2 problems found
Let \(A,B,C\) be three non-collinear points in the plane. Explain briefly why it is possible to choose an origin equidistant from the three points. Let \(O\) be such an origin, let \(G\) be the centroid of the triangle \(ABC,\) let \(Q\) be a point such that \(\overrightarrow{GQ}=2\overrightarrow{OG},\) and let \(N\) be the midpoint of \(OQ.\)
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