Problems

Filters
Clear Filters

17 problems found

1992 Paper 3 Q7
D: 1700.0 B: 1474.8
vectors geometric proof parallelogram concurrency collinearity position vectors

The points \(P\) and \(R\) lie on the sides \(AB\) and \(AD,\) respectively, of the parallelogram \(ABCD.\) The point \(Q\) is the fourth vertex of the parallelogram \(APQR.\) Prove that \(BR,CQ\) and \(DP\) meet in a point.

Solution: Let \(\overrightarrow{AX} = \mathbf{x}\) for all points, so: \begin{align*} \mathbf{p} &= p\mathbf{b}\\ \mathbf{r} &= r\mathbf{d}\\ \mathbf{q} &= \mathbf{p}+\mathbf{r} \\ &= p\mathbf{b} + r\mathbf{d} \end{align*} Therefore \begin{align*} BR: && \mathbf{b} + \lambda(\mathbf{r}-\mathbf{b}) \\ &&= (1-\lambda) \mathbf{b}+ \lambda r \mathbf{d} \\ CQ: && \mathbf{c} + \mu(\mathbf{q} - \mathbf{c}) \\ &&= \mathbf{b}+\mathbf{d} + \mu(p\mathbf{b}+r\mathbf{d} - (\mathbf{b}+\mathbf{d}) ) \\ &&= (1+\mu(p-1))\mathbf{b} + (1+\mu(r-1))\mathbf{d} \\ DP: && \mathbf{d} + \nu (\mathbf{p} - \mathbf{d}) \\ &&= \nu p\mathbf{b} +(1-\nu) \mathbf{d} \end{align*} So we need \(1-\nu = \lambda r, \nu p = 1-\lambda, \) so lets say \(1 = \nu + \lambda r, 1 = \lambda + \nu p \Rightarrow \lambda(pr-1) = p-1 \Rightarrow \lambda = \frac{p-1}{pr-1}\) so they intersect at \(\frac{rp-r}{pr-1} \mathbf{d} + \frac{pr-p}{pr-1}\mathbf{b}\). If we take \(\mu = -\frac{\lambda}{p-1} = 1-pr\) this is clearly also on \(CQ\) hence they all meet at a point

View
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 \(\mathbf{a}, 2\mathbf{a-b}, 2\mathbf{a-c}\).
    TikZ diagram

View