AMC10 2007 10B, Problem 23

A pyramid with a square base is cut by a plane that is parallel to its base and $2$ units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?

Solution: Since the two pyramids are similar, the ratio of the altitudes is the square root of the ratio of the surface areas. If $a$ is the altitude of the larger pyramid, then $a-2$ is the altitude of the smaller pyramid. \[\frac{a}{a-2}=\sqrt{\frac21}=\frac{\sqrt{2}}{1} \longrightarrow a= a\sqrt{2} - 2\sqrt{2} \longrightarrow a\sqrt{2}-a=2\sqrt{2}\] \[a=\frac{2\sqrt{2}}{\sqrt{2}-1} = \frac{4+2\sqrt{2}}{2-1} = \fbox{4+2\sqrt{2}}.\]

Answer: 4+2\sqrt{2}