# 2007 AMC 10B Problems/Problem 23

## Problem

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? $\textbf{(A) } 2 \qquad\textbf{(B) } 2+\sqrt{2} \qquad\textbf{(C) } 1+2\sqrt{2} \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 4+2\sqrt{2}$

## 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. $$\frac{a}{a-2}=\frac{\sqrt{2}}{1} \longrightarrow a= a\sqrt{2} - 2\sqrt{2} \longrightarrow a\sqrt{2}-a=2\sqrt{2}$$ $$a=\frac{2\sqrt{2}}{\sqrt{2}-1} = \frac{4+2\sqrt{2}}{2-1} = \boxed{\mathrm{(E) \ } 4+2\sqrt{2}}$$

## Solution 2 (Playing with the answers)

Instead of actually solving this problem, we can play with the answers. The ratios of the altitudes squared is the ratio of the surface areas, so the answer choice has to be larger than 4 because a original height of 4 would give the ratio 1:4 for the surface area and only choice $E$ is larger than 4, so the answer is $E$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 