Three balls are placed inside a cone such that each ball is in contact with the edge of the cone and the next ball.
If the radii of the balls are 20 cm, 12 cm, and r cm from top to bottom, what is the value of r?
\(\begin{array}{|llllll|} \hline &\mathbf{ \dfrac{x+r}{r} } &=& \mathbf{ \dfrac{x+2r+12}{12} } &=& \mathbf{ \dfrac{x+2r+2*12+20}{20} } \\\\ &\dfrac{x+r}{r} &=& \dfrac{x+2r+12}{12} &=& \dfrac{x+2r+44}{20} \\ \hline \end{array}\\ \begin{array}{|lrcl|} \hline 1. & \mathbf{ \dfrac{x+r}{r}} &=& \mathbf{ \dfrac{x+2r+12}{12} } \\\\ & 12(x+r) &=& r(x+2r+12) \\ & 12x+12r &=& rx+2r^2+12r \\ & 12x &=& rx+2r^2 \\ & 12x-rx &=& 2r^2 \\ & x(12-r) &=& 2r^2 \\ & \mathbf{ x } &=& \mathbf{ \dfrac{2r^2} {12-r} } \\ \hline \end{array} \begin{array}{|lrcl|} \hline 2. & \mathbf{ \dfrac{x+r}{r}} &=& \mathbf{ \dfrac{x+2r+44}{20} } \\\\ & 20(x+r) &=& r(x+2r+44) \\ & 20x+20r &=& rx+2r^2+44r \\ & 20x-rx &=& 2r^2+24r \\ & x(20-r) &=& 2r^2+24r \\ & \mathbf{ x } &=& \mathbf{ \dfrac{2r^2+24r } {20-r} } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{ x = \dfrac{2r^2} {12-r} } &=& \mathbf{ \dfrac{2r^2+24r } {20-r} } \\\\ \dfrac{2r^2} {12-r} &=& \dfrac{2r^2+24r } {20-r} \\\\ (2r^2) (20-r) &=& (12-r) (2r^2+24r) \\\\ 40r^2-2r^3 &=& 24r^2+12*24r-2r^3-24r^2 \\ 40r^2 &=& 12*24r \quad | \quad :r \\ 40r &=& 12*24 \quad | \quad :40 \\\\ r &=& \dfrac{12*24} {40} \\\\ \mathbf{ r } &=& \mathbf{7.2\ \text{cm}} \\ \hline \end{array}\)
