2009 AMC 10B Problem 22

The cuts are vertical, so every piece is a straight-sided column standing on its top-view shape. Everything comes down to the area of triangle $B$.

Step 1: set up the top view.

1. Put the top of the cake on a grid with corners $(0,0)$, $(2,0)$, $(2,2)$, $(0,2)$. The midpoint $M$ is $(0,1)$ on the left edge.
2. The first cut runs from the top-right corner $Q = (2,2)$ to $M$.
3. The second cut runs from the bottom-right corner $R = (2,0)$ and meets the first cut at a right angle, at a point $P$.
4. Triangle $B$ is $PQR$. It has a right angle at $P$, and its long side is the right edge $QR = 2$.

Step 2: find the area of $B$ using matching shapes.

5. Mark the point $N = (2, 1)$ on the right edge, level with $M$. Triangle $QNM$ is a right triangle (right angle at $N$) with legs $QN = 1$ and $NM = 2$, so its long side is $QM = \sqrt{1^2 + 2^2} = \sqrt5$.
6. Triangles $QNM$ and $QPR$ both have a right angle, and they share the angle at $Q$ (both use the right edge and the cut $QM$). So they have the same shape. In $QNM$ the leg next to $Q$ is $1$ and the leg across from $Q$ is $2$. Scaling the long side from $\sqrt5$ up to $QR = 2$ multiplies every side by $\frac{2}{\sqrt5}$:
$$PQ = 1 \times \tfrac{2}{\sqrt5} = \tfrac{2}{\sqrt5}, \qquad PR = 2 \times \tfrac{2}{\sqrt5} = \tfrac{4}{\sqrt5}.$$
7. Check with the Pythagorean theorem: $\frac45 + \frac{16}{5} = 4 = QR^2$. Good.
8. Area of $B$: $\frac12 \times \frac{2}{\sqrt5} \times \frac{4}{\sqrt5} = \frac12 \times \frac85 = \frac45$.

Step 3: cake and icing.

1. Cake: the piece is a column of height $2$ on base $B$, so $c = \frac45 \times 2 = \frac85$.
2. Icing on top: the top of the piece is triangle $B$, area $\frac45$.
3. Icing on the sides: the only outside side of this piece is the face below edge $QR$, a $2 \times 2$ square with area $4$. The other two sides are cut faces, and cut faces have no icing.
4. So $s = \frac45 + 4 = \frac{24}{5}$.
5. Total: $c + s = \frac85 + \frac{24}{5} = \frac{32}{5}$.

The answer is $\boxed{\textbf{(B)}\ \frac{32}{5}}$.

Vertical cuts turn a 3-D question into a 2-D one: volume is the top-view area times the height, and side icing is the outer edge length times the height. The only real work is one flat area, and a right triangle that shares an angle with a simple $1$-$2$-$\sqrt5$ triangle gives it quickly.

Find $PR$ as a height instead. Triangle $MQR$ has base $QR = 2$ on the right edge and its third corner $M$ is $2$ away from that edge, so its area is $\frac12 \times 2 \times 2 = 2$. Using $QM = \sqrt5$ as the base instead, the height is $PR$ (because $RP$ meets $QM$ at a right angle), so $\frac12 \times \sqrt5 \times PR = 2$ and $PR = \frac{4}{\sqrt5}$. Then $PQ = \sqrt{4 - \frac{16}{5}} = \frac{2}{\sqrt5}$, and the area of $B$ is again $\frac45$.