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

(New page: == Problem == A cubical cake with edge length <math>2</math> inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where <math>M</math...)
 
(→‎Solution)
Line 56: Line 56:
  
 
Let's label the points as in the picture above. Let <math>[RNQ]</math> be the area of <math>\triangle RNQ</math>. Then the volume of the corresponding piece is <math>c=2[RNQ]</math>. This cake piece has icing on the top and on the vertical side that contains the edge <math>QR</math>. Hence the total area with icing is <math>[RNQ]+2^2 = [RNQ]+4</math>. Thus the answer to our problem is <math>3[RNQ]+4</math>, and all we have to do now is to determine <math>[RNQ]</math>.
 
Let's label the points as in the picture above. Let <math>[RNQ]</math> be the area of <math>\triangle RNQ</math>. Then the volume of the corresponding piece is <math>c=2[RNQ]</math>. This cake piece has icing on the top and on the vertical side that contains the edge <math>QR</math>. Hence the total area with icing is <math>[RNQ]+2^2 = [RNQ]+4</math>. Thus the answer to our problem is <math>3[RNQ]+4</math>, and all we have to do now is to determine <math>[RNQ]</math>.
 +
 +
=== Solution 1 ===
  
 
Introduce a coordinate system where <math>Q=(0,0)</math>, <math>P=(2,0)</math> and <math>R=(0,2)</math>.  
 
Introduce a coordinate system where <math>Q=(0,0)</math>, <math>P=(2,0)</math> and <math>R=(0,2)</math>.  
Line 61: Line 63:
 
In this coordinate system we have <math>M=(2,1)</math>, and the line <math>QM</math> has the equation <math>2y-x=0</math>.  
 
In this coordinate system we have <math>M=(2,1)</math>, and the line <math>QM</math> has the equation <math>2y-x=0</math>.  
  
−
As the line <math>RN</math> is orthogonal to <math>QM</math>, it must have the equation <math>y+2x+c=0</math> for some suitable constant <math>c</math>. As this line contains the point <math>R=(0,2)</math>, we have <math>c=-2</math>.
+
As the line <math>RN</math> is orthogonal to <math>QM</math>, it must have the equation <math>y+2x+q=0</math> for some suitable constant <math>q</math>. As this line contains the point <math>R=(0,2)</math>, we have <math>q=-2</math>.
  
 
Substituting <math>x=2y</math> into <math>y+2x-2=0</math>, we get <math>y=\frac 25</math>, and then <math>x=\frac 45</math>.  
 
Substituting <math>x=2y</math> into <math>y+2x-2=0</math>, we get <math>y=\frac 25</math>, and then <math>x=\frac 45</math>.  
  
 
We can note that in <math>\triangle RNQ</math> <math>x</math> is the height from <math>N</math> onto <math>RQ</math>, hence its area is <math>[RNQ] = \frac{x \cdot RQ} 2 = \frac{2x}2 = x = \frac 45</math>, and therefore the answer is <math>3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5}</math>.
 
We can note that in <math>\triangle RNQ</math> <math>x</math> is the height from <math>N</math> onto <math>RQ</math>, hence its area is <math>[RNQ] = \frac{x \cdot RQ} 2 = \frac{2x}2 = x = \frac 45</math>, and therefore the answer is <math>3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5}</math>.
 +
 +
=== Solution 2 ===
 +
 +
Extend <math>RN</math> to intersect <math>PQ</math> at <math>O</math>:
 +
 +
<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 <math>O</math> is the midpoint of <math>PQ</math>. (Imagine rotating the square <math>PQRS</math> by <math>90^\circ</math> clockwise around its center. This rotation will map the segment <math>MQ</math> to a segment that is orthogonal to <math>MQ</math>, contains <math>R</math> and contains the midpoint of <math>PQ</math>.)
 +
 +
From <math>\triangle PQM</math> we can compute that <math>QM = \sqrt{1^2 + 2^2} = \sqrt 5</math>.
 +
 +
Observe that <math>\triangle PQM</math> and <math>\triangle NQO</math> have the same angles and therefore they are similar. The ratio of their sides is <math>\frac{QM}{OQ} = \frac{\sqrt 5}1 = \sqrt 5</math>.
 +
 +
Hence we have <math>ON = \frac{PM}{\sqrt 5} = \frac 1{\sqrt 5}</math>, and <math>NQ = \frac{PQ}{\sqrt 5} = \frac 2{\sqrt 5}</math>.
 +
 +
Knowing this, we can compute the area of <math>\triangle NQO</math> as <math>[NQO] = \frac{ON \cdot NQ}2 = \frac 15</math>.
 +
 +
Finally, we compute <math>[RNQ] = [ROQ] - [NQO] = 1 - \frac 15 = \frac 45</math>, and conclude that the answer is <math>3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2009|ab=B|num-b=21|num-a=23}}
 
{{AMC10 box|year=2009|ab=B|num-b=21|num-a=23}}

Revision as of 12:59, 10 March 2009

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

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

See Also

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