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IAL 2025 Oct A Q1

A Level / Edexcel / P1

IAL 2025 Oct A Paper · Question 1

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 a sketch of a triangle LMNLMN and a sketch of a triangle PQRPQR.

Figure 1

Triangle LMNLMN is an equilateral triangle with sides (x+7)(x+7) cm.

Triangle PQRPQR is a right-angled isosceles triangle with PQ=PR=2xPQ=PR=2x cm and angle QPR=90QPR=90^\circ.

(a) Find, in its simplest form in terms of xx, the length QRQR.

(2)

Given that the perimeter of triangle LMNLMN is equal to the perimeter of triangle PQRPQR,

(b) find the value of xx, writing your answer in the form a2+ba\sqrt2+b, where aa and bb are integers.

(5)

解答

(a)

解法一

思路

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PQRPQR 是直角等腰三角形,QRQR 是斜边。用勾股定理即可。

答题过程

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By Pythagoras’ theorem,

QR2=(2x)2+(2x)2=4x2+4x2=8x2.\begin{align*} QR^2=&\,(2x)^2+(2x)^2\\ =&\,4x^2+4x^2\\ =&\,8x^2. \end{align*}

Since lengths are positive,

QR=8x2=22x.\begin{align*} QR=&\,\sqrt{8x^2}\\ =&\,2\sqrt2x. \end{align*}

(b)

解法一

思路

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先写两个三角形的周长并令它们相等。整理后会得到 x=211+22x=\dfrac{21}{1+2\sqrt2},再把分母理性化,化成题目要求的 a2+ba\sqrt2+b 形式。

答题过程

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The perimeter of triangle LMNLMN is

3(x+7).\begin{align*} 3(x+7). \end{align*}

The perimeter of triangle PQRPQR is

2x+2x+22x=4x+22x.\begin{align*} 2x+2x+2\sqrt2x =4x+2\sqrt2x. \end{align*}

Equating the perimeters,

3(x+7)=4x+22x3x+21=x(4+22)21=x(1+22).\begin{align*} 3(x+7)=&\,4x+2\sqrt2x\\ 3x+21=&\,x(4+2\sqrt2)\\ 21=&\,x(1+2\sqrt2). \end{align*}

So

x=211+22=211+22×221221=21(221)(22)21=21(221)7=3(221)=623.\begin{align*} x=&\,\frac{21}{1+2\sqrt2}\\ =&\,\frac{21}{1+2\sqrt2} \times\frac{2\sqrt2-1}{2\sqrt2-1}\\ =&\,\frac{21(2\sqrt2-1)}{(2\sqrt2)^2-1}\\ =&\,\frac{21(2\sqrt2-1)}{7}\\ =&\,3(2\sqrt2-1)\\ =&\,6\sqrt2-3. \end{align*}

解法二

思路

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也可以从周长方程出发,把含根式的一项单独留在一边,然后平方消去根式。由于 xx 是长度参数,最后只取正值。

答题过程

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From the perimeter equation,

3(x+7)=4x+22x3x+21=4x+22x21x=22x.\begin{align*} 3(x+7)=&\,4x+2\sqrt2x\\ 3x+21=&\,4x+2\sqrt2x\\ 21-x=&\,2\sqrt2x. \end{align*}

Square both sides:

(21x)2=(22x)244142x+x2=8x27x2+42x441=0.\begin{align*} (21-x)^2=&\,(2\sqrt2x)^2\\ 441-42x+x^2=&\,8x^2\\ 7x^2+42x-441=&\,0. \end{align*}

Divide by 77:

x2+6x63=0.\begin{align*} x^2+6x-63=0. \end{align*}

Using the quadratic formula,

x=6±624(1)(63)2=6±2882=6±1222=3±62.\begin{align*} x=&\,\frac{-6\pm\sqrt{6^2-4(1)(-63)}}{2}\\ =&\,\frac{-6\pm\sqrt{288}}{2}\\ =&\,\frac{-6\pm12\sqrt2}{2}\\ =&\,-3\pm6\sqrt2. \end{align*}

Since x>0x>0,

x=623.\begin{align*} x=6\sqrt2-3. \end{align*}