2009 AMC 10B Problems/Problem 22

Revision as of 10:07, 20 May 2021 by Whitelisted (talk | contribs) (Solution 3)

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(best solution)

Look at the person beside you's paper. Just cheat, man. Google aops amc 10 2009 b click on link go to problem easy

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 5

[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 (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions

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

Invalid username
Login to AoPS