3 problems found
The three points \(A\), \(B\) and \(C\) have coordinates \(\l p_1 \, , \; q_1 \r\), \(\l p_2 \, , \; q_2 \r\) and \(\l p_3 \, , \; q_3 \r\,\), respectively. Find the point of intersection of the line joining \(A\) to the midpoint of \(BC\), and the line joining~\(B\) to the midpoint of \(AC\). Verify that this point lies on the line joining \(C\) to the midpoint of~\(AB\). The point \(H\) has coordinates \(\l p_1 + p_2 + p_3 \, , \; q_1 + q_2 + q_3 \r\,\). Show that if the line \(AH\) intersects the line \(BC\) at right angles, then \(p_2^2 + q_2^2 = p_3^2 + q_3^2\,\), and write down a similar result if the line \(BH\) intersects the line \(AC\) at right angles. Deduce that if \(AH\) is perpendicular to \(BC\) and also \(BH\) is perpendicular to \(AC\), then \(CH\) is perpendicular to \(AB\).
Points \(\mathbf{A},\mathbf{B},\mathbf{C}\) in three dimensions have coordinate vectors \(\mathbf{a},\mathbf{b},\mathbf{c}\), respectively. Show that the lines joining the vertices of the triangle \(ABC\) to the mid-points of the opposite sides meet at a point \(R\). \(P\) is a point which is {\bf not} in the plane \(ABC\). Lines are drawn through the mid-points of \(BC\), \(CA\) and \(AB\) parallel to \(PA\), \(PB\) and \(PC\) respectively. Write down the vector equations of the lines and show by inspection that these lines meet at a common point \(Q\). Prove further that the line \(PQ\) meets the plane \(ABC\) at \(R\).
Find the equations of the tangent and normal to the parabola \(y^{2}=4ax\) at the point \((at^{2},2at).\) For \(i=1,2,\) and 3, let \(P_{i}\) be the point \((at_{i}^{2},2at_{i}),\) where \(t_{1},t_{2}\) and \(t_{3}\) are all distinct. Let \(A_{1}\) be the area of the triangle formed by the tangents at \(P_{1},P_{2}\) and \(P_{3},\) and let \(A_{2}\) be the area of the triangle formed by the normals at \(P_{1},P_{2}\) and \(P_{3}.\) Using the fact that the area of the triangle with vertices at \((x_{1},y_{1}),(x_{2},y_{2})\) and \((x_{3},y_{3})\) is the absolute value of \[ \tfrac{1}{2}\det\begin{pmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{pmatrix}, \] show that \(A_{3}=(t_{1}+t_{2}+t_{3})^{2}A_{1}.\) Deduce a necessary and sufficient condition in terms of \(t_{1},t_{2}\) and \(t_{3}\) for the normals at \(P_{1},P_{2}\) and \(P_{3}\) to be concurrent.
Solution: \(\frac{dy}{dt} = 2a, \frac{dx}{dt} = 2at \Rightarrow \frac{dy}{dx} = \frac{1}{t}\). Therefore the equation of the tangent will be \(\frac{y - 2at}{x-at^2} = \frac{1}{t} \Rightarrow y = \frac1tx +at\) and normal will be \(\frac{y-2at}{x-at^2} = -t \Rightarrow y = t(at^2-x+2a)\). The tangents will meet when: \begin{align*} && \begin{cases} t_iy -x &= at_i^2 \\ t_j y - x &= at_j^2 \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)y &= a(t_i-t_j)(t_i+t_j) \\ \Rightarrow && y &= a(t_i+t_j) \\ && x &= at_it_j \end{align*} The normals will meet when: \begin{align*} && \begin{cases} y +t_i x &= at_i^3+2at_i \\ y +t_j x &= at_j^3+2at_j \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)x &= a(t_i-t_j)(t_i^2+t_it_j+t_j^2+2) \\ \Rightarrow && x&= a(t_i^2+t_it_j+t_j^2+2) \\ && y &= -at_it_j(t_i+t_j) \end{align*} Therefore the area of our triangles will be: \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}at_1t_2 & a(t_1+t_2) & 1\\ at_2t_3 & a(t_2+t_3) & 1\\ at_3t_1 & a(t_3+t_1) & 1 \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2t_3 & (t_2+t_3) & 1\\ t_3t_1 & (t_3+t_1) & 1 \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2(t_3-t_1) & (t_3-t_1) & 0\\ t_1(t_3-t_2) & (t_3-t_2) & 0 \end{pmatrix} \\ &= \frac{a^2}{2}|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} and \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}a(t_1^2+t_1t_2+t_2^2+2) & -at_1t_2(t_1+t_2) & 1\\ a(t_2^2+t_2t_3+t_3^2+2) & -at_2t_3(t_2+t_3) & 1\\ a(t_3^2+t_3t_1+t_1^2+2) & -at_3t_1(t_3+t_1) & 1\\ \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_2^2+t_2t_3+t_3^2+2) & -t_2t_3(t_2+t_3) & 1\\ (t_3^2+t_3t_1+t_1^2+2) & -t_3t_1(t_3+t_1) & 1\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ t_3^2-t_1^2+t_2(t_3-t_1) & t_2(t_1^2+t_1t_2-t_2t_3-t_3^2) & 0\\ t_3^2-t_2^2+t_1(t_3-t_2) & t_1(t_2^2+t_2t_1-t_1t_3-t_3^2) & 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_3-t_1)(t_3+t_2+t_1) & t_2(t_1-t_3)(t_1+t_3+t_2) & 0\\ (t_3-t_2)(t_3+t_2+t_1) & t_1(t_2-t_3)(t_1+t_2+t_3)& 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}(t_1+t_2+t_3)^2|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} as required. The normals will be concurrent iff the area of their triangle is \(0\). This is certainly true if \(t_1+t_2+t_3 = 0\). In fact the only if is also true, since no \(3\) tangents can be concurrent.