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2008 Paper 2 Q8
D: 1600.0 B: 1484.0
vectors angle bisector section formula position vectors scalar product geometric proof distance formula

The points \(A\) and \(B\) have position vectors \(\bf a\) and \(\bf b\), respectively, relative to the origin \(O\). The points \(A\), \(B\) and \(O\) are not collinear. The point \(P\) lies on \(AB\) between \(A\) and \(B\) such that \[ AP : PB = (1-\lambda):\lambda\,. \] Write down the position vector of \(P\) in terms of \(\bf a\), \(\bf b\) and \(\lambda\). Given that \(OP\) bisects \(\angle AOB\), determine \(\lambda\) in terms of \(a\) and \(b\), where \(a=\vert \bf a\vert\) and $b=\vert \mathbf{b}\vert\(. The point \)Q\( also lies on \)AB\( between \)A\( and \)B\(, and is such that \)AP=BQ$. Prove that $$OQ^2-OP^2=(b-a)^2\,.$$

Solution:

TikZ diagram
\(\mathbf{p} = \lambda \mathbf{a} +(1- \lambda) \mathbf{b}\) Applying the Sine rule, we can see that: \begin{align*} && \frac{OA}{\sin \angle APO} = \frac{AP}{\sin \angle AOP} \\ && \frac{OB}{\sin \angle BPO} = \frac{BP}{\sin \angle BOP} \\ \end{align*} But \(\angle AOP = \angle BOP = \frac12 \angle AOP\) (since \(OP\) bisects the angle) and \(\angle APO = 180^{\circ} -\angle BPO\), so their sines are also equal. Therefore \begin{align*} && \frac{a}{AP} &= \frac{b}{BP} \\ \Rightarrow && \frac{a}{b} &= \frac{1-\lambda}{\lambda} \\ \Rightarrow && \lambda &= \frac{b}{a+b} \end{align*} Note that \(\mathbf{p} = \frac{b\mathbf{a} + a \mathbf{b}}{a+b} = \mathbf{a} + \frac{a}{a+b}(\mathbf{b}-\mathbf{a})\) and \(\mathbf{q} = \mathbf{b} +\frac{a}{a+b}(\mathbf{a}-\mathbf{b}) = \frac{a\mathbf{a} +b \mathbf{b}}{a+b}\) Therefore \begin{align*} && OQ^2 &= \frac{1}{(a+b)^2} \left (a\mathbf{a} + b \mathbf{b} \right)\cdot \left (a\mathbf{a} + b \mathbf{b} \right) \\ &&&= \frac{a^4+b^4+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ &&OP^2 &= \frac{1}{(a+b)^2} \left (b\mathbf{a} + a \mathbf{b} \right)\cdot \left (b\mathbf{a} + a \mathbf{b} \right) \\ &&&= \frac{2a^2b^2+2ab\mathbf{a}\cdot\mathbf{b}}{(a+b)^2} \\ \\ && OQ^2 - OP^2 &= \frac{a^4+b^4-2a^2b^2}{(a+b)^2} \\ &&&= \frac{(a^2-b^2)^2}{(a+b)^2} \\ &&&= (a-b)^2 \end{align*}

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2005 Paper 1 Q6
D: 1500.0 B: 1490.2
coordinate geometry circle distance formula locus Apollonius circle simultaneous equations ratio of distances

  1. The point \(A\) has coordinates \(\l 5 \, , 16 \r\) and the point \(B\) has coordinates \(\l -4 \, , 4 \r\). The variable point \(P\) has coordinates \(\l x \, , y \r\,\) and moves on a path such that \(AP=2BP\). Show that the Cartesian equation of the path of \(P\) is \[ \displaystyle \l x+7 \r^2 + y^2 =100 \;. \]
  2. The point \(C\) has coordinates \(\l a \, , 0 \r\) and the point \(D\) has coordinates \(\l b \, , 0 \r\), where \(a\ne b\). The variable point \(Q\) moves on a path such that \[ QC = k \times QD\;, \] where \(k>1\,\). Given that the path of \(Q\) is the same as the path of \(P\), show that \[ \frac{a+7}{b+7}=\frac{a^2+51}{b^2+51}\;. \] Show further that \((a+7)(b+7)=100\,\).

Solution:

  1. Since \(AP = 2BP\) we also have \(|AP|^2 = 4|BP|^2\) ie \begin{align*} && (x-5)^2 + (y-16)^2 &= 4(x+4)^2 + 4(y-4)^2 \\ \Rightarrow && x^2 - 10x+25 + y^2 -32y + 256 &= 4x^2+32x+64+4y^2-32y+64 \\ \Rightarrow && 281 &= 3x^2+42x+3y^2+128\\ && 281 &= 3(x+7)^2-147+3y^2+128 \\ \Rightarrow && 300 &= 3(x+7)^2 + 3y^2 \\ && 100 &= (x+7)^2 + y^2 \end{align*}
  2. Since \(|QC|^2 = k^2 |QD|^2\), \begin{align*} && (x-a)^2 + y^2 &= k^2 (x-b)^2 + k^2y^2 \\ \Rightarrow && x^2-2ax+a^2 &= k^2x^2-2k^2bx+k^2b^2 + (k^2-1)y^2 \\ && a^2-k^2b^2 &= (k^2-1)x^2-2(k^2b-a)x + (k^2-1)y^2 \\ && a^2-k^2b^2&= (k^2-1)\left(x-\frac{k^2b-a}{k^2-1}\right)^2-(k^2-1)\left(\frac{k^2b-a}{k^2-1}\right)^2+(k^2-1)y^2 \\ && \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2&= \left(x-\frac{k^2b-a}{k^2-1}\right)^2+y^2 \\ \Rightarrow && -7 &= \frac{k^2b-a}{k^2-1} \tag{*} \\ && 100 &= \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2 \\ &&&= \frac{a^2-k^2b^2}{k^2-1}+7^2 \\ \Rightarrow && 51 &= \frac{a^2-k^2b^2}{k^2-1} \tag{**} \\ (*) \Rightarrow && k^2(b+7)&= a+7 \\ (**) \Rightarrow && k^2(51+b^2)&= a^2+51 \\ \Rightarrow && \frac{a^2+51}{b^2+51} &= \frac{a+7}{b+7} \\ \\ \Rightarrow && a^2b+51b+7a^2 &= ab^2+51a+7b^2 \\ && 0 &= ab(b-a)-51(b-a)+7(b-a)(b+a) \\ &&&= (b-a)(ab+7(b+a)-51) \\ &&&= (b-a)((a+7)(b+7)-100) \\ \Rightarrow && 100 &= (a+7)(b+7) \end{align*} Since \(a \neq b\)

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1997 Paper 1 Q5
D: 1500.0 B: 1484.0
vectors geometric proof quadrilateral midpoints distance formula scalar product invariant

Four rigid rods \(AB\), \(BC\), \(CD\) and \(DA\) are freely jointed together to form a quadrilateral in the plane. Show that if \(P\), \(Q\), \(R\), \(S\) are the mid-points of the sides \(AB\), \(BC\), \(CD\), \(DA\), respectively, then \[|AB|^{2}+|CD|^{2}+2|PR|^{2}=|AD|^{2}+|BC|^{2}+2|QS|^{2}.\] Deduce that \(|PR|^{2}-|QS|^{2}\) remains constant however the vertices move. (Here \(|PR|\) denotes the length of \(PR\).)

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