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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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 | + | ==Solution 3 (Pythagorean Theorem only)== |
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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== |
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Latest revision as of 13:09, 2 July 2021
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 .
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.