Skip to content
CalcGospel 國際數學圖譜
返回

IAL 2021 June Q9

A Level / Edexcel / P4

IAL 2021 June Paper · Question 9

题目

Problem

(i) Relative to a fixed origin OO, the points AA, BB and CC have position vectors aa, bb and cc respectively.

Points AA, BB and CC lie in a straight line, with BB lying between AA and CC.

Given that AB:AC=1:3AB:AC=1:3 show that

c=3b2ac=3b-2a
(3)

(ii) Given that nNn\in\mathbb{N}, prove by contradiction that if n2n^2 is a multiple of 3 then nn is a multiple of 3

(5)
题目中文翻译

(i) 相对于固定原点 OO,点 AABBCC 的位置向量分别为 a,b,ca,b,c

AABBCC 共线,且 BB 在线段 ACAC 之间。

已知 AB:AC=1:3AB:AC=1:3,证明

c=3b2ac=3b-2a

(ii) 已知 nNn\in\mathbb{N},用反证法证明:如果 n2n^2 是 3 的倍数,那么 nn 也是 3 的倍数。

解答

(i)

解法一

思路

展开

因为 AB:AC=1:3AB:AC=1:3BB 位于 A,CA,C 之间,所以 BC=2ABBC=2AB。分别用位置向量写出 AB\overrightarrow{AB}BC\overrightarrow{BC},再整理即可得到目标式。

答题过程

展开

Since AB:AC=1:3AB:AC=1:3, we have

BC=2AB.BC=2AB.

In vector form,

BC=cb\overrightarrow{BC}=\mathbf{c}-\mathbf{b}

and

AB=ba.\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.

Therefore,

cb=2(ba)=2b2a.\begin{align*} \mathbf{c}-\mathbf{b} =&\,2(\mathbf{b}-\mathbf{a})\\ =&\,2\mathbf{b}-2\mathbf{a}. \end{align*}

Hence,

c=3b2a,\boxed{\mathbf{c}=3\mathbf{b}-2\mathbf{a}},

as required.

(ii)

解法一

思路

展开

反设 n2n^2 是 3 的倍数,但 nn 不是。任何不是 3 的倍数的自然数都可写成 3p+13p+13p+23p+2;分别平方后,结果除以 3 都余 1,与假设矛盾。

答题过程

展开

Assume, for a contradiction, that n2n^2 is a multiple of 33 but nn is not a multiple of 33.

Then, for some integer pp, either

n=3p+1n=3p+1

or

n=3p+2.n=3p+2.

In the first case,

n2=(3p+1)2=9p2+6p+1=3(3p2+2p)+1.\begin{align*} n^2 =&\,(3p+1)^2\\ =&\,9p^2+6p+1\\ =&\,3(3p^2+2p)+1. \end{align*}

Thus n2n^2 leaves remainder 11 when divided by 33.

In the second case,

n2=(3p+2)2=9p2+12p+4=3(3p2+4p+1)+1.\begin{align*} n^2 =&\,(3p+2)^2\\ =&\,9p^2+12p+4\\ =&\,3(3p^2+4p+1)+1. \end{align*}

Again, n2n^2 leaves remainder 11 when divided by 33.

In neither case is n2n^2 a multiple of 33, contradicting the assumption. Therefore,

n2 is a multiple of 3n is a multiple of 3.\boxed{ n^2\text{ is a multiple of }3 \qquad\Longrightarrow\qquad n\text{ is a multiple of }3 }.

解法二

思路

展开

官方资料也接受用模 3 算术表达同一个反证。若 nn 不是 3 的倍数,则它模 3 只能为 1 或 2;两种余数平方后都模 3 余 1。

答题过程

展开

Assume, for a contradiction, that

n20(mod3)n^2\equiv0\pmod3

but

n≢0(mod3).n\not\equiv0\pmod3.

Then n1(mod3)n\equiv1\pmod3 or n2(mod3)n\equiv2\pmod3. However,

121(mod3)1^2\equiv1\pmod3

and

22=41(mod3).2^2=4\equiv1\pmod3.

Thus n21(mod3)n^2\equiv1\pmod3, contradicting n20(mod3)n^2\equiv0\pmod3. Hence,

3n2  3n.\boxed{3\mid n^2\ \Longrightarrow\ 3\mid n}.