Problems

Filters
Clear Filters

1 problem found

2016 Paper 1 Q5
D: 1484.0 B: 1516.0
coordinate geometry circles tangent Pythagorean theorem algebraic identity geometric proof radii relationship

  1. TikZ diagram
    The diagram shows three touching circles \(A\), \(B\) and \(C\), with a common tangent \(PQR\). The radii of the circles are \(a\), \(b\) and \(c\), respectively. Show that \[ \frac 1 {\sqrt b} = \frac 1 {\sqrt{a}} + \frac1{\sqrt{c}} \tag{\(*\)} \] and deduce that \[ 2\left(\frac1{a^2} + \frac1 {b^2} + \frac1 {c^2} \right) = \left(\frac1 a + \frac1 {b} + \frac1 {c} \right)^{\!2} . \tag{\(**\)} \]
  2. Instead, let \(a\), \(b\) and \(c\) be positive numbers, with \(b < c < a\), which satisfy \((**)\). Show that they also satisfy \((*)\).

Solution:

  1. \(\,\)
    TikZ diagram
    Notice that \begin{align*} && (a+b)^2 &= PQ^2 + (a-b)^2 \\ \Rightarrow && PQ^2 &= 4ab \\ && (b+c)^2 &= QR^2 + (c-b)^2 \\ \Rightarrow && QR^2 &= 4bc \\ && (a+c)^2 &= PR^2 + (a-c)^2 \\ \Rightarrow && PR^2 &= 4ac \\ \Rightarrow && 2\sqrt{ac} &= 2\sqrt{ab}+2\sqrt{bc} \\ \Rightarrow && \frac{1}{\sqrt{b}} &= \frac{1}{\sqrt{c}} + \frac1{\sqrt{a}} \\ \end{align*} Let \(x, y, z = \frac{1}{\sqrt{a}}, \frac1{\sqrt{b}}, \frac{1}{\sqrt{z}}\) so we would like to prove that \(2(x^4+y^4+z^4) = (x^2+y^2+z^2)^2\) or \(x^4+y^4+z^4 = 2x^2y^2+2y^2z^2+2z^2x^2\). We also have \begin{align*} && y &= x+z \\ \Rightarrow &&y^2 &= x^2+z^2+2xz \\ \Rightarrow && (y^2-x^2-z^2)^2 &= 4x^2z^2 \\ \Rightarrow && y^4+x^4+z^4 - 2x^2y^2-2y^2z^2+2x^2z^2 &= 4x^2z^2\\ \Rightarrow && y^4+x^4+z^4 &= 2x^2y^2+2y^2z^2+2z^2x^2 \end{align*}
  2. Notice that subject to \(y > z > x\) all these steps are reversible, so we must have the equality we desire

View