Difference between revisions of "2018 AMC 10B Problems/Problem 10"
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==Solution 1== | ==Solution 1== | ||
Consider the cross-sectional plane, and label it as b. Note that <math>bh/2=3</math> and we want <math>bh/3</math>, so the answer is <math>\boxed{2}</math>. (AOPS12142015) | Consider the cross-sectional plane, and label it as b. Note that <math>bh/2=3</math> and we want <math>bh/3</math>, so the answer is <math>\boxed{2}</math>. (AOPS12142015) | ||
+ | |||
+ | IMPORTANT: This is assuming the parallelepiped is a rectangular prism, which isn't correct. All we know is that each side is a parallelogram, so this solution doesn't work. | ||
==Solution 2== | ==Solution 2== |
Revision as of 18:38, 17 February 2018
Contents
[hide]Problem
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, and label it as b. Note that and we want
, so the answer is
. (AOPS12142015)
IMPORTANT: This is assuming the parallelepiped is a rectangular prism, which isn't correct. All we know is that each side is a parallelogram, so this solution doesn't work.
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
Deducing that the altitude from vertex to base
is
and calling the point of intersection between the altitude and the base as point
, we get the area of the rectangular pyramid to be
Written by: Adharshk
Solution 3
We can start by finding the total volume of the parallelepiped. It is , because a rectangular parallelepiped is a rectangular prism.
Next, we can consider the wedge-shaped section made when the plane cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is
. Since BC is given to be
, we have that FM is
. Using the formula for the volume of a triangular pyramid, we have
. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume
as well.
The original wedge we considered in the last step has volume , because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have
. Thus, the volume of the figure we are trying to find is
. This means that the correct answer choice is
.
Written by: Archimedes15
Solution 4 (Vectors)
By the Pythagorean theorem, . Because
, the area of the base is
. Now, we need to find the height.
Define as the midpoint of
and
as the midpoint of
. Consider a vector coordinate system with origin
with
and
axes parallel to
and
respectively (positive
direction is towards
, positive
direction is towards
, positive
direction is towards
). Then,
The dot product of
and
is the length of the projection of
onto
multiplied by the length of
, so dividing the dot product of
and
by the length of
should give the length of the projection of
onto
. Doing this calculation, we get that the length of the projection is
. Notice that this projection onto
is the same as projecting
onto the plane.
Denote as the foot of the projection of
onto
. Then
is right, so
is a right triangle. Applying the Pythagorean theorem on
and calling
(which is actually the height of the pyramid)
, we get
. Therefore,
.
Now since we have the base and the height of the pyramid, we can find its volume. , so the answer is
.
Written by: SS4
Solution 5 (slicker method)
Rotate the rectangular pyramid so that rectangle is the base of our rectangular pyramid. Now our height becomes
We know that the volume of our rectangular pyramid is
(MathloverMC)
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.