Difference between revisions of "2009 AMC 10B Problems/Problem 22"
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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 | + | 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 determine <math>[RNQ]</math>. |
=== Solution 1 === | === Solution 1 === | ||
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</asy> | </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>. | + | 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>. | From <math>\triangle PQM</math> we can compute that <math>QM = \sqrt{1^2 + 2^2} = \sqrt 5</math>. | ||
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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> | 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> | ||
− | *You could also notice that the two triangles in the original figure are similar. | + | *You could also notice that the two triangles <math>\triangle PMQ</math> and <math>\triangle NQR</math> in the original figure are similar. |
− | === Solution 3 | + | ===Solution 3 (Pythagorean Theorem only)=== |
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− | Since <math>PQ = SR = 2</math> and <math>PM = MS = 1</math>, we know that <math>MQ = MR = \sqrt{2^{ | + | Since <math>PQ = SR = 2</math> and <math>PM = MS = 1</math>, we know that <math>MQ = MR = \sqrt{2^{2} + 1^{2}} = \sqrt{5}</math>. If we let <math>NQ = x</math>, then <math>MN = \sqrt{5} - x</math>. Now, by the Pythagorean Theorem, we have: |
<cmath>x^{2} + NR^{2} = 2^{2} = 4</cmath> | <cmath>x^{2} + NR^{2} = 2^{2} = 4</cmath> | ||
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Therefore, the area of triangle <math>RNQ</math> is <math>\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}</math>. Since the solution to the problem is <math>3[RNQ] + 4</math>, the answer is <math>3(\frac{4}{5}) + 4 = \frac{12}{5} + \frac{20}{5} = \boxed{(B) \frac{32}{5}}</math>. | Therefore, the area of triangle <math>RNQ</math> is <math>\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}</math>. Since the solution to the problem is <math>3[RNQ] + 4</math>, the answer is <math>3(\frac{4}{5}) + 4 = \frac{12}{5} + \frac{20}{5} = \boxed{(B) \frac{32}{5}}</math>. | ||
− | ==Solution | + | ===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> | ||
+ | |||
+ | <math>MQ = \sqrt{2^{2} + 1^{2}} = \sqrt{5}</math> | ||
− | + | since <math>\angle PQM + \angle PMQ = 90 = \angle PQM + \angle NQR</math> | |
− | + | therefore <math>\angle PMQ = \angle NQR</math> | |
− | + | and since <math>\angle MPQ = \angle QNR = 90</math> | |
− | + | therefore <math>\triangle MPQ \sim \triangle QNR</math> | |
− | therefore <math>\ | + | therefore <math>\frac {[QNR]}{[MPQ]} = (\frac{QR}{QM})^{2}, [QNR] = [MPQ] \cdot (\frac{QR}{QM})^{2}</math> |
− | + | <math>[MPQ] = \frac{1}{2} \cdot 2 \cdot 1 = 1</math> | |
− | \frac {2}{\sqrt{5} | + | <math>[QNR] = 1 \cdot (\frac{2}{\sqrt{5}})^{2} = 1 \cdot \frac{4}{5} = \frac{4}{5}</math> |
− | + | Since the solution to the problem is <math>3[QNR] + 4</math>, the answer is <math>3(\frac{4}{5}) + 4 = \frac{12}{5} + \frac{20}{5} = \boxed{(B) \frac{32}{5}}</math>. | |
+ | |||
+ | |||
+ | |||
+ | ===Solution 5 (Similarity)=== | ||
+ | <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> | ||
+ | |||
+ | All units of length in the following solution are in inches, or inches squared, or inches cubed. Units of angles are in degrees. | ||
+ | |||
+ | |||
+ | <math>PQ = 2</math>. Since <math>M</math> is the midpoint of <math>\overline{SP}</math> which measures <math>2</math>, <math>MP = 1</math>. | ||
+ | |||
+ | |||
+ | Since angle MNR is right, angle QNR is also right. Let <math>m\angle PQM = x</math>. Then <math>m\angle PMQ = 90 - x</math>. Notice also since <math>\angle PQR</math> is right, <math>m \angle NQR = 90 - x</math>. Since <math>\angle QNR</math> is right, <math>m\angle QRN = x</math>. Therefore, <math>\triangle PQM \sim \triangle NRQ</math>. | ||
+ | |||
+ | |||
+ | Let <math>QN = a</math>. By the Pythagorean theorem, <math>MQ = \sqrt{5}</math>. By similarity, <math>\frac{PM}{MQ} = \frac{NQ}{QR} \longrightarrow \frac{1}{\sqrt{5}} = \frac{a}{2}</math>, so <math>a=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}</math>. By the Pythagorean theorem, <math>NR^2+NQ^2=QR^2</math>. Substituting known values in and solving for <math>NR</math>, we get <math>NR=\frac{4\sqrt{5}}{5}</math>. (Alternatively, use the fact that <math>\triangle PQM \sim \triangle NRQ</math>). Since <math>\triangle NQR</math> is a right triangle, the area is just <math>NQ \cdot NR \cdot \frac{1}{2}</math> which, substituting values, is equal to <math>\frac{4}{5}</math>. But remember that <math>s</math> also consists of the side of the cake, so we have to add <math>2^2=4</math>. So <math>s=\frac{4}{5}+4=\frac{24}{5}</math>. | ||
+ | |||
+ | |||
+ | Meanwhile, <math>c</math> is the volume of the slice (a triangular prism) which is found by the base area times height. We already calculated the base area to be <math>\frac{4}{5}</math>, so simply multiply by <math>2</math> to get the volume <math>=\frac{8}{5}</math>. This is the value of <math>c</math>. | ||
+ | |||
+ | Sum <math>c+s</math>: <math>c+s=\frac{8}{5}+\frac{24}{5}=\frac{32}{5} \Longrightarrow \boxed{\textbf{(B) } \frac{32}{5}}</math>. | ||
+ | |||
+ | |||
+ | ~JH. L | ||
+ | |||
+ | ===Solution 6 (only Pythagorean Theorem, no algebra)=== | ||
+ | Label the vertices of the square, <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> and draw line segment <math>MA</math> (as shown below): | ||
+ | <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((1, 1)--(1,0)); | ||
+ | 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); | ||
+ | </asy> | ||
− | + | <math>PM=MS=1</math> and <math>SR=2</math>, so by the pythagorean theorem, <cmath>MR = \sqrt{1^2+2^2} = \sqrt{5}.</cmath> By the same logic, <math>MQ=\sqrt{5}</math>. The area of <math>\triangle QMR</math> is the area of the whole square, minus the combined areas of <math>\triangle PMQ</math> and <math>\triangle MSR</math>, so <cmath>[QMR] = 4-1-1=2.</cmath> Since <math>QM</math> is the base of <math>\triangle QMR</math> and <math>RN</math> is the height, <cmath>\frac{QM(RN)}{2} = 2</cmath> <cmath>\frac{\sqrt5(RN)}{2} = 2,</cmath> so <math>RN=\frac{3\sqrt5}{5}</math>. We also know that <cmath>QN = QM-MN = \sqrt{5} - \frac{3\sqrt5}{5} = \frac{2\sqrt5}{5}.</cmath> Now, we can find the area of <math>\triangle QNR</math>. <cmath>[QNR] = \frac{1}{2}(QN)(RN) = \frac{1}{2}(\frac{4\sqrt5}{5})(\frac{2\sqrt5}{5}) = \frac{4}{5}.</cmath> The area of icing is the area of <math>\triangle QNR</math>, plus the area of the 2x2 square on <math>QR</math>, so <cmath>s=\frac{4}{5}+4 =\frac{4}{5} + \frac{20}{5} = \frac{24}{5}.</cmath> The cubic inches of cake is the volume of the piece, which is the area of <math>\triangle QNR</math> times <math>2</math> (the height of the cake), so <math>c = \frac{8}{5}</math>. Hence, <math>c+s = \frac{32}{5}</math>, and the answer is <math>\boxed{\textbf{(B) } \frac{32}{5}}</math> ~azc1027 | |
== See Also == | == See Also == |
Latest revision as of 03:36, 30 September 2023
Contents
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 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
- You could also notice that the two triangles
and
in the original figure are similar.
Solution 3 (Pythagorean Theorem only)
Since and
, we know that
. If we let
, then
. Now, by the Pythagorean Theorem, we have:
Expanding and rearranging the second equation gives:
Since , we have that:
Knowing , we can solve for the height
:
Therefore, the area of triangle is
. Since the solution to the problem is
, the answer is
.
Solution 4
since
therefore
and since
therefore
therefore
Since the solution to the problem is , the answer is
.
Solution 5 (Similarity)
All units of length in the following solution are in inches, or inches squared, or inches cubed. Units of angles are in degrees.
. Since
is the midpoint of
which measures
,
.
Since angle MNR is right, angle QNR is also right. Let . Then
. Notice also since
is right,
. Since
is right,
. Therefore,
.
Let . By the Pythagorean theorem,
. By similarity,
, so
. By the Pythagorean theorem,
. Substituting known values in and solving for
, we get
. (Alternatively, use the fact that
). Since
is a right triangle, the area is just
which, substituting values, is equal to
. But remember that
also consists of the side of the cake, so we have to add
. So
.
Meanwhile, is the volume of the slice (a triangular prism) which is found by the base area times height. We already calculated the base area to be
, so simply multiply by
to get the volume
. This is the value of
.
Sum :
.
~JH. L
Solution 6 (only Pythagorean Theorem, no algebra)
Label the vertices of the square, ,
,
, and
and draw line segment
(as shown below):
and
, so by the pythagorean theorem,
By the same logic,
. The area of
is the area of the whole square, minus the combined areas of
and
, so
Since
is the base of
and
is the height,
so
. We also know that
Now, we can find the area of
.
The area of icing is the area of
, plus the area of the 2x2 square on
, so
The cubic inches of cake is the volume of the piece, which is the area of
times
(the height of the cake), so
. Hence,
, and the answer is
~azc1027
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.