smc

Final Scenario For the narrow cone and the wide cone, let their base radii be $3x$ and $6y$ (for some $x,y>1$), respectively. By the similar triangles discussed above, their heights must be $h_{1}x$ and $h_{2}y,$ respectively. We have the following table: $$\begin{array}{cccccc}& \text{Base Radius}& \text{Height}& & \text{Volume}& \\ [2ex]\text{Narrow Cone}& 3x& h_{1}x& & \frac{1}{3}\pi (3x{)}^2(h_{1}x)=3\pi h_1{x}^3& \\ [2ex]\text{Wide Cone}& 6y& h_{2}y& & \frac{1}{3}\pi (6y{)}^2(h_{2}y)=12\pi h_2{y}^3& \end{array}$$ Recall that $\frac{h_1}{h_2}=4.$ Equating the volumes gives $3\pi h_1{x}^3=12\pi h_2{y}^3,$ which simplifies to $x^3=y^3,$ or $x=y.$ Finally, the requested ratio is $$\frac{h_{1}x-h_1}{h_{2}y-h_2}=\frac{h_1(x-1)}{h_2(y-1)}=\frac{h_1}{h_2}=\boxed{4:1}.$$ Remarks 1. This solution uses the following property of fractions: For unequal positive numbers $a,b,c$ and $d,$ if $\frac{a}{b}=\frac{c}{d}=k,$ then $\frac{a\pm c}{b\pm d}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.$ 2. This solution shows that, regardless of the shape or the volume of the solid dropped into each cone, the requested ratio stays the same as long as the solid sinks to the bottom and is completely submerged without spilling any liquid.