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...) |
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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+ | + | 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 13:59, 10 March 2009
Problem
A cubical cake with edge length inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where
is the midpoint of a top edge. The piece whose top is triangle
contains
cubic inches of cake and
square inches of icing. What is
?
Solution
Let's label the points as in the picture above. Let be the area of
. Then the volume of the corresponding piece is
. This cake piece has icing on the top and on the vertical side that contains the edge
. Hence the total area with icing is
. Thus the answer to our problem is
, and all we have to do now is to determine
.
Solution 1
Introduce a coordinate system where ,
and
.
In this coordinate system we have , and the line
has the equation
.
As the line is orthogonal to
, it must have the equation
for some suitable constant
. As this line contains the point
, we have
.
Substituting into
, we get
, and then
.
We can note that in
is the height from
onto
, hence its area is
, and therefore the answer is
.
Solution 2
Extend to intersect
at
:
It is now obvious that is the midpoint of
. (Imagine rotating the square
by
clockwise around its center. This rotation will map the segment
to a segment that is orthogonal to
, contains
and contains the midpoint of
.)
From we can compute that
.
Observe that and
have the same angles and therefore they are similar. The ratio of their sides is
.
Hence we have , and
.
Knowing this, we can compute the area of as
.
Finally, we compute , and conclude that the answer is
.
See Also
2009 AMC 10B (Problems • Answer Key • 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 | ||
All AMC 10 Problems and Solutions |