CalcGospel 國際數學圖譜
返回

IAL 2021 June Q3

A Level / Edexcel / P1

IAL 2021 June Paper · Question 3

题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Figure 1 shows the plan view of a flower bed.

Figure 1

The flowerbed is in the shape of a triangle ABCABC with

  • AB=pAB=p metres
  • AC=qAC=q metres
  • BC=22BC=2\sqrt2 metres
  • angle BAC=60∘BAC=60^\circ

(a) Show that

p2+q2−pq=8.\begin{align*} p^2+q^2-pq=8. \end{align*}
(2)

Given that side ACAC is 22 metres longer than side ABAB, use algebra to find

(b)

(i) the exact value of pp,

(ii) the exact value of qq.

(5)

Using the answers to part (b),

(c) calculate the exact area of the flower bed.

(2)

解答

(a)

解法一

思路

展开

已知两边 p,qp,q 和夹角 60∘60^\circ，要求第三边 222\sqrt2，直接使用余弦定理。

答题过程

展开

By the cosine rule,

BC2= AB2+AC2−2(AB)(AC)cos⁡60∘(22)2= p2+q2−2pq(12)8= p2+q2−pq.\begin{align*} BC^2=&\,AB^2+AC^2-2(AB)(AC)\cos60^\circ\\ (2\sqrt2)^2=&\,p^2+q^2-2pq\left(\frac12\right)\\ 8=&\,p^2+q^2-pq. \end{align*}

Therefore

p2+q2−pq=8.\begin{align*} p^2+q^2-pq=8. \end{align*}

解法二

思路

展开

作高线构造直角三角形，使用勾股定理推导余弦定理的方法（Otherwise）。 我们可以从几何第一原理出发来证明该等式。 过顶点 BB 向对边 ACAC 作高线，设交点为 XX（即 BX⊥ACBX \perp AC）。 在直角三角形 ABXABX 中，由于角 ∠BAC=60∘\angle BAC = 60^\circ 且 AB=pAB = p，利用锐角三角函数可得：

AX=pcos⁡60∘=12p,BX=psin⁡60∘=32p.AX = p \cos 60^\circ = \frac{1}{2}p, \qquad BX = p \sin 60^\circ = \frac{\sqrt{3}}{2}p.

因为 AC=qAC = q，所以 XC=AC−AX=q−12pXC = AC - AX = q - \frac{1}{2}p。 在直角三角形 BCXBCX 中，利用勾股定理：

BC2=BX2+XC2\begin{align*} BC^2 = BX^2 + XC^2 \end{align*}

代入已知量展开并化简，即可直接得出 p2+q2−pq=8p^2 + q^2 - pq = 8。该方法直观展示了余弦定理的几何来源，逻辑严密且易于理解。

答题过程

展开

Draw a perpendicular line from BB to ACAC, meeting ACAC at point XX (so BX⊥ACBX \perp AC).

In the right-angled triangle ABXABX:

AX=  ABcos⁡60∘=p(12)=12p,BX=  ABsin⁡60∘=p(32)=32p.\begin{align*} AX =&\,\, AB \cos 60^\circ = p \left(\frac{1}{2}\right) = \frac{1}{2}p,\\[3mm] BX =&\,\, AB \sin 60^\circ = p \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}p. \end{align*}

Since the length of ACAC is qq, the length of XCXC is:

XC=AC−AX=q−12p.XC = AC - AX = q - \frac{1}{2}p.

\begin{align*} In the right-angled triangle BCXBCX, apply Pythagoras’ theorem: \end{align*}

BC2=BX2+XC2BC^2 = BX^2 + XC^2

\begin{align*} Substitute BC=22BC = 2\sqrt{2} and the expressions for BXBX and XCXC: \end{align*}

(22)2=  (32p)2+(q−12p)28=  34p2+q2−pq+14p28=  p2+q2−pq.\begin{align*} (2\sqrt{2})^2 =&\,\, \left(\frac{\sqrt{3}}{2}p\right)^2 + \left(q - \frac{1}{2}p\right)^2\\[3mm] 8 =&\,\, \frac{3}{4}p^2 + q^2 - pq + \frac{1}{4}p^2\\[3mm] 8 =&\,\, p^2 + q^2 - pq. \end{align*}

Therefore:

p2+q2−pq=8.p^2 + q^2 - pq = 8.

(b)

解法一

思路

展开

ACAC 比 ABAB 长 22，所以 q=p+2q=p+2。代入 (a) 的等式，得到关于 pp 的二次方程。长度必须为正，所以舍去负值。

答题过程

展开

Since ACAC is 22 metres longer than ABAB,

q=p+2.\begin{align*} q=p+2. \end{align*}

Substitute into p2+q2−pq=8p^2+q^2-pq=8:

p2+(p+2)2−p(p+2)= 8p2+p2+4p+4−p2−2p= 8p2+2p−4= 0.\begin{align*} p^2+(p+2)^2-p(p+2)=&\,8\\ p^2+p^2+4p+4-p^2-2p=&\,8\\ p^2+2p-4=&\,0. \end{align*}

Use the quadratic formula:

p= −2±22−4(1)(−4)2= −2±202= −1±5.\begin{align*} p =&\,\frac{-2\pm\sqrt{2^2-4(1)(-4)}}{2}\\ =&\,\frac{-2\pm\sqrt{20}}2\\ =&\,-1\pm\sqrt5. \end{align*}

Since pp is a length,

p=−1+5.\begin{align*} p=-1+\sqrt5. \end{align*}

Then

q=p+2=1+5.\begin{align*} q=p+2=1+\sqrt5. \end{align*}

(c)

解法一

思路

展开

三角形面积公式为 12absin⁡C\frac12ab\sin C。这里两边是 p,qp,q，夹角是 60∘60^\circ。

答题过程

展开 Area= 12pqsin⁡60∘= 12(−1+5)(1+5)⋅32.\begin{align*} \text{Area} =&\,\frac12pq\sin60^\circ\\ =&\,\frac12(-1+\sqrt5)(1+\sqrt5)\cdot\frac{\sqrt3}{2}. \end{align*}

Now

(−1+5)(1+5)= 5−1= 4.\begin{align*} (-1+\sqrt5)(1+\sqrt5) =&\,5-1\\ =&\,4. \end{align*}

Therefore

Area= 12⋅4⋅32= 3.\begin{align*} \text{Area} =&\,\frac12\cdot4\cdot\frac{\sqrt3}{2}\\ =&\,\sqrt3. \end{align*}