n-1}x.$

Dividing by $2+\sin x$ and integrating:

$I_n = \int_{0}^{\frac{1}{2}\pi} \sin^{n-1}x\,\mathrm{d}x - 2I_{n-1}.$

Rearranging: $I_n + 2I_{n-1} = \int_{0}^{\frac{1}{2}\pi} \sin^{n-1}x\,\mathrm{d}x$, or equivalently, replacing $n$ by $n+1$:

$I_{n+1} + 2I_n = \int_{0}^{\frac{1}{2}\pi} \sin^n x\,\mathrm{d}x. \qquad (\star)$

Let $J_n = \int_{0}^{\frac{1}{2}\pi}\sin^n x\,\mathrm{d}x$. We need $J_0, J_1, J_2$:

• $J_0 = \frac{\pi}{2}$.
• $J_1 = [-\cos x]_0^{\pi/2} = 1$.
• $J_2 = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$ (using the Wallis formula or $\sin^2 x = \frac{1-\cos 2x}{2}$).

From $(\star)$ with $n = 0$: $I_1 + 2I_0 = J_0 = \frac{\pi}{2}$.

To find $I_0$, set $a = \frac{1}{2}$ in the integral result (since $\frac{1}{2+\sin x} = \frac{1}{2}\cdot\frac{1}{1+\frac{1}{2}\sin x}$):

$I_0 = \int_{0}^{\frac{1}{2}\pi}\frac{1}{2+\sin x}\,\mathrm{d}x = \frac{1}{2}\cdot\frac{2}{\sqrt{1-\frac{1}{4}}}\arctan\sqrt{\frac{\frac{1}{2}}{\frac{3}{2}}} = \frac{1}{\frac{\sqrt{3}}{2}}\arctan\frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}\cdot\frac{\pi}{6} = \frac{\pi}{3\sqrt{3}}.$

From $n=0$: $I_1 = \frac{\pi}{2} - 2I_0 = \frac{\pi}{2} - \frac{2\pi}{3\sqrt{3}}$.

From $n=1$: $I_2 + 2I_1 = J_1 = 1$, so $I_2 = 1 - 2I_1 = 1 - 2\!\left(\frac{\pi}{2} - \frac{2\pi}{3\sqrt{3}}\right) = 1 - \pi + \frac{4\pi}{3\sqrt{3}}$.

From $n=2$: $I_3 + 2I_2 = J_2 = \frac{\pi}{4}$, so:

$I_3 = \frac{\pi}{4} - 2I_2 = \frac{\pi}{4} - 2\!\left(1 - \pi + \frac{4\pi}{3\sqrt{3}}\right) = \frac{\pi}{4} - 2 + 2\pi - \frac{8\pi}{3\sqrt{3}}.$

$I_3 = \frac{9\pi}{4} - 2 - \frac{8\pi}{3\sqrt{3}} = \frac{9\pi}{4} - \frac{8\pi}{3\sqrt{3}} - 2.$

Examiner Notes

Most candidates attempted this question, making it the most popular and it was also the most successful with a mean score of about two thirds marks. The first two standard results caused few problems, nor did the integration, but some struggled to simplify to the single inverse tan form. In the final part, common errors were failure to reduce to the $n = 0$ case, confusion with the index e.g. $I_n + 2I_{n-1} = \int_0^{\frac{1}{2}\pi} \sin^n x \, dx$ instead of the correct result, or for those that were more successful, algebraic inaccuracies let them down. Some attempted a recursive formula to evaluate $\int_0^{\frac{1}{2}\pi} \sin^n x \, dx$ with varying success. Most attempting the last part saw the connection between $I_0$ and the main result of the question.

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Question 2

Topic: 级数 Series  |  Difficulty: Hard  |  Marks: 20

Problem

2 In this question, you may ignore questions of convergence.

Let $y = \frac{\arcsin x}{\sqrt{1 - x^2}}$. Show that

$(1 - x^2) \frac{\mathrm{d}y}{\mathrm{d}x} - xy - 1 = 0$

and prove that, for any positive integer $n$,

$(1 - x^2) \frac{\mathrm{d}^{n+2}y}{\mathrm{d}x^{n+2}} - (2n + 3)x \frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}} - (n + 1)^2 \frac{\mathrm{d}^n y}{\mathrm{d}x^n} = 0.$

Hence obtain the Maclaurin series for $\frac{\arcsin x}{\sqrt{1 - x^2}}$, giving the general term for odd and for even powers of $x$.

Evaluate the infinite sum

$1 + \frac{1}{3!} + \frac{2^2}{5!} + \frac{2^2 \times 3^2}{7!} + \dots + \frac{2^2 \times 3^2 \times \dots \times n^2}{(2n + 1)!} + \dots.$

Hint

2. It is elegant to multiply by the denominator, then differentiate implicitly, and finally multiply by the same factor again to achieve the desired first result. The general result can be proved by then using induction, or by Leibnitz, if known. The general result can be used alongside the expression for $y$, and the first derived result with the substitution $x = 0$ to show that the general term of the Maclaurin series for even powers of $x$ is zero, and for odd powers of $x$ is $\frac{2^{2r}(r!)^2}{(2r+1)!} x^{2r+1}$. Thus, as

$y = x + \frac{2^2}{3!} x^3 + \frac{4^2 2^2}{5!} x^5 + \dots$ the required infinite sum is $\frac{y}{x}$ with $x = \frac{1}{2}$, that is $\frac{2\pi\sqrt{3}}{9}$.

Model Solution

Showing that $(1-x^2)\frac{\mathrm{d}y}{\mathrm{d}x} - xy - 1 = 0$:

Let $y = \frac{\arcsin x}{\sqrt{1-x^2}}$. Multiply both sides by $\sqrt{1-x^2}$:

$y\sqrt{1-x^2} = \arcsin x.$

Differentiate both sides with respect to $x$:

$\frac{\mathrm{d}y}{\mathrm{d}x}\sqrt{1-x^2} + y \cdot \frac{-x}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}}.$

Multiply through by $\sqrt{1-x^2}$:

$\frac{\mathrm{d}y}{\mathrm{d}x}(1-x^2) - xy = 1.$

Rearranging:

$(1-x^2)\frac{\mathrm{d}y}{\mathrm{d}x} - xy - 1 = 0. \qquad \blacksquare$

Proving the general recurrence by induction:

We need to show that for any positive integer $n$:

$(1-x^2)\frac{\mathrm{d}^{n+2}y}{\mathrm{d}x^{n+2}} - (2n+3)x\frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}} - (n+1)^2\frac{\mathrm{d}^{n}y}{\mathrm{d}x^{n}} = 0. \qquad (\star)$

We use Leibniz’s rule: $\frac{\mathrm{d}^n}{\mathrm{d}x^n}[fg] = \sum_{r=0}^{n}\binom{n}{r}\frac{\mathrm{d}^r f}{\mathrm{d}x^r}\frac{\mathrm{d}^{n-r}g}{\mathrm{d}x^{n-r}}$.

Differentiate $(1-x^2)y' - xy = 1$ a total of $n$ times using Leibniz’s rule.

For the term $(1-x^2)y'$, differentiating $n$ times:

$\frac{\mathrm{d}^n}{\mathrm{d}x^n}[(1-x^2)y'] = (1-x^2)y^{(n+1)} + n(-2x)y^{(n)} + \frac{n(n-1)}{2}(-2)y^{(n-1)}$

$= (1-x^2)y^{(n+1)} - 2nxy^{(n)} - n(n-1)y^{(n-1)}.$

For the term $xy$, differentiating $n$ times:

$\frac{\mathrm{d}^n}{\mathrm{d}x^n}[xy] = xy^{(n)} + ny^{(n-1)}.$

Since the right-hand side is the constant 1, its $n$th derivative is 0 for $n \geq 1$. So for $n \geq 1$:

$(1-x^2)y^{(n+1)} - 2nxy^{(n)} - n(n-1)y^{(n-1)} - xy^{(n)} - ny^{(n-1)} = 0.$

Collecting terms:

$(1-x^2)y^{(n+1)} - (2n+1)xy^{(n)} - n^2 y^{(n-1)} = 0. \qquad (\dagger)$

Now replace $n$ by $n+1$ (valid since $n \geq 1$ means $n+1 \geq 2$):

$(1-x^2)y^{(n+2)} - (2(n+1)+1)xy^{(n+1)} - (n+1)^2 y^{(n)} = 0,$

$(1-x^2)y^{(n+2)} - (2n+3)xy^{(n+1)} - (n+1)^2 y^{(n)} = 0. \qquad \blacksquare$

Obtaining the Maclaurin series:

Let $y = \sum_{k=0}^{\infty} a_k x^k$. We need $a_k = \frac{y^{(k)}(0)}{k!}$.

From the original equation $(1-x^2)y' - xy = 1$, set $x = 0$:

$y'(0) = 1.$

Also $y(0) = \frac{\arcsin 0}{\sqrt{1}} = 0$, so $a_0 = 0$ and $a_1 = 1$.

From $(\dagger)$ with $n \geq 1$, set $x = 0$:

$y^{(n+1)}(0) = n^2 y^{(n-1)}(0).$

This means $a_{n+1} = \frac{y^{(n+1)}(0)}{(n+1)!} = \frac{n^2 y^{(n-1)}(0)}{(n+1)!} = \frac{n^2}{(n+1)n} \cdot \frac{y^{(n-1)}(0)}{(n-1)!} = \frac{n}{n+1} a_{n-1}$.

Wait, let me be more careful. We have $y^{(n+1)}(0) = n^2 y^{(n-1)}(0)$ from $(\dagger)$ with $x=0$: $(1)\cdot y^{(n+1)}(0) - 0 - n^2 y^{(n-1)}(0) = 0$.

So $a_{n+1} = \frac{y^{(n+1)}(0)}{(n+1)!}$ and $a_{n-1} = \frac{y^{(n-1)}(0)}{(n-1)!}$, giving:

$a_{n+1} = \frac{n^2}{(n+1)!} y^{(n-1)}(0) = \frac{n^2}{(n+1)!} \cdot (n-1)! \cdot a_{n-1} = \frac{n^2}{(n+1)n} a_{n-1} = \frac{n}{n+1} a_{n-1}.$

Hmm, that doesn’t seem right for producing the expected coefficients. Let me recheck.

Actually, from $(\star)$ directly with $x=0$: $y^{(n+2)}(0) = (n+1)^2 y^{(n)}(0)$.

So $a_{n+2} = \frac{(n+1)^2}{(n+2)!} y^{(n)}(0) = \frac{(n+1)^2}{(n+2)!} \cdot n! \cdot a_n = \frac{(n+1)^2}{(n+2)(n+1)} a_n = \frac{n+1}{n+2} a_n$.

For even indices: $a_0 = 0$, so $a_{2k} = 0$ for all $k$ (each even coefficient is a multiple of the previous even coefficient, which starts at 0).

For odd indices: $a_1 = 1$, and $a_{n+2} = \frac{n+1}{n+2} a_n$.

$a_3 = \frac{2}{3} a_1 = \frac{2}{3}$.

$a_5 = \frac{4}{5} a_3 = \frac{4}{5}\cdot\frac{2}{3} = \frac{8}{15}$.

$a_7 = \frac{6}{7} a_5 = \frac{6}{7}\cdot\frac{8}{15} = \frac{48}{105} = \frac{16}{35}$.

In general, for $n = 2r+1$ (where $r \geq 0$), we get by iterating $a_{2r+1} = \frac{2r}{2r+1} a_{2r-1}$:

$a_{2r+1} = \frac{2r}{2r+1} \cdot \frac{2r-2}{2r-1} \cdots \frac{2}{3} \cdot a_1 = \frac{(2r)!!}{(2r+1)!!}.$

Here $(2r)!! = 2 \cdot 4 \cdot 6 \cdots (2r) = 2^r r!$ and $(2r+1)!! = 1 \cdot 3 \cdot 5 \cdots (2r+1) = \frac{(2r+1)!}{2^r r!}$.

Therefore:

$a_{2r+1} = \frac{2^r r!}{\frac{(2r+1)!}{2^r r!}} = \frac{2^{2r}(r!)^2}{(2r+1)!}.$

So the Maclaurin series is:

$y = \sum_{r=0}^{\infty} \frac{2^{2r}(r!)^2}{(2r+1)!} x^{2r+1} = x + \frac{2^2}{3!}x^3 + \frac{2^2 \cdot 4^2}{5!}x^5 + \frac{2^2 \cdot 4^2 \cdot 6^2}{7!}x^7 + \cdots$

and all even-power coefficients are zero.

Evaluating the infinite sum:

The given sum is:

$S = 1 + \frac{1}{3!} + \frac{2^2}{5!} + \frac{2^2 \times 3^2}{7!} + \cdots + \frac{2^2 \times 3^2 \times \cdots \times n^2}{(2n+1)!} + \cdots$

Note that $\frac{2^2 \times 3^2 \times \cdots \times n^2}{(2n+1)!} = \frac{(n!)^2}{(2n+1)!}$ (since the numerator $2^2 \cdot 3^2 \cdots n^2 = (n!)^2 / 1^2 = (n!)^2$… wait, let me reconsider).

Actually $2^2 \times 3^2 \times \cdots \times n^2 = (2 \cdot 3 \cdots n)^2 = (n!)^2$ (since $1^2 = 1$ is omitted). Hmm, but the first term is 1, which corresponds to $n=0$: $\frac{(0!)^2}{1!} = 1$. And $\frac{1}{3!}$ corresponds to $n=1$: $\frac{(1!)^2}{3!} = \frac{1}{6}$. And $\frac{2^2}{5!}$ corresponds to $n=2$: $\frac{(2!)^2}{5!} = \frac{4}{120}$.

So $S = \sum_{n=0}^{\infty} \frac{(n!)^2}{(2n+1)!}$.

Now compare with the Maclaurin series $\frac{y}{x} = \sum_{r=0}^{\infty} \frac{2^{2r}(r!)^2}{(2r+1)!} x^{2r}$.

At $x = \frac{1}{2}$:

$\frac{y}{x}\bigg|_{x=\frac{1}{2}} = \sum_{r=0}^{\infty} \frac{2^{2r}(r!)^2}{(2r+1)!} \cdot \frac{1}{2^{2r}} = \sum_{r=0}^{\infty} \frac{(r!)^2}{(2r+1)!} = S.$

Now $y = \frac{\arcsin x}{\sqrt{1-x^2}}$, so:

$S = \frac{y(1/2)}{1/2} = 2 \cdot \frac{\arcsin\frac{1}{2}}{\sqrt{1-\frac{1}{4}}} = \frac{2 \cdot \frac{\pi}{6}}{\frac{\sqrt{3}}{2}} = \frac{\frac{\pi}{3}}{\frac{\sqrt{3}}{2}} = \frac{2\pi}{3\sqrt{3}} = \frac{2\pi\sqrt{3}}{9}.$

Examiner Notes

This was the second most popular question, attempted by six out of every seven candidates, with only marginally less success than its predecessor. The first differential equation was proved correctly and many successfully completed the general result by induction, although there were some problems with the initial case. Some had difficulty finding the correct coefficients for the odd powers of $x$ in the Maclaurin series but the last part produced a variety of errors and few correct answers. Such errors included $\sin^{-1} \frac{1}{2} = \frac{\pi}{3}$, forgetting to divide $y$ by $x$, and attempting to evaluate the series using $x = 1$.

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Question 3

Topic: 向量与几何 Vectors and Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

3 The four vertices $P_i$ ($i = 1, 2, 3, 4$) of a regular tetrahedron lie on the surface of a sphere with centre at $O$ and of radius 1. The position vector of $P_i$ with respect to $O$ is $\mathbf{p}_i$ ($i = 1, 2, 3, 4$). Use the fact that $\mathbf{p}_1 + \mathbf{p}_2 + \mathbf{p}_3 + \mathbf{p}_4 = \mathbf{0}$ to show that $\mathbf{p}_i \cdot \mathbf{p}_j = -\frac{1}{3}$ for $i \neq j$.

Let $X$ be any point on the surface of the sphere, and let $XP_i$ denote the length of the line joining $X$ and $P_i$ ($i = 1, 2, 3, 4$).

(i) By writing $(XP_i)^2$ as $(\mathbf{p}_i - \mathbf{x}) \cdot (\mathbf{p}_i - \mathbf{x})$, where $\mathbf{x}$ is the position vector of $X$ with respect to $O$, show that $\sum_{i=1}^{4} (XP_i)^2 = 8.$

(ii) Given that $P_1$ has coordinates $(0, 0, 1)$ and that the coordinates of $P_2$ are of the form $(a, 0, b)$, where $a > 0$, show that $a = 2\sqrt{2}/3$ and $b = -1/3$, and find the coordinates of $P_3$ and $P_4$.

(iii) Show that $\sum_{i=1}^{4} (XP_i)^4 = 4 \sum_{i=1}^{4} (1 - \mathbf{x} \cdot \mathbf{p}_i)^2.$

By letting the coordinates of $X$ be $(x, y, z)$, show further that $\sum_{i=1}^{4} (XP_i)^4$ is independent of the position of $X$.

Hint

3. The scalar product of $p_i$ with $\sum p_r$, which is of course zero, can be expanded giving $p_i \cdot p_i = 1$ and three products $p_i \cdot p_j$ which are equal by symmetry, giving the required result. Expanding the expression suggested in (i), gives $\sum_{i=1}^{4} (p_i \cdot p_i - 2x \cdot p_i + x \cdot x)$, which, bearing in mind that $p_i \cdot p_i = 1$, $x \cdot x = 1$, and that $x \cdot \sum_{i=1}^{4} p_i = 0$, gives the correct result. Considering that $p_1 \cdot p_2 = -\frac{1}{3}$, $p_2 \cdot p_2 = 1$, and that $a$ is positive, enables the given values to be found. Similarly $p_1 \cdot p_3 = -\frac{1}{3}$, $p_2 \cdot p_3 = -\frac{1}{3}$, and $p_3 \cdot p_3 = 1$ yields $P_3, P_4 = \left( -\frac{\sqrt{2}}{3}, \pm \frac{\sqrt{2}}{\sqrt{3}}, -\frac{1}{3} \right)$. In (iii), using the logic of (i), $(XP_i)^4 = ((p_i - x) \cdot (p_i - x))^2 = 4(1 - x \cdot p_i)^2$, as required. Expanding this, and using the coordinates of $X$ and those of $P_i$ that have been found,

$\sum_{i=1}^{4} (XP_i)^4 = 16 + 4 \left( z^2 + \left( \frac{2\sqrt{2}}{3} x - \frac{1}{3} z \right)^2 + \left( -\frac{\sqrt{2}}{3} x + \frac{\sqrt{2}}{\sqrt{3}} y - \frac{1}{3} z \right)^2 + \left( -\frac{\sqrt{2}}{3} x - \frac{\sqrt{2}}{\sqrt{3}} y - \frac{1}{3} z \right)^2 \right)$

$= 16 + 4 \left( \frac{4}{3} x^2 + \frac{4}{3} y^2 + \frac{4}{3} z^2 \right) = \frac{64}{3}$ which is sufficient.

Model Solution

Showing $\mathbf{p}_i \cdot \mathbf{p}_j = -\frac{1}{3}$ for $i \neq j$:

Since $P_1, P_2, P_3, P_4$ form a regular tetrahedron on a sphere of radius 1, by symmetry the dot product $\mathbf{p}_i \cdot \mathbf{p}_j$ is the same for all $i \neq j$. Taking the dot product of $\mathbf{p}_1$ with $\mathbf{p}_1 + \mathbf{p}_2 + \mathbf{p}_3 + \mathbf{p}_4 = \mathbf{0}$:

$\mathbf{p}_1 \cdot \mathbf{p}_1 + \mathbf{p}_1 \cdot \mathbf{p}_2 + \mathbf{p}_1 \cdot \mathbf{p}_3 + \mathbf{p}_1 \cdot \mathbf{p}_4 = 0$

Since $|\mathbf{p}_1|^2 = 1$ and $\mathbf{p}_1 \cdot \mathbf{p}_j$ is the same for $j = 2, 3, 4$:

$1 + 3(\mathbf{p}_1 \cdot \mathbf{p}_2) = 0 \implies \mathbf{p}_i \cdot \mathbf{p}_j = -\frac{1}{3} \text{ for } i \neq j. \qquad \square$

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Part (i)

Expanding $(XP_i)^2 = (\mathbf{p}_i - \mathbf{x}) \cdot (\mathbf{p}_i - \mathbf{x})$:

$(XP_i)^2 = \mathbf{p}_i \cdot \mathbf{p}_i - 2\mathbf{x} \cdot \mathbf{p}_i + \mathbf{x} \cdot \mathbf{x} = 1 - 2\mathbf{x} \cdot \mathbf{p}_i + 1 = 2 - 2\mathbf{x} \cdot \mathbf{p}_i$

Summing over $i = 1, 2, 3, 4$:

$\sum_{i=1}^{4} (XP_i)^2 = \sum_{i=1}^{4} (2 - 2\mathbf{x} \cdot \mathbf{p}_i) = 8 - 2\mathbf{x} \cdot \sum_{i=1}^{4} \mathbf{p}_i = 8 - 2\mathbf{x} \cdot \mathbf{0} = 8. \qquad \square$

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Part (ii)

Given $P_1 = (0, 0, 1)$ and $P_2 = (a, 0, b)$ with $a > 0$.

From $\mathbf{p}_1 \cdot \mathbf{p}_2 = -\frac{1}{3}$: $\quad 0 + 0 + b = -\frac{1}{3}$, so $b = -\frac{1}{3}$.

From $|\mathbf{p}_2|^2 = 1$: $\quad a^2 + \frac{1}{9} = 1$, so $a^2 = \frac{8}{9}$, giving $a = \frac{2\sqrt{2}}{3}$ (since $a > 0$). $\qquad \square$

For $P_3 = (c, d, e)$, we need $|\mathbf{p}_3|^2 = 1$, $\mathbf{p}_1 \cdot \mathbf{p}_3 = -\frac{1}{3}$, and $\mathbf{p}_2 \cdot \mathbf{p}_3 = -\frac{1}{3}$.

From $\mathbf{p}_1 \cdot \mathbf{p}_3 = -\frac{1}{3}$: $\quad e = -\frac{1}{3}$.

From $|\mathbf{p}_3|^2 = 1$: $\quad c^2 + d^2 = \frac{8}{9}$.

From $\mathbf{p}_2 \cdot \mathbf{p}_3 = -\frac{1}{3}$: $\quad \frac{2\sqrt{2}}{3}c + (-\frac{1}{3})(-\frac{1}{3}) = -\frac{1}{3}$, so $\frac{2\sqrt{2}}{3}c + \frac{1}{9} = -\frac{1}{3}$, giving $\frac{2\sqrt{2}}{3}c = -\frac{4}{9}$, hence $c = -\frac{\sqrt{2}}{3}$.

From $c^2 + d^2 = \frac{8}{9}$: $\quad \frac{2}{9} + d^2 = \frac{8}{9}$, so $d^2 = \frac{2}{3}$, giving $d = \pm\frac{\sqrt{6}}{3}$.

By the symmetry of the regular tetrahedron, we take:

$P_3 = \left(-\frac{\sqrt{2}}{3},\; \frac{\sqrt{6}}{3},\; -\frac{1}{3}\right), \qquad P_4 = \left(-\frac{\sqrt{2}}{3},\; -\frac{\sqrt{6}}{3},\; -\frac{1}{3}\right).$

One can verify $\mathbf{p}_3 \cdot \mathbf{p}_4 = \frac{2}{9} - \frac{6}{9} + \frac{1}{9} = -\frac{3}{9} = -\frac{1}{3}$. $\qquad \square$

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Part (iii)

Since $(XP_i)^2 = 2 - 2\mathbf{x} \cdot \mathbf{p}_i = 2(1 - \mathbf{x} \cdot \mathbf{p}_i)$, we have:

$(XP_i)^4 = 4(1 - \mathbf{x} \cdot \mathbf{p}_i)^2.$

Therefore:

$\sum_{i=1}^{4} (XP_i)^4 = 4\sum_{i=1}^{4} (1 - \mathbf{x} \cdot \mathbf{p}_i)^2. \qquad \square$

Expanding:

$\sum_{i=1}^{4} (1 - \mathbf{x} \cdot \mathbf{p}_i)^2 = 4 - 2\mathbf{x} \cdot \sum_{i=1}^{4} \mathbf{p}_i + \sum_{i=1}^{4} (\mathbf{x} \cdot \mathbf{p}_i)^2 = 4 + \sum_{i=1}^{4} (\mathbf{x} \cdot \mathbf{p}_i)^2$

since $\sum \mathbf{p}_i = \mathbf{0}$. With $\mathbf{x} = (x, y, z)$:

$\mathbf{x} \cdot \mathbf{p}_1 = z$

$\mathbf{x} \cdot \mathbf{p}_2 = \frac{2\sqrt{2}}{3}x - \frac{1}{3}z$

$\mathbf{x} \cdot \mathbf{p}_3 = -\frac{\sqrt{2}}{3}x + \frac{\sqrt{6}}{3}y - \frac{1}{3}z$

$\mathbf{x} \cdot \mathbf{p}_4 = -\frac{\sqrt{2}}{3}x - \frac{\sqrt{6}}{3}y - \frac{1}{3}z$

Squaring and summing. The cross terms in $xz$ cancel between $P_2$ and $P_3 + P_4$, and the cross terms in $yz$ cancel between $P_3$ and $P_4$:

$(\mathbf{x} \cdot \mathbf{p}_3)^2 + (\mathbf{x} \cdot \mathbf{p}_4)^2 = 2\left(\frac{2}{9}x^2 + \frac{2}{3}y^2 + \frac{1}{9}z^2\right) = \frac{4}{9}x^2 + \frac{4}{3}y^2 + \frac{2}{9}z^2$

$(\mathbf{x} \cdot \mathbf{p}_2)^2 = \frac{8}{9}x^2 - \frac{4\sqrt{2}}{9}xz + \frac{1}{9}z^2$

Adding all four terms (the $xz$ terms from $P_2$ and $P_3 + P_4$ cancel):

$\sum_{i=1}^{4} (\mathbf{x} \cdot \mathbf{p}_i)^2 = \left(\frac{8}{9} + \frac{4}{9}\right)x^2 + \frac{4}{3}y^2 + \left(1 + \frac{1}{9} + \frac{2}{9}\right)z^2 = \frac{4}{3}x^2 + \frac{4}{3}y^2 + \frac{4}{3}z^2 = \frac{4}{3}(x^2 + y^2 + z^2) = \frac{4}{3}$

since $X$ lies on the unit sphere. Therefore:

$\sum_{i=1}^{4} (XP_i)^4 = 4\left(4 + \frac{4}{3}\right) = 4 \cdot \frac{16}{3} = \frac{64}{3}$

which is independent of the position of $X$. $\qquad \square$

Examiner Notes

3. Most that attempted this question managed to achieve the first two results successfully, unless they got the diagram wrong. However, the final result was found trickier as some forgot to include the gravitational potential energy, some failed to evaluate the correct elastic potential energy and there were many mistakes made handling the surds.

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Question 4

Topic: 复数 Complex Numbers  |  Difficulty: Challenging  |  Marks: 20

Problem

4 Show that $(z - \mathrm{e}^{\mathrm{i}\theta})(z - \mathrm{e}^{-\mathrm{i}\theta}) = z^2 - 2z \cos \theta + 1$.

Write down the $(2n)$th roots of $-1$ in the form $\mathrm{e}^{\mathrm{i}\theta}$, where $-\pi < \theta \leqslant \pi$, and deduce that $z^{2n} + 1 = \prod_{k=1}^{n} \left( z^2 - 2z \cos \left( \frac{(2k-1)\pi}{2n} \right) + 1 \right).$

Here, $n$ is a positive integer, and the $\prod$ notation denotes the product.

(i) By substituting $z = \mathrm{i}$ show that, when $n$ is even, $\cos \left( \frac{\pi}{2n} \right) \cos \left( \frac{3\pi}{2n} \right) \cos \left( \frac{5\pi}{2n} \right) \cdots \cos \left( \frac{(2n-1)\pi}{2n} \right) = (-1)^{\frac{1}{2}n} 2^{1-n}.$

(ii) Show that, when $n$ is odd, $\cos^2 \left( \frac{\pi}{2n} \right) \cos^2 \left( \frac{3\pi}{2n} \right) \cos^2 \left( \frac{5\pi}{2n} \right) \cdots \cos^2 \left( \frac{(n-2)\pi}{2n} \right) = n 2^{1-n}.$

You may use without proof the fact that $1 + z^{2n} = (1 + z^2)(1 - z^2 + z^4 - \cdots + z^{2n-2})$ when $n$ is odd.

Hint

4. The initial result is obtained by expanding the brackets and expressing the exponentials in trigonometric form. The $(2n)$th roots of -1 are $e^{i \frac{2m+1}{2n} \pi}$, $-n \le m \le n - 1$, which lead to the factors of $z^{2n} + 1$ and these paired using the initial result give the required result. Part (i) follows directly from substituting $z = i$ in the previous result, and as $n$ is even, $z^{2n} + 1 = 2$. Using the given factorisation in part (ii), the general result can be simplified by the factor

$z^2 - 2z \cos \frac{n}{2n} \pi + 1 = z^2 + 1$. Again substituting $z = i$, and that $\cos \frac{2n-r}{2n} \pi = -\cos \frac{r}{2n} \pi$ gives the evaluation required.

Model Solution

Showing $(z - \mathrm{e}^{\mathrm{i}\theta})(z - \mathrm{e}^{-\mathrm{i}\theta}) = z^2 - 2z\cos\theta + 1$:

$(z - \mathrm{e}^{\mathrm{i}\theta})(z - \mathrm{e}^{-\mathrm{i}\theta}) = z^2 - z(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta}) + \mathrm{e}^{\mathrm{i}\theta}\mathrm{e}^{-\mathrm{i}\theta} = z^2 - 2z\cos\theta + 1. \qquad \square$

Finding the $(2n)$th roots of $-1$:

We solve $z^{2n} = -1 = \mathrm{e}^{\mathrm{i}(\pi + 2m\pi)}$ for integer $m$, giving:

$z = \mathrm{e}^{\mathrm{i}(2m+1)\pi/(2n)}.$

To obtain $2n$ distinct roots with $-\pi < \theta \leqslant \pi$, we take $m = -n, -n+1, \ldots, n-1$. These give angles $\theta_m = \frac{(2m+1)\pi}{2n}$ ranging from $-\pi + \frac{\pi}{2n}$ to $\pi - \frac{\pi}{2n}$, all lying in $(-\pi, \pi)$.

Since $z^{2n} + 1$ has real coefficients, its roots come in conjugate pairs. The root for $m$ and the root for $m' = -(m+1)$ are conjugates: $\theta_{m'} = -\theta_m$. Pairing $m = 0$ with $m = -1$, $m = 1$ with $m = -2$, and in general $m = k$ with $m = -(k+1)$ for $k = 0, 1, \ldots, n-1$, we obtain $n$ pairs with angles $\frac{(2k+1)\pi}{2n}$ for $k = 0, 1, \ldots, n-1$, i.e. $\frac{\pi}{2n}, \frac{3\pi}{2n}, \ldots, \frac{(2n-1)\pi}{2n}$.

Each conjugate pair gives a quadratic factor $(z - \mathrm{e}^{\mathrm{i}\theta})(z - \mathrm{e}^{-\mathrm{i}\theta}) = z^2 - 2z\cos\theta + 1$. Therefore:

$z^{2n} + 1 = \prod_{k=1}^{n}\left(z^2 - 2z\cos\frac{(2k-1)\pi}{2n} + 1\right). \qquad \square$

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Part (i): $n$ even

Substituting $z = \mathrm{i}$ into the factorization:

Left side: $\mathrm{i}^{2n} + 1 = (\mathrm{i}^2)^n + 1 = (-1)^n + 1 = 2$ since $n$ is even.

Right side: each factor becomes $\mathrm{i}^2 - 2\mathrm{i}\cos\theta_k + 1 = -2\mathrm{i}\cos\theta_k$.

Using the exponential form $2\cos\theta_k = \mathrm{e}^{\mathrm{i}\theta_k} + \mathrm{e}^{-\mathrm{i}\theta_k}$:

$\mathrm{i}^2 - 2\mathrm{i}\cos\theta_k + 1 = -\mathrm{i}(\mathrm{e}^{\mathrm{i}\theta_k} + \mathrm{e}^{-\mathrm{i}\theta_k}) = -\mathrm{i}\,\mathrm{e}^{-\mathrm{i}\theta_k}(1 + \mathrm{e}^{2\mathrm{i}\theta_k}).$

Therefore:

$\prod_{k=1}^{n}(-\mathrm{i}\,\mathrm{e}^{-\mathrm{i}\theta_k}) = (-\mathrm{i})^n \cdot \mathrm{e}^{-\mathrm{i}\sum\theta_k} = (-\mathrm{i})^n \cdot 1$

since $\sum_{k=1}^{n}\theta_k = \frac{\pi}{2n}\sum_{k=1}^{n}(2k-1) = \frac{\pi}{2n} \cdot n^2 = \frac{n\pi}{2}$, and $\mathrm{e}^{-\mathrm{i}n\pi/2} = (\mathrm{e}^{-\mathrm{i}\pi/2})^n = (-\mathrm{i})^n$, so this combines with $(-\mathrm{i})^n$ to give $(-\mathrm{i})^{2n} = ((-\mathrm{i})^2)^n = (-1)^n = 1$.

Wait, let me recompute. The product of all factors is:

$\prod_{k=1}^{n}(-2\mathrm{i}\cos\theta_k) = (-2\mathrm{i})^n \prod_{k=1}^{n}\cos\theta_k.$

For even $n = 2m$: $(-2\mathrm{i})^{2m} = ((-2\mathrm{i})^2)^m = (-4)^m = (-1)^m \cdot 4^m = (-1)^m \cdot 2^{2m} = (-1)^{n/2} \cdot 2^n$.

Setting left side equal to right side:

$2 = (-1)^{n/2} \cdot 2^n \prod_{k=1}^{n}\cos\frac{(2k-1)\pi}{2n}$

$\prod_{k=1}^{n}\cos\frac{(2k-1)\pi}{2n} = \frac{2}{(-1)^{n/2} \cdot 2^n} = (-1)^{n/2} \cdot 2^{1-n}. \qquad \square$

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Part (ii): $n$ odd

Using the given factorization $1 + z^{2n} = (1 + z^2)(1 - z^2 + z^4 - \cdots + z^{2n-2})$.

At $z = \mathrm{i}$: left side $= 1 + (-1)^n = 0$ (since $n$ is odd), and $1 + \mathrm{i}^2 = 0$, so both sides are zero.

The right side factor $z^2 - 2z\cos\frac{(2k-1)\pi}{2n} + 1$ equals $z^2 + 1$ when $\cos\frac{(2k-1)\pi}{2n} = 0$, which occurs when $\frac{(2k-1)\pi}{2n} = \frac{\pi}{2}$, i.e. $k = \frac{n+1}{2}$ (an integer since $n$ is odd).

So the factor with $k = \frac{n+1}{2}$ is exactly $z^2 + 1$. Removing this factor:

$Q(z) := \frac{z^{2n}+1}{z^2+1} = 1 - z^2 + z^4 - \cdots + z^{2n-2} = \prod_{\substack{k=1 \\ k \neq (n+1)/2}}^{n}\left(z^2 - 2z\cos\frac{(2k-1)\pi}{2n} + 1\right).$

The remaining $n - 1$ factors pair into conjugate pairs (since the quadratic factors for $k$ and $n - k + 1$ are conjugates, as $\cos\frac{(2(n-k+1)-1)\pi}{2n} = -\cos\frac{(2k-1)\pi}{2n}$). There are $\frac{n-1}{2}$ such pairs.

Substituting $z = \mathrm{i}$ into $Q$:

Left side: $Q(\mathrm{i}) = 1 - (-1) + 1 - (-1) + \cdots + 1 = n$ (since $n$ is odd, there are $n$ terms all equal to $+1$).

Right side: each quadratic factor at $z = \mathrm{i}$ gives $-2\mathrm{i}\cos\frac{(2k-1)\pi}{2n}$. The product over the pair $(k, n-k+1)$ gives:

$(-2\mathrm{i}\cos\theta_k)(-2\mathrm{i}\cos(\pi - \theta_k)) = (-2\mathrm{i}\cos\theta_k)(2\mathrm{i}\cos\theta_k) = 4\cos^2\theta_k.$

The product over all $\frac{n-1}{2}$ pairs is:

$\prod_{k=1}^{(n-1)/2} 4\cos^2\frac{(2k-1)\pi}{2n} = 2^{n-1}\prod_{k=1}^{(n-1)/2}\cos^2\frac{(2k-1)\pi}{2n}.$

Setting left side equal to right side:

$n = 2^{n-1}\prod_{k=1}^{(n-1)/2}\cos^2\frac{(2k-1)\pi}{2n}$

$\prod_{k=1}^{(n-1)/2}\cos^2\frac{(2k-1)\pi}{2n} = n \cdot 2^{1-n}. \qquad \square$

Examiner Notes

4. Many candidates tried to write $z = x + iy$ or similar and likewise for $w$ and then tried to expand which involved a lot more work than dealing with conjugates directly. Some tried to use the cosine rule rather than the triangle inequality from the diagram. In general, the first result and parts (i) and (ii) were well done but only the strongest candidates did better than pick up the odd mark here and there in trying to obtain the inequality. A lot of mistakes were made mishandling inequalities, but even those who could do this correctly overlooked the necessity of substantiating that th4 square roots are positive and that the denominator is non-zero.

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Question 5

Topic: 数论 Number Theory  |  Difficulty: Hard  |  Marks: 20

Problem

5 In this question, you may assume that, if $a$, $b$ and $c$ are positive integers such that $a$ and $b$ are coprime and $a$ divides $bc$, then $a$ divides $c$. (Two positive integers are said to be coprime if their highest common factor is 1.)

(i) Suppose that there are positive integers $p$, $q$, $n$ and $N$ such that $p$ and $q$ are coprime and $q^n N = p^n$. Show that $N = k p^n$ for some positive integer $k$ and deduce the value of $q$.

Hence prove that, for any positive integers $n$ and $N$, $\sqrt[n]{N}$ is either a positive integer or irrational.

(ii) Suppose that there are positive integers $a$, $b$, $c$ and $d$ such that $a$ and $b$ are coprime and $c$ and $d$ are coprime, and $a^a d^b = b^a c^b$. Prove that $d^b = b^a$ and deduce that, if $p$ is a prime factor of $d$, then $p$ is also a prime factor of $b$.

If $p^m$ and $p^n$ are the highest powers of the prime number $p$ that divide $d$ and $b$, respectively, express $b$ in terms of $a$, $m$ and $n$ and hence show that $p^n \leqslant n$. Deduce the value of $b$. (You may assume that if $x > 0$ and $y \geqslant 2$ then $y^x > x$.)

Hence prove that, if $r$ is a positive rational number such that $r^r$ is rational, then $r$ is a positive integer.

Hint

5. Writing $q^n N$ as $q q^{n-1} N$, and employing the permitted assumption, as $p$ and $q$ are coprime, $p$ divides $q^{n-1} N$. Repetitions of this argument imply finally that $p$ divides $N$. Letting $N = p Q_1$, $q^n Q_1 = p^{n-1}$. Continuing this argument similarly gives the result $N = k p^n$. As a consequence, $q^n k = 1$, and thus $q$ and $k$ must both be

Model Solution

Part (i)

We are given positive integers $p, q, n, N$ with $\gcd(p, q) = 1$ and $q^n N = p^n$.

Showing $N = kp^n$:

Since $q^n N = p^n$, we have $p^n \mid q^n N$. Because $\gcd(p, q) = 1$, we know $\gcd(p^n, q^n) = 1$ (any prime dividing $p^n$ divides $p$, and since $\gcd(p,q) = 1$ it does not divide $q$, hence does not divide $q^n$). By the assumed result (applied with $a = p^n$, $b = q^n$, $c = N$), since $\gcd(p^n, q^n) = 1$ and $p^n \mid q^n N$, we conclude $p^n \mid N$.

Therefore $N = kp^n$ for some positive integer $k$.

Deducing $q$:

Substituting $N = kp^n$ into $q^n N = p^n$:

$q^n \cdot kp^n = p^n$

Dividing both sides by $p^n > 0$:

$q^n k = 1$

Since $q$ and $k$ are positive integers, $q^n \geqslant 1$ and $k \geqslant 1$. The only way their product equals 1 is if $q^n = 1$ and $k = 1$. Since $q$ is a positive integer, $q^n = 1$ gives $q = 1$.

Proving $\sqrt[n]{N}$ is either a positive integer or irrational:

We prove the contrapositive: if $\sqrt[n]{N}$ is a positive rational number that is not an integer, we reach a contradiction.

Suppose $\sqrt[n]{N} = \frac{p}{q}$ where $p, q$ are positive integers with $\gcd(p, q) = 1$ and $q > 1$ (i.e.\ $\sqrt[n]{N}$ is rational but not an integer). Raising both sides to the $n$-th power:

$N = \frac{p^n}{q^n} \implies q^n N = p^n$

This satisfies the conditions of the first part, so we deduce $q = 1$. But we assumed $q > 1$, a contradiction.

Therefore $\sqrt[n]{N}$ is either a positive integer or irrational. $\blacksquare$

Part (ii)

We are given positive integers $a, b, c, d$ with $\gcd(a, b) = 1$ and $\gcd(c, d) = 1$, and $a^a d^b = b^a c^b$.

Proving $d^b = b^a$:

From $a^a d^b = b^a c^b$, since $b^a$ divides the right-hand side, $b^a \mid a^a d^b$.

Since $\gcd(a, b) = 1$, we have $\gcd(a^a, b^a) = 1$ (any prime dividing $b^a$ divides $b$, and since $\gcd(a,b) = 1$ it does not divide $a$, hence does not divide $a^a$). By the assumed result (with $a = b^a$, $b = a^a$, $c = d^b$), since $\gcd(b^a, a^a) = 1$ and $b^a \mid a^a d^b$, we get $b^a \mid d^b$.

So $d^b = k \cdot b^a$ for some positive integer $k$.

Similarly, from $a^a d^b = b^a c^b$, since $a^a$ divides the left-hand side, $a^a \mid b^a c^b$. By the same argument (with roles swapped), since $\gcd(a^a, b^a) = 1$ and $a^a \mid b^a c^b$, we get $a^a \mid c^b$, so $c^b = k' \cdot a^a$ for some positive integer $k'$.

Substituting into $a^a d^b = b^a c^b$:

$a^a \cdot k b^a = b^a \cdot k' a^a$

Dividing both sides by $a^a b^a$:

$k = k'$

So $d^b = k b^a$ and $c^b = k a^a$.

Now we use $\gcd(c, d) = 1$. Any prime $p$ dividing $k$ must divide $d^b = kb^a$ and $c^b = ka^a$. Since $p \mid d^b$ and $p$ is prime, $p \mid d$. Similarly $p \mid c^b$ implies $p \mid c$. But $\gcd(c, d) = 1$, so no prime can divide both $c$ and $d$. Therefore $k$ has no prime factors, meaning $k = 1$.

Hence $d^b = b^a$ (and $c^b = a^a$). $\blacksquare$

Deducing that prime factors of $d$ are prime factors of $b$:

If $p$ is a prime factor of $d$, then $p \mid d$, so $p^b \mid d^b = b^a$. Since $p \mid b^a$ and $p$ is prime, $p \mid b$.

Therefore every prime factor of $d$ is also a prime factor of $b$. $\blacksquare$

Expressing $b$ and showing $p^n \leqslant n$:

Let $p^m \| d$ and $p^n \| b$ (meaning $p^m$ is the highest power of $p$ dividing $d$, and $p^n$ is the highest power of $p$ dividing $b$, where $m \geqslant 1$ since $p \mid d$). Then $p^{mb} \| d^b$ and $p^{na} \| b^a$. Since $d^b = b^a$:

$mb = na \qquad \qquad (\dagger)$

Since $\gcd(a, b) = 1$ and $b \mid na$, we have $b \mid n$ (because $b$ and $a$ share no common factor). So $n = lb$ for some positive integer $l$, giving:

$b = \frac{n}{l} \leqslant n$

Since $p^n \| b$, we have $p^n \leqslant b$, hence:

$p^n \leqslant n$

Deducing $b$:

We have $p^n \leqslant n$ where $p$ is prime (so $p \geqslant 2$).

If $n \geqslant 1$: by the given result with $x = n > 0$ and $y = 2 \geqslant 2$, we have $2^n > n$. Since $p \geqslant 2$, $p^n \geqslant 2^n > n$, contradicting $p^n \leqslant n$.

Therefore $n = 0$, meaning $p^n \| b$ with $n = 0$ for every prime $p$. But $p^n \| b$ with $n = 0$ means $p \nmid b$. Since this holds for every prime $p$ (recall every prime factor of $d$ is a prime factor of $b$, and we have shown no such prime can divide $b$ with a positive power), $b$ has no prime factors. Since $b$ is a positive integer, $b = 1$.

Proving that if $r$ is a positive rational with $r^r$ rational, then $r$ is a positive integer:

Let $r = \frac{a}{b}$ in lowest terms (so $\gcd(a, b) = 1$, $a, b$ positive integers), with $r^r = \frac{c}{d}$ also in lowest terms (so $\gcd(c, d) = 1$). Then:

$\left(\frac{a}{b}\right)^{a/b} = \frac{c}{d}$

Raising both sides to the power $b$:

$\left(\frac{a}{b}\right)^{a} = \left(\frac{c}{d}\right)^{b} \implies \frac{a^a}{b^a} = \frac{c^b}{d^b} \implies a^a d^b = b^a c^b$

This satisfies the conditions of part (ii) ($\gcd(a,b) = 1$, $\gcd(c,d) = 1$, $a^a d^b = b^a c^b$), so by part (ii), $b = 1$.

Therefore $r = a$ is a positive integer. $\blacksquare$

Examiner Notes

Nearly as many attempted this as question 4, but only achieving a quarter of the marks making it the least successfully answered question. Almost all missed the point of the question given in the first sentence, and made other assumptions, which frequently only applied to primes rather than integers in general. As a consequence, most did not satisfactorily justify their results.

They generally fared better tackling the second part of (i), though some tried to prove the statement in the wrong direction. They approached (ii) better though few gave a valid argument why $p^n \le n$.

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Question 6

Topic: 复数 Complex Numbers  |  Difficulty: Challenging  |  Marks: 20

Problem

6 Let $z$ and $w$ be complex numbers. Use a diagram to show that $|z - w| \leqslant |z| + |w|$.

For any complex numbers $z$ and $w$, $E$ is defined by

$E = z w^* + z^* w + 2 |z w| .$

(i) Show that $|z - w|^2 = (|z| + |w|)^2 - E$, and deduce that $E$ is real and non-negative.

(ii) Show that $|1 - z w^*|^2 = (1 + |z w|)^2 - E$.

Hence show that, if both $|z| > 1$ and $|w| > 1$, then

$\frac{|z - w|}{|1 - z w^*|} \leqslant \frac{|z| + |w|}{1 + |z w|} .$

Does this inequality also hold if both $|z| < 1$ and $|w| < 1$?

Hint

6. The opening result is the triangle inequality applied to OW, OZ, and WZ where OW and OZ are represented by the complex numbers w and z.

Part (i) relies on using $|z - w|^2 = (z - w)(z - w)^*$, $(z - w)^* = (z^* - w^*)$, $|zw| = |z| |w|$, and substituting $w z^* + z w^* = (E - 2|zw|)$. Having obtained the desired equation, the reality of E is apparent from the reality of the other terms and its non-negativity is obtained from the opening result of the question. Part (ii) relies on the same principles as part (i).

The inequality can be most easily obtained by squaring it, and substituting for both numerator and denominator on the left hand side using parts (i) and (ii), and algebraic rearrangement leads to $E(1 - |z|^2)(1 - |w|^2) \ge 0$ which is certainly true. The argument is fully reversible as $|z| > 1$, and $|w| > 1$, $|zw^*| > 1$, and so $1 - zw^* \ne 0$ so the division is permissible, and the square rooting of the inequality causes no problem as the quantities are positive. The working follows identically if $|z| < 1$, and $|w| < 1$.

Model Solution

Triangle inequality $|z - w| \leqslant |z| + |w|$:

In the Argand diagram, let $O$ be the origin, $Z$ the point representing $z$, and $W$ the point representing $w$. Then $|z|$ is the distance $OZ$, $|w|$ is the distance $OW$, and $|z - w|$ is the distance $ZW$. By the triangle inequality applied to triangle $OZW$:

$|z - w| \leqslant |z| + |w|$

with equality when $O$, $Z$, $W$ are collinear with $Z$ and $W$ on opposite sides of $O$. $\blacksquare$

Part (i)

We expand $|z - w|^2$ using the property $|u|^2 = uu^*$:

$|z - w|^2 = (z - w)(z - w)^* = (z - w)(z^* - w^*)$

$= zz^* - zw^* - wz^* + ww^* = |z|^2 - zw^* - z^*w + |w|^2$

Now expand $(|z| + |w|)^2$:

$(|z| + |w|)^2 = |z|^2 + 2|z||w| + |w|^2 = |z|^2 + 2|zw| + |w|^2$

since $|zw| = |z||w|$. Therefore:

$(|z| + |w|)^2 - |z - w|^2 = 2|zw| + zw^* + z^*w = zw^* + z^*w + 2|zw| = E$

which gives:

$|z - w|^2 = (|z| + |w|)^2 - E$

Deducing $E$ is real and non-negative:

Since $|z - w|^2$ and $(|z| + |w|)^2$ are both real (being squares of real moduli), their difference $E$ is real.

From the triangle inequality, $|z - w| \leqslant |z| + |w|$, so $|z - w|^2 \leqslant (|z| + |w|)^2$, hence $E \geqslant 0$. $\blacksquare$

Part (ii)

We expand $|1 - zw^*|^2$ using the same property:

$|1 - zw^*|^2 = (1 - zw^*)(1 - zw^*)^* = (1 - zw^*)(1 - z^*w)$

$= 1 - z^*w - zw^* + zw^*z^*w = 1 - zw^* - z^*w + |z|^2|w|^2$

$= 1 - zw^* - z^*w + |zw|^2$

Now expand $(1 + |zw|)^2$:

$(1 + |zw|)^2 = 1 + 2|zw| + |zw|^2$

Therefore:

$(1 + |zw|)^2 - |1 - zw^*|^2 = 2|zw| + zw^* + z^*w = zw^* + z^*w + 2|zw| = E$

which gives:

$|1 - zw^*|^2 = (1 + |zw|)^2 - E$

Proving the inequality for $|z| > 1$ and $|w| > 1$:

From parts (i) and (ii), the inequality

$\frac{|z - w|}{|1 - zw^*|} \leqslant \frac{|z| + |w|}{1 + |zw|}$

is equivalent (after squaring both sides, which preserves the inequality since both sides are non-negative) to:

$\frac{(|z| + |w|)^2 - E}{(1 + |zw|)^2 - E} \leqslant \frac{(|z| + |w|)^2}{(1 + |zw|)^2}$

Let $A = (|z| + |w|)^2$ and $B = (1 + |zw|)^2$, both positive. We need $\frac{A - E}{B - E} \leqslant \frac{A}{B}$.

Cross-multiplying (valid since $B > 0$ and $B - E = |1 - zw^*|^2 > 0$ because $|zw^*| = |z||w| > 1$ means $zw^* \neq 1$):

$B(A - E) \leqslant A(B - E)$

$AB - BE \leqslant AB - AE$

$AE \leqslant BE$

$E(A - B) \leqslant 0$

Since $E \geqslant 0$, we need $A - B \leqslant 0$, i.e.\ $(|z| + |w|)^2 \leqslant (1 + |zw|)^2$. This is equivalent to $|z| + |w| \leqslant 1 + |z||w|$ (both sides positive), which rearranges to:

$0 \leqslant 1 - |z| - |w| + |z||w| = (1 - |z|)(1 - |w|)$

Since $|z| > 1$ and $|w| > 1$, both factors $(1 - |z|)$ and $(1 - |w|)$ are negative, so their product is positive. Hence $A - B \leqslant 0$ holds, and therefore $E(A - B) \leqslant 0$, confirming the inequality.

All steps are reversible (the squaring step is valid since both sides are non-negative, and the denominator $1 - zw^* \neq 0$ since $|zw^*| > 1$). $\blacksquare$

Does the inequality hold if $|z| < 1$ and $|w| < 1$?

Yes. The entire argument above depends only on $(1 - |z|)(1 - |w|) \geqslant 0$. When $|z| < 1$ and $|w| < 1$, both factors $(1 - |z|)$ and $(1 - |w|)$ are positive, so their product is positive. The denominator $1 - zw^*$ is non-zero since $|zw^*| = |z||w| < 1$, so $zw^* \neq 1$. All other steps carry through identically.

Therefore the inequality also holds when both $|z| < 1$ and $|w| < 1$. $\blacksquare$

Examiner Notes

Question 6

About half attempted this with marginally more success than question

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Question 7

Topic: 微分方程 Differential Equations  |  Difficulty: Hard  |  Marks: 20

Problem

7 (i) Let $y(x)$ be a solution of the differential equation $\frac{\text{d}^2y}{\text{d}x^2} + y^3 = 0$ with $y = 1$ and $\frac{\text{d}y}{\text{d}x} = 0$ at $x = 0$, and let $\text{E}(x) = \left( \frac{\text{d}y}{\text{d}x} \right)^2 + \frac{1}{2}y^4 .$ Show by differentiation that E is constant and deduce that $|y(x)| \leqslant 1$ for all $x$.

(ii) Let $v(x)$ be a solution of the differential equation $\frac{\text{d}^2v}{\text{d}x^2} + x \frac{\text{d}v}{\text{d}x} + \sinh v = 0$ with $v = \ln 3$ and $\frac{\text{d}v}{\text{d}x} = 0$ at $x = 0$, and let $\text{E}(x) = \left( \frac{\text{d}v}{\text{d}x} \right)^2 + 2 \cosh v .$ Show that $\frac{\text{dE}}{\text{d}x} \leqslant 0$ for $x \geqslant 0$ and deduce that $\cosh v(x) \leqslant \frac{5}{3}$ for $x \geqslant 0$.

(iii) Let $w(x)$ be a solution of the differential equation $\frac{\text{d}^2w}{\text{d}x^2} + (5 \cosh x - 4 \sinh x - 3) \frac{\text{d}w}{\text{d}x} + (w \cosh w + 2 \sinh w) = 0$ with $\frac{\text{d}w}{\text{d}x} = \frac{1}{\sqrt{2}}$ and $w = 0$ at $x = 0$. Show that $\cosh w(x) \leqslant \frac{5}{4}$ for $x \geqslant 0$.

Hint

7. As $\frac{dE}{dx} = 2 \frac{dy}{dx} \left( \frac{d^2y}{dx^2} + y^3 \right)$ is zero for all $x$, $E(x)$ is constant, and $E(x) = \frac{1}{2}$ using the initial conditions. The deduction follows from the non-negativity of $\left( \frac{dy}{dx} \right)^2$. In part (ii), it can be shown that $\frac{dE}{dx} = -2x \left( \frac{dv}{dx} \right)^2 \le 0$ for $x \ge 0$, and as initially $E(x) = \frac{10}{3}$, the deduction for $\cosh v(x)$ follows in the same way as that in part (i). In part (iii), the choice of $E(x)$ relies on $2 \int (w \cosh w + 2 \sinh w) dw$ so $E(x) = \left( \frac{dw}{dx} \right)^2 + 2(w \sinh w + \cosh w)$. Then

$\frac{dE}{dx} = -2 \left( \frac{dw}{dx} \right)^2 (5 \cosh x - 4 \sinh x - 3) = -2 \left( \frac{dw}{dx} \right)^2 \frac{e^{-x}}{2} (e^x - 3)^2$, and initially $E(x) = \frac{5}{2}$.

The final result can be deduced as in the previous parts, with the additional consideration that $w \sinh w \ge 0$.

Model Solution

Part (i)

We differentiate $\text{E}(x) = \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + \frac{1}{2}y^4$ with respect to $x$:

$\frac{\mathrm{dE}}{\mathrm{d}x} = 2\frac{\mathrm{d}y}{\mathrm{d}x}\cdot\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2y^3\frac{\mathrm{d}y}{\mathrm{d}x} = 2\frac{\mathrm{d}y}{\mathrm{d}x}\left(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + y^3\right).$

Since $y$ satisfies $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + y^3 = 0$, the bracket vanishes, so $\frac{\mathrm{dE}}{\mathrm{d}x} = 0$ for all $x$. Therefore $\text{E}$ is constant.

Using the initial conditions $y(0) = 1$ and $y'(0) = 0$:

$\text{E}(x) = \text{E}(0) = 0^2 + \frac{1}{2}(1)^4 = \frac{1}{2}.$

Since $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \geqslant 0$, we have $\frac{1}{2}y^4 \leqslant \text{E} = \frac{1}{2}$, giving $y^4 \leqslant 1$, hence $|y(x)| \leqslant 1$ for all $x$.

Part (ii)

We differentiate $\text{E}(x) = \left(\frac{\mathrm{d}v}{\mathrm{d}x}\right)^2 + 2\cosh v$:

$\frac{\mathrm{dE}}{\mathrm{d}x} = 2\frac{\mathrm{d}v}{\mathrm{d}x}\cdot\frac{\mathrm{d}^2v}{\mathrm{d}x^2} + 2\sinh v\cdot\frac{\mathrm{d}v}{\mathrm{d}x} = 2\frac{\mathrm{d}v}{\mathrm{d}x}\left(\frac{\mathrm{d}^2v}{\mathrm{d}x^2} + \sinh v\right).$

From the ODE, $\frac{\mathrm{d}^2v}{\mathrm{d}x^2} = -x\frac{\mathrm{d}v}{\mathrm{d}x} - \sinh v$, so $\frac{\mathrm{d}^2v}{\mathrm{d}x^2} + \sinh v = -x\frac{\mathrm{d}v}{\mathrm{d}x}$.

Substituting:

$\frac{\mathrm{dE}}{\mathrm{d}x} = 2\frac{\mathrm{d}v}{\mathrm{d}x}\left(-x\frac{\mathrm{d}v}{\mathrm{d}x}\right) = -2x\left(\frac{\mathrm{d}v}{\mathrm{d}x}\right)^2.$

For $x \geqslant 0$, since $\left(\frac{\mathrm{d}v}{\mathrm{d}x}\right)^2 \geqslant 0$, we have $\frac{\mathrm{dE}}{\mathrm{d}x} \leqslant 0$.

Using the initial conditions $v(0) = \ln 3$ and $v'(0) = 0$, and the identity $\cosh(\ln 3) = \frac{3 + \frac{1}{3}}{2} = \frac{5}{3}$:

$\text{E}(0) = 0 + 2 \cdot \frac{5}{3} = \frac{10}{3}.$

Since $\text{E}$ is non-increasing for $x \geqslant 0$, we have $\text{E}(x) \leqslant \frac{10}{3}$ for all $x \geqslant 0$. Therefore $2\cosh v \leqslant \left(\frac{\mathrm{d}v}{\mathrm{d}x}\right)^2 + 2\cosh v = \text{E}(x) \leqslant \frac{10}{3}$, giving $\cosh v(x) \leqslant \frac{5}{3}$ for $x \geqslant 0$.

Part (iii)

We need to find an energy function. The last term in the ODE, $w\cosh w + 2\sinh w$, is the derivative of $w\sinh w + \cosh w$ with respect to $w$, since

$\frac{\mathrm{d}}{\mathrm{d}w}(w\sinh w + \cosh w) = \sinh w + w\cosh w + \sinh w = w\cosh w + 2\sinh w.$

We choose $\text{E}(x) = \left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2 + 2(w\sinh w + \cosh w)$.

Differentiating:

$\frac{\mathrm{dE}}{\mathrm{d}x} = 2\frac{\mathrm{d}w}{\mathrm{d}x}\cdot\frac{\mathrm{d}^2w}{\mathrm{d}x^2} + 2\frac{\mathrm{d}w}{\mathrm{d}x}(w\cosh w + 2\sinh w) = 2\frac{\mathrm{d}w}{\mathrm{d}x}\left(\frac{\mathrm{d}^2w}{\mathrm{d}x^2} + w\cosh w + 2\sinh w\right).$

From the ODE, $\frac{\mathrm{d}^2w}{\mathrm{d}x^2} + w\cosh w + 2\sinh w = -(5\cosh x - 4\sinh x - 3)\frac{\mathrm{d}w}{\mathrm{d}x}$, so

$\frac{\mathrm{dE}}{\mathrm{d}x} = -2\left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2(5\cosh x - 4\sinh x - 3).$

We simplify the coefficient:

$5\cosh x - 4\sinh x - 3 = \frac{5(e^x + e^{-x})}{2} - \frac{4(e^x - e^{-x})}{2} - 3 = \frac{e^x + 9e^{-x} - 6}{2} = \frac{e^{-x}(e^{2x} - 6e^x + 9)}{2} = \frac{e^{-x}(e^x - 3)^2}{2}.$

Since $e^{-x} > 0$ and $(e^x - 3)^2 \geqslant 0$, the coefficient is non-negative. With $\left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2 \geqslant 0$, we get $\frac{\mathrm{dE}}{\mathrm{d}x} \leqslant 0$ for all $x$, so $\text{E}$ is non-increasing.

Using the initial conditions $w(0) = 0$ and $w'(0) = \frac{1}{\sqrt{2}}$:

$\text{E}(0) = \frac{1}{2} + 2(0 \cdot \sinh 0 + \cosh 0) = \frac{1}{2} + 2 = \frac{5}{2}.$

Since $\text{E}$ is non-increasing, $\text{E}(x) \leqslant \frac{5}{2}$ for $x \geqslant 0$.

Now $w\sinh w \geqslant 0$ for all real $w$ (since $w$ and $\sinh w$ have the same sign), so $2\cosh w \leqslant 2(w\sinh w + \cosh w) \leqslant \text{E}(x) \leqslant \frac{5}{2}$, giving $\cosh w(x) \leqslant \frac{5}{4}$ for $x \geqslant 0$.

Examiner Notes

7. Common errors were false attempts for the volume at the beginning using hemisphere and cones, and in the last part approximating $x$ small rather than $x - \frac{1}{2}R$ small. Many candidates were successful as far as the equilibrium but couldn’t deal with the small oscillations successfully.

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Question 8

Topic: 复数与几何 Complex Numbers and Geometry  |  Difficulty: Hard  |  Marks: 20

Problem

8 Evaluate $\sum_{r=0}^{n-1} \mathrm{e}^{2\mathrm{i}(\alpha+r\pi/n)}$ where $\alpha$ is a fixed angle and $n \geqslant 2$.

The fixed point $O$ is a distance $d$ from a fixed line $D$. For any point $P$, let $s$ be the distance from $P$ to $D$ and let $r$ be the distance from $P$ to $O$. Write down an expression for $s$ in terms of $d$, $r$ and the angle $\theta$, where $\theta$ is as shown in the diagram below.

The curve $E$ shown in the diagram is such that, for any point $P$ on $E$, the relation $r = ks$ holds, where $k$ is a fixed number with $0 < k < 1$.

Each of the $n$ lines $L_1, \dots, L_n$ passes through $O$ and the angle between adjacent lines is $\frac{\pi}{n}$. The line $L_j$ ($j = 1, \dots, n$) intersects $E$ in two points forming a chord of length $l_j$. Show that, for $n \geqslant 2$,

$\sum_{j=1}^{n} \frac{1}{l_j} = \frac{(2 - k^2)n}{4kd} .$

Hint

8. The sum is evaluated by recognising that it is a geometric progression with common ratio $e^{2i\pi/n}$ which may be summed using the standard formula and as $1 - e^{2i\pi/n} \ne 0$, the denominator

is non-zero so the sum is zero. By simple trigonometry, $s = d - r \cos \theta$. As $r = ks$, $r = \frac{kd}{1+k \cos \theta}$. Thus $l_j = \frac{kd}{1+k \cos \theta} + \frac{kd}{1+k \cos(\theta+\pi)}$ where $\theta = \alpha + (j - 1) \pi/n$. Simplifying, $l_j = \frac{2kd}{1-k^2 \cos^2 \theta}$. The summation of the reciprocals of this expression is simply found using a double angle formula and then by expressing the trigonometric terms as the real part of the sum at the start of the question.

Model Solution

Evaluating the sum

$\sum_{r=0}^{n-1} \mathrm{e}^{2\mathrm{i}(\alpha+r\pi/n)} = \mathrm{e}^{2\mathrm{i}\alpha} \sum_{r=0}^{n-1} \left(\mathrm{e}^{2\mathrm{i}\pi/n}\right)^r.$

This is a geometric series with common ratio $\omega = \mathrm{e}^{2\mathrm{i}\pi/n}$. Since $n \geqslant 2$, $\omega \neq 1$ (as $2\pi/n$ is not a multiple of $2\pi$), so the sum equals

$\mathrm{e}^{2\mathrm{i}\alpha} \cdot \frac{1 - \omega^n}{1 - \omega} = \mathrm{e}^{2\mathrm{i}\alpha} \cdot \frac{1 - \mathrm{e}^{2\mathrm{i}\pi}}{1 - \mathrm{e}^{2\mathrm{i}\pi/n}} = \mathrm{e}^{2\mathrm{i}\alpha} \cdot \frac{1 - 1}{1 - \mathrm{e}^{2\mathrm{i}\pi/n}} = 0.$

Expression for $s$

From the geometry, the distance from $P$ to the line $D$ is

$s = d - r\cos\theta.$

Finding the chord lengths

On the curve $E$, we have $r = ks$. Substituting $s = d - r\cos\theta$:

$r = k(d - r\cos\theta) = kd - kr\cos\theta,$

so $r(1 + k\cos\theta) = kd$, giving $r = \frac{kd}{1 + k\cos\theta}$.

The line $L_j$ passes through $O$ at angle $\theta_j = \alpha + (j-1)\pi/n$ to the horizontal. It meets $E$ at two points: one at angle $\theta_j$ and the other at angle $\theta_j + \pi$. The distances from $O$ are

$r_1 = \frac{kd}{1 + k\cos\theta_j} \qquad \text{and} \qquad r_2 = \frac{kd}{1 + k\cos(\theta_j + \pi)} = \frac{kd}{1 - k\cos\theta_j}.$

The chord length is

$l_j = r_1 + r_2 = kd\left(\frac{1}{1 + k\cos\theta_j} + \frac{1}{1 - k\cos\theta_j}\right) = kd \cdot \frac{(1 - k\cos\theta_j) + (1 + k\cos\theta_j)}{(1 + k\cos\theta_j)(1 - k\cos\theta_j)} = \frac{2kd}{1 - k^2\cos^2\theta_j}.$

Computing the sum of reciprocals

$\frac{1}{l_j} = \frac{1 - k^2\cos^2\theta_j}{2kd}.$

Therefore

$\sum_{j=1}^{n} \frac{1}{l_j} = \frac{1}{2kd} \sum_{j=1}^{n} (1 - k^2\cos^2\theta_j) = \frac{1}{2kd}\left(n - k^2\sum_{j=1}^{n}\cos^2\theta_j\right).$

Using the double angle formula $\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)$:

$\sum_{j=1}^{n}\cos^2\theta_j = \frac{n}{2} + \frac{1}{2}\sum_{j=1}^{n}\cos 2\theta_j.$

Now $\sum_{j=1}^{n}\cos 2\theta_j = \sum_{j=1}^{n}\cos(2\alpha + 2(j-1)\pi/n)$ is the real part of $\sum_{j=1}^{n}\mathrm{e}^{\mathrm{i}(2\alpha + 2(j-1)\pi/n)}$.

Setting $r = j - 1$, this is the real part of $\sum_{r=0}^{n-1}\mathrm{e}^{2\mathrm{i}(\alpha + r\pi/n)}$, which we showed equals $0$ in the first part. Therefore $\sum_{j=1}^{n}\cos 2\theta_j = 0$, and

$\sum_{j=1}^{n}\cos^2\theta_j = \frac{n}{2}.$

Substituting back:

$\sum_{j=1}^{n} \frac{1}{l_j} = \frac{1}{2kd}\left(n - k^2 \cdot \frac{n}{2}\right) = \frac{n(2 - k^2)}{4kd}.$

Examiner Notes

8. The immediate problem was many made no mention of probabilities in order to calculate expectations. Throughout, there was very poor justification, which included treating the random variables as though they were independent and compensating errors which led to given results. Most progressed no further than part (a) of (ii) at best and many had $E(X_i) = \frac{ab}{n^2}$.