Difference between revisions of "2018 AMC 10B Problems/Problem 10"
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==Solution 2== | ==Solution 2== | ||
− | We | + | We start by finding side <math>\overline{BE}</math> of base <math>BCHE</math> by using the Pythagorean theorem on <math>\triangle ABE</math>. Doing this, we get |
− | + | <cmath>\overline{BE}^2 = \overline{AB}^2 + \overline{AE}^2 = 3^2 + 2^2 = 11.</cmath> | |
− | The original wedge we considered in the last step has volume <math>3< | + | Taking the square root of both sides of the equation, we get <math>\overline{BE} = \sqrt {11}</math>. We can then find the area of rectangle <math>BCHE</math>, noting that |
+ | |||
+ | <cmath>[BCHE] = \overline{BE} \cdot \overline{BC} = \sqrt {11} \cdot 1 = \sqrt {11}.</cmath> | ||
+ | |||
+ | |||
+ | Taking the vertical cross-sectional plane of the rectangular prism, we see that the distance from point <math>M</math> to base <math>BCHE</math> is the same as the distance from point <math>F</math> to side <math>\overline{BE}</math>. Calling the point where the altitude from vertex <math>F</math> touches side <math>\overline{BE}</math> as point <math>K</math>, we can easily find this altitude using the area of right <math>\triangle BFE</math>, as | ||
+ | |||
+ | <cmath>\frac{\overline{BF} \cdot \overline{FE}}{2} = \frac{\overline{BE} \cdot FK}{2}.</cmath> | ||
+ | |||
+ | Multiplying both sides of the equation by 2 and substituting in known values, we get | ||
+ | |||
+ | <math></math>2 \cdot 3 = FK \cdot \sqrt {11} \Rightarrow FK = \frac{6\sqrt {11}{11}.<math> | ||
+ | |||
+ | Deducing that the altitude from vertex </math>M<math> to base </math>BCHE<math> is </math>\frac{6\sqrt {11}{11}<math> and calling the point of intersection between the altitude and the base as point </math>N<math>, we get the area of the rectangular pyramid to be | ||
+ | |||
+ | <cmath>\frac{1}{3}([BCHE] \cdot MN) = \frac{1}{3}\left(\sqrt {11} \cdot \frac{6\sqrt {11}{11}\right) = \frac{66}{33} = \boxed{2}.</cmath> | ||
+ | |||
+ | Written by: Adharshk | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | We can start by finding the total volume of the parallelepiped. It is </math>2 \cdot 3 \cdot 1 = 6<math>, because a rectangular parallelepiped is a rectangular prism. | ||
+ | |||
+ | Next, we can consider the wedge-shaped section made when the plane </math>BCHE<math> cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is </math>\frac{1}{2} \cdot 2 \cdot 3 = 3<math>. Since BC is given to be </math>1<math>, we have that FM is </math>\frac{1}{2}<math>. Using the formula for the volume of a triangular pyramid, we have </math>V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}<math>. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume </math>\frac{1}{2}<math> as well. | ||
+ | |||
+ | The original wedge we considered in the last step has volume </math>3<math>, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have </math>3 - \frac{1}{2} \cdot 2 = 2<math>. Thus, the volume of the figure we are trying to find is </math>2<math>. This means that the correct answer choice is </math>\boxed{E}$. | ||
Written by: Archimedes15 | Written by: Archimedes15 |
Revision as of 18:29, 16 February 2018
In the rectangular parallelpiped shown, =
,
=
, and
=
. Point
is the midpoint of
. What is the volume of the rectangular pyramid with base
and apex
?
Solution 1
Consider the cross-sectional plane. Note that and we want
, so the answer is
. (AOPS12142015)
Solution 2
We start by finding side of base
by using the Pythagorean theorem on
. Doing this, we get
Taking the square root of both sides of the equation, we get . We can then find the area of rectangle
, noting that
Taking the vertical cross-sectional plane of the rectangular prism, we see that the distance from point to base
is the same as the distance from point
to side
. Calling the point where the altitude from vertex
touches side
as point
, we can easily find this altitude using the area of right
, as
Multiplying both sides of the equation by 2 and substituting in known values, we get
$$ (Error compiling LaTeX. Unknown error_msg)2 \cdot 3 = FK \cdot \sqrt {11} \Rightarrow FK = \frac{6\sqrt {11}{11}.M
BCHE
\frac{6\sqrt {11}{11}
N$, we get the area of the rectangular pyramid to be
<cmath>\frac{1}{3}([BCHE] \cdot MN) = \frac{1}{3}\left(\sqrt {11} \cdot \frac{6\sqrt {11}{11}\right) = \frac{66}{33} = \boxed{2}.</cmath>
Written by: Adharshk
==Solution 3==
We can start by finding the total volume of the parallelepiped. It is$ (Error compiling LaTeX. Unknown error_msg)2 \cdot 3 \cdot 1 = 6$, because a rectangular parallelepiped is a rectangular prism.
Next, we can consider the wedge-shaped section made when the plane$ (Error compiling LaTeX. Unknown error_msg)BCHE\frac{1}{2} \cdot 2 \cdot 3 = 3
1
\frac{1}{2}
V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}
\frac{1}{2}$as well.
The original wedge we considered in the last step has volume$ (Error compiling LaTeX. Unknown error_msg)33 - \frac{1}{2} \cdot 2 = 2
2
\boxed{E}$.
Written by: Archimedes15
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.