AMC10 2009 10B, Problem 22

A cubical  cake with edge length $2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $M$ is the midpoint of a top edge. The piece whose top is triangle $B$ contains $c$ cubic inches of cake and $s$ square inches of icing. What is $c+s$? [asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt));  draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4));  label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4)); [/asy]

Solution: Introduce a coordinate system where $Q=(0,0)$, $P=(2,0)$ and $R=(0,2)$. In this coordinate system we have $M=(2,1)$, and the line $QM$ has the equation $2y-x=0$. As the line $RN$ is orthogonal to $QM$, it must have the equation $y+2x+q=0$ for some suitable constant $q$. As this line contains the point $R=(0,2)$, we have $q=-2$. Substituting $x=2y$ into $y+2x-2=0$, we get $y=\frac 25$, and then $x=\frac 45$. We can note that in $\triangle RNQ$ $x$ is the height from $N$ onto $RQ$, hence its area is $[RNQ] = \frac{x \cdot RQ} 2 = \frac{2x}2 = x = \frac 45$, and therefore the answer is $3[RNQ]+4 = 3\cdot \frac 45 + 4 = \fbox{\frac{32}{5}}$.

Answer: \frac{32}{5}