Difference between revisions of "2009 AMC 10B Problems/Problem 22"
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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 (Pythagorean Theorem only)== | ==Solution 3 (Pythagorean Theorem only)== | ||
Line 202: | Line 202: | ||
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>. | 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 | ||
== See Also == | == See Also == |
Latest revision as of 07:23, 1 July 2022
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
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 |
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