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# Difference between revisions of "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 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 $3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5 \Longrightarrow B}$.

### Solution 2

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

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); draw(P -- (0,1)); label("M",(-1,0),W); label("C",(-0.1,-0.3)); label("A",(-0.4,0.7)); label("B",(0.7,0.4)); label("P",(-1,1),NW); label("Q",(1,1),NE); label("R",(1,-1),SE); label("S",(-1,-1),SW); label("N",P,1.5*WNW); label("O",(0,1),N); [/asy]$

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 $\triangle NPMQ$ and $\triangle MQR$ in the original figure are similar.

## Solution 3 (Pythagorean Theorem only)

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); draw((-1,0)--(1,-1)); label("M",(-1,0),W); label("C",(-0.1,-0.3)); label("A",(-0.4,0.7)); label("B",(0.7,0.4)); label("P",(-1,1),NW); label("Q",(1,1),NE); label("R",(1,-1),SE); label("S",(-1,-1),SW); label("N",P,NW); label("x", (0.65, 0.7)); label("\sqrt{5} - x", (-0.3, 0.15)); [/asy]$

Since $PQ = SR = 2$ and $PM = MS = 1$, we know that $MQ = MR = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$. If we let $NQ = x$, then $MN = \sqrt{5} - x$. Now, by the Pythagorean Theorem, we have:

$$x^{2} + NR^{2} = 2^{2} = 4$$ $$(\sqrt{5} - x)^{2} + NR^{2} = (\sqrt{5})^{2} = 5$$

Expanding and rearranging the second equation gives:

$$5 - 2x\sqrt{5} + x^{2} + NR^{2} = 5$$ $$x^{2} + NR^{2} - 2x\sqrt{5} = 0$$ $$x^{2} + NR^{2} = 2x\sqrt{5}$$

Since $x^{2} + NR^{2} = 4$, we have that:

$$2x\sqrt{5} = 4$$ $$x\sqrt{5} = 2$$ $$x = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$

Knowing $x$, we can solve for the height $NR$:

$$NR^{2} = 2^{2} - x^{2} = 4 - ({\frac{2\sqrt{5}}{5}})^{2} = 4 - \frac{4}{5} = \frac{16}{5}$$ $$NR = \sqrt{\frac{16}{5}} = \frac{4}{\sqrt{5}} = \frac{4\sqrt{5}}{5}$$

Therefore, the area of triangle $RNQ$ is $\frac{1}{2} \cdot \frac{2\sqrt{5}}{5} \cdot\frac{4\sqrt{5}}{5} = \frac{1}{2} \cdot\frac{40}{25} = \frac{4}{5}$. Since the solution to the problem is $3[RNQ] + 4$, the answer is $3(\frac{4}{5}) + 4 = \frac{12}{5} + \frac{20}{5} = \boxed{(B) \frac{32}{5}}$.

## Solution 4

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

$MQ = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$

since $\angle PQM + \angle PMQ = 90 = \angle PQM + \angle NQR$

therefore $\angle PMQ = \angle NQR$

and since $\angle MPQ = \angle QNR = 90$

therefore $\triangle MPQ \sim \triangle QNR$

therefore $\frac {[QNR]}{[MPQ]} = (\frac{QR}{QM})^{2}, [QNR] = [MPQ] \cdot (\frac{QR}{QM})^{2}$

$[MPQ] = \frac{1}{2} \cdot 2 \cdot 1 = 1$

$[QNR] = 1 \cdot (\frac{2}{\sqrt{5}})^{2} = 1 \cdot \frac{4}{5} = \frac{4}{5}$

Since the solution to the problem is $3[QNR] + 4$, the answer is $3(\frac{4}{5}) + 4 = \frac{12}{5} + \frac{20}{5} = \boxed{(B) \frac{32}{5}}$.

## See Also

 2009 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 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 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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