2009 AMC 10B Problems/Problem 22

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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 $c+s$ ?

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

Solution

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 $[RNQ]$ .

Solution 1

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

Solution 2

Extend $RN$ to intersect $PQ$ at $O$ :

It is now obvious that $O$ is the midpoint of $PQ$ . (Imagine rotating the square $PQRS$ by $90^\circ$ clockwise around its center. This rotation will map the segment $MQ$ to a segment that is orthogonal to $MQ$ , contains $R$ and contains the midpoint of $PQ$ .)

From $\triangle PQM$ we can compute that $QM = \sqrt{1^2 + 2^2} = \sqrt 5$ .

Observe that $\triangle PQM$ and $\triangle NQO$ have the same angles and therefore they are similar. The ratio of their sides is $\frac{QM}{OQ} = \frac{\sqrt 5}1 = \sqrt 5$ .

Hence we have $ON = \frac{PM}{\sqrt 5} = \frac 1{\sqrt 5}$ , and $NQ = \frac{PQ}{\sqrt 5} = \frac 2{\sqrt 5}$ .

Knowing this, we can compute the area of $\triangle NQO$ as $[NQO] = \frac{ON \cdot NQ}2 = \frac 15$ .

Finally, we compute $[RNQ] = [ROQ] - [NQO] = 1 - \frac 15 = \frac 45$ , and conclude that the answer is $3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5}$ .

You could also notice that the two triangles in the original figure are similar.

2009 AMC 10B (Problems • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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