2007 AMC 10B Problem 23

The cut is parallel to the base, so the small pyramid on top is an exact shrunken copy of the original. Let the original height be $H$. The top pyramid's height is $H - 2$, because the cut is $2$ units above the base.

1. How areas shrink. If every length of a shape is multiplied by $k$, every area is multiplied by $k \times k = k^2$. For example, a square of side $2$ has area $4$, and shrinking its side to $1$ (so $k = \frac12$) gives area $1$, a quarter of the original.
2. Here the area is halved, so
$$k^2 = \frac12, \qquad k = \frac{1}{\sqrt2}.$$
3. The height is a length, so it also shrinks by $k$:
$$H - 2 = \frac{H}{\sqrt2}.$$
4. Multiply both sides by $\sqrt2$: $\sqrt2\,H - 2\sqrt2 = H$. So $(\sqrt2 - 1)H = 2\sqrt2$ and
$$H = \frac{2\sqrt2}{\sqrt2 - 1}.$$
5. Clear the square root from the bottom by multiplying top and bottom by $\sqrt2 + 1$. The bottom becomes $(\sqrt2)^2 - 1 = 1$:
$$H = 2\sqrt2(\sqrt2 + 1) = 4 + 2\sqrt2.$$
6. Check: $H \approx 6.83$, so $H - 2 \approx 4.83$, and $4.83 \div 6.83 \approx 0.707$, which is $\frac{1}{\sqrt2}$.

The answer is $\boxed{\textbf{(E)}\ 4+2\sqrt{2}}$.

For shapes of the same form, lengths scale by $k$, areas by $k^2$ and volumes by $k^3$. So "half the area" means lengths are multiplied by $\frac{1}{\sqrt2}$. The only number given, $2$, is the difference between the two heights, so write it as $H$ minus the smaller height.

Test the choices. The ratio $\frac{H - 2}{H}$ must be $\frac{1}{\sqrt2} \approx 0.707$.

• $H = 4$: ratio $\frac24 = 0.5$. No.
• $H = 2 + \sqrt2 \approx 3.41$: ratio $\approx \frac{1.41}{3.41} \approx 0.41$. No.
• $H = 4 + 2\sqrt2 \approx 6.83$: ratio $\approx \frac{4.83}{6.83} \approx 0.707$. Yes.