2009 AMC 10B Problems/Problem 22

Problem

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$",(0.7,0.4)); [/asy]$

$\text{(A) } \frac{24}{5} \qquad \text{(B) } \frac{32}{5} \qquad \text{(C) } 8+\sqrt5 \qquad \text{(D) } 5+\frac{16\sqrt5}{5} \qquad \text{(E) } 10+5\sqrt5$

Solution

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4)); label("$P$",(-1,1),NW); label("$Q$",(1,1),NE); label("$R$",(1,-1),SE); label("$S$",(-1,-1),SW); label("$N$",P,NW); [/asy]$

Let's label the points as in the picture above. Let $[RNQ]$ be the area of $\triangle RNQ$ . Then the volume of the corresponding piece is $c=2[RNQ]$ . This cake piece has icing on the top and on the vertical side that contains the edge $QR$ . Hence the total area with icing is $[RNQ]+2^2 = [RNQ]+4$ . Thus the answer to our problem is $3[RNQ]+4$ , and all we have to do now is to determine $[RNQ]$ .

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 $y+2x+c=0$ for some suitable constant $c$ . As this line contains the point $R=(0,2)$ , we have $c=-2$ .

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 $3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5}$ .

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2009 AMC 10B Problems/Problem 22

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