2009 AMC 10B Problem 22

Put the top face on coordinates with corners $(0,0), (2,0), (2,2), (0,2)$ and $M = (0,1)$. The first cut runs from the corner $Q = (2,2)$ to $M$; the second runs from the corner $R = (2,0)$ perpendicular to that cut, meeting it at a point $P$. Triangle $B$ is $PQR$, the piece against the right edge $QR$.

Find $P$. The line $QM$ has direction $(-2, -1)$ and length $\sqrt5$. The distance from $R$ to that line is the height of triangle $MQR$ (base $QM$): its area is $\frac12 \cdot 2 \cdot 2 = 2$, so the height is $\frac{2 \cdot 2}{\sqrt5} = \frac{4}{\sqrt5}$. In right triangle $PQR$ with hypotenuse $QR = 2$ and leg $PR = \frac{4}{\sqrt5}$, the other leg is $PQ = \sqrt{4 - \frac{16}{5}} = \frac{2}{\sqrt5}$. Hence
$$[B] = \frac12 \cdot PQ \cdot PR = \frac12 \cdot \frac{2}{\sqrt5} \cdot \frac{4}{\sqrt5} = \frac45.$$

Because the cuts are vertical, the piece is a prism of height $2$ on base $B$: $c = 2 \cdot \frac45 = \frac85$.

Icing: the top of the piece (area $\frac45$) and the one outer side face along edge $QR$, a $2 \times 2$ square of area $4$. The two cut faces are bare. So $s = \frac45 + 4 = \frac{24}{5}$.

Therefore $c + s = \frac85 + \frac{24}{5} = \frac{32}{5}$.

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

Vertical cuts through a box create prisms, so a 3-D question collapses onto the top view: volume is base area times height, and side icing is edge length times height. The only real work is one planar area, found here through the altitude of a triangle with a known area.

Triangles $PQR$ and $QMR'$, where $R' = (0,2)$ is the top-left corner, are similar right triangles (they share the angle at $Q$). Triangle $QMR'$ has legs $1$ and $2$, hypotenuse $\sqrt5$; scaling to hypotenuse $QR = 2$ gives legs $\frac{2}{\sqrt5}$ and $\frac{4}{\sqrt5}$ and area $\frac45$ immediately.

• Icing the cut faces, or forgetting that the piece has an entire outer side face of area 4 iced in addition to its top.
• Reporting $s$ alone, $\frac{24}{5}$ (choice A), instead of $c + s$.