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1994 Paper 3 Q14
D: 1700.0 B: 1516.0
probability geometric probability circle triangle inclusion-exclusion inequality uniform distribution

Three points, \(P,Q\) and \(R\), are independently randomly chosen on the perimeter of a circle. Prove that the probability that at least one of the angles of the triangle \(PQR\) will exceed \(k\pi\) is \(3(1-k)^{2}\) if \(\frac{1}{2}\leqslant k\leqslant1.\) Find the probability if \(\frac{1}{3}\leqslant k\leqslant\frac{1}{2}.\)

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1989 Paper 1 Q3
D: 1516.0 B: 1516.0
vectors scalar product geometric proof triangle perpendicular inequality optimisation intersection point

In the triangle \(OAB,\) \(\overrightarrow{OA}=\mathbf{a},\) \(\overrightarrow{OB}=\mathbf{b}\) and \(OA=OB=1\). Points \(C\) and \(D\) trisect \(AB\) (i.e. \(AC=CD=DB=\frac{1}{3}AB\)). \(X\) and \(Y\) lie on the line-segments \(OA\) and \(OB\) respectively, in such a way that \(CY\) and \(DX\) are perpendicular, and \(OX+OY=1\). Denoting \(OX\) by \(x\), obtain a condition relating \(x\) and \(\mathbf{a\cdot b}\), and prove that \[ \frac{8}{17}\leqslant\mathbf{a\cdot b}\leqslant1. \] If the angle \(AOB\) is as large as possible, determine the distance \(OE,\) where \(E\) is the point of intersection of \(CY\) and \(DX\).

Solution:

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Denoting \(\overrightarrow{OY}\) by \(\mathbf{y}\) and \(\overrightarrow{OC}\) by \(\mathbf{c}\) etc, we have: \begin{align*} \mathbf{c} &= \frac23 \mathbf{a} + \frac13 \mathbf{b} \\ \mathbf{d} &= \frac13 \mathbf{a} + \frac23 \mathbf{b} \\ \mathbf{x} &= \lambda \mathbf{a} \\ \mathbf{y} &= (1-\lambda) \mathbf{b} \\ 0 &= (\mathbf{d}-\mathbf{x}) \cdot (\mathbf{c} - \mathbf{y}) \\ &=((\frac13 -\lambda)\mathbf{a} + \frac23 \mathbf{b})\cdot(\frac23 \mathbf{a} + (\frac13-1+\lambda) \mathbf{b} ) \\ &= \frac{2}{3} \cdot (\frac13-\lambda) +\frac23 \cdot(\lambda - \frac23)+(\frac{4}{9}+(\frac13-\lambda)(-\frac23+\lambda))\mathbf{a}\cdot\mathbf{b} \\ &= -\frac{2}{9} + (\frac{4}{9} - \frac{2}{9}+\lambda-\lambda^2)\mathbf{a}\cdot \mathbf{b} \\ &= - \frac{2}{9} + (\frac{2}{9} + \lambda - \lambda^2)\mathbf{a}\cdot \mathbf{b} \\ \mathbf{a}\cdot \mathbf{b} &= \frac{2/9}{2/9+\lambda - \lambda^2} \end{align*} Since \(0 \leq \lambda - \lambda^2 \leq \frac14\), \(\frac{\frac29}{\frac29+\frac14} = \frac8{17} \leq \mathbf{a}\cdot\mathbf{b} \leq 1\) If \(\angle AOB\) is as large as possible, \(\mathbf{a}\cdot\mathbf{b}\) is as small as possible, ie \(\lambda = \frac12\) and \(\mathbf{a}\cdot \mathbf{b} = \frac{8}{17}\) First notice that the length \(OM\) to the midpoint of \(AB\) is \(\sqrt{\frac14 (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})} = \sqrt{\frac14 (2 + 2\mathbf{a}\cdot \mathbf{b})} = \sqrt{\frac12 + \frac4{17}} = \sqrt{\frac{25}{34}} = \frac{5}{\sqrt{34}}\) Notice that \(XYE\) and \(DCE\) are similar triangles, and so the heights satisfy \(\frac{h_1}{h_2} = \frac{\frac12}{\frac13} = \frac32\). Therefore the length \(OE\) is \(\frac12 \frac{5}{\sqrt{34}} + \frac{3}{5} \frac12 \frac{5}{\sqrt{34}} = \frac{8}{10} \frac{5}{\sqrt{34}} = \frac{4}{\sqrt{34}}\)

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1987 Paper 1 Q9
D: 1500.0 B: 1500.0
vectors position vectors barycentric coordinates triangle convex combination geometric proof locus

\(ABC\) is a triangle whose vertices have position vectors \(\mathbf{a,b,c}\)brespectively, relative to an origin in the plane \(ABC\). Show that an arbitrary point \(P\) on the segment \(AB\) has position vector \[ \rho\mathbf{a}+\sigma\mathbf{b}, \] where \(\rho\geqslant0\), \(\sigma\geqslant0\) and \(\rho+\sigma=1\). Give a similar expression for an arbitrary point on the segment \(PC\), and deduce that any point inside \(ABC\) has position vector \[ \lambda\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c}, \] where \(\lambda\geqslant0\), \(\mu\geqslant0\), \(\nu\geqslant0\) and \(\lambda+\mu+\nu=1\). Sketch the region of the plane in which the point \(\lambda\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c}\) lies in each of the following cases:

  1. \(\lambda+\mu+\nu=-1\), \(\lambda\leqslant0\), \(\mu\leqslant0\), \(\nu\leqslant0\);
  2. \(\lambda+\mu+\nu=1\), \(\mu\leqslant0\), \(\nu\leqslant0\).

Solution:

TikZ diagram
Suppose \(P\) is a fraction \(0 \leq k\leq 1\) along \(AB\), then \(\overrightarrow{OP} = \overrightarrow{OA} +\overrightarrow{AP} = \overrightarrow{OA} +k\overrightarrow{AB} = \mathbf{a} + k(\mathbf{b} - \mathbf{a}) = \lambda \mathbf{b} + (1-k) \mathbf{a}\), ie an arbitrary point on \(AB\) has the position vector where \((1-k) = \rho \geq 0\) and \(k= \sigma \geq 0\) and \((1-\lambda) + \lambda = 1\). Any point on the segment \(PC\) will be of the form \(l\mathbf{c} + (1-l) (k \mathbf{b} + (1-k) \mathbf{a})\) which has the form \(\lambda \mathbf{a} + \mu \mathbf{b} + \nu \mathbf{c}\) where \(\lambda + \mu + \nu = (1-l)(1-k) + (1-l)k + l = 1\) and all coefficients are positive.
  1. This is equivalent to the area inside the triangle where every point (\(\mathbf{a}, \mathbf{b}, \mathbf{c}\)) has been send to it's negative (\(-\mathbf{a}, -\mathbf{b}, -\mathbf{c}\)), ie
    TikZ diagram
  2. Writing points as: \begin{align*} \lambda\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c} &= (1-\mu - \nu)\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c} \\ &= \mathbf{a} + (-\mu)(\mathbf{a}-\mathbf{b}) + (-\nu)(\mathbf{a} - \mathbf{c}) \\ &= \mathbf{a} + (-\mu)(\mathbf{a}-\mathbf{b}) + (-\nu)(\mathbf{a} - \mathbf{c}) + (1+\mu+\nu)\mathbf{0}\\ \end{align*} So this is a translation of \(\mathbf{a}\) of the triangle with vertices at \(\mathbf{0}, \mathbf{a-b}, \mathbf{a-c}\), or a triangle with vertices \(\mathbf{a}, 2\mathbf{a-b}, 2\mathbf{a-c}\).
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