Difference between revisions of "2009 AMC 10B Problems/Problem 22"

Revision as of 22:47, 15 January 2020

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 $[RNQ]$ .

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 in the original figure are similar.

Solution 3

Use trigonometry.

The length of $PM$ and $PQ$ is $1$ and $2$ respectively. So $\angle PQM = \arctan \frac 12$ , and $\angle NQR = 90 - \angle PQM = 90 - \arctan \frac 12$ .

From the right-angled triangle $\triangle NQR$ , the hypotenuse, $QR = 2$ So $NR = 2 \sin (90 - \arctan \frac 12)$ , and $NQ = 2 \cos (90 - \arctan \frac 12)$

Knowing this, $[RNQ] = \frac 12 \cdot NR \cdot NQ$ . So we proceed as follows:

$[RNQ] = \frac 12 \cdot NR \cdot NQ$

$[RNQ] = \frac 12 \cdot 2 \sin (90 - \arctan \frac 12) \cdot 2 \cos (90 - \arctan \frac 12)$

$[RNQ] = 2 \sin (90 - \arctan \frac 12) \cos (90 - \arctan \frac 12)$

$[RNQ] = \sin [2(90 - \arctan \frac 12)] = \sin (2 \arctan \frac 12)$

$[RNQ] = \frac{2 \cdot \frac 12}{1+ (\frac 12)^2}$

$[RNQ] = \frac 45$

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

Note that we didn't use a calculator, but we used trigonometric identities

Solution 4 (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 4b

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); label("$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 $\angle PQM + \angle PMQ = 90 = \angle PQM + \angle MQR$

therefore $\angle PMQ = \angle MQR$

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

therefore $\triangle MPQ$ and $\triangle QNR$ are similar

therefore $\frac {QR}{QM} = \frac {NR}{PQ} = \frac {QN}{PM}$ $\frac {2}{\sqrt{5}} = \frac {NR}{2} = \frac {QN}{1}$

$NR = \frac{4}{\sqrt{5}} = \frac{4\sqrt{5}}{5}$

$QN = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$

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 $3(\frac{4}{5}) + 4 = \frac{12}{5} + \frac{20}{5} = \boxed{(B) \frac{32}{5}}$ .

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