Difference between revisions of "2018 AMC 10B Problems/Problem 10"
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Written by: Archimedes15 | Written by: Archimedes15 | ||
+ | ==Solution 4 (Vectors)== | ||
+ | By the Pythagorean theorem, <math>EB=\sqrt{13}</math>. Because <math>EH=1</math>, the area of the base is <math>\sqrt{13}</math>. Now, we need to find the height. | ||
+ | |||
+ | Define <math>X</math> as the midpoint of <math>BC</math> and <math>Y</math> as the midpoint of <math>EH</math>. Consider a vector coordinate system with origin <math>X</math> with <math>x, y,</math> and <math>z</math> axes parallel to <math>AB, BC,</math> and <math>CG</math> respectively (positive <math>x</math> direction is towards <math>A</math>, positive <math>y</math> direction is towards <math>C</math>, positive <math>z</math> direction is towards <math>G</math>). Then, | ||
+ | <cmath> | ||
+ | M=\begin{bmatrix} | ||
+ | 0\\ | ||
+ | 0\\ | ||
+ | 2\\ | ||
+ | \end{bmatrix}, | ||
+ | Y=\begin{bmatrix} | ||
+ | 3\\ | ||
+ | 0\\ | ||
+ | 2\\ | ||
+ | \end{bmatrix} | ||
+ | </cmath> | ||
+ | The dot product of <math>M</math> and <math>Y</math> is the length of the projection of <math>M</math> onto <math>Y</math> multiplied by the length of <math>Y</math>, so dividing the dot product of <math>M</math> and <math>Y</math> by the length of <math>Y</math> should give the length of the projection of <math>M</math> onto <math>Y</math>. Doing this calculation, we get that the length of the projection is <math>\frac4{\sqrt{13}}</math>. Notice that this projection onto <math>Y</math> is the same as projecting <math>M</math> onto the plane. | ||
+ | |||
+ | Denote <math>P</math> as the foot of the projection of <math>M</math> onto <math>Y</math>. Then <math>\angle MPY</math> is right, so <math>\triangle MPY</math> is a right triangle. Applying the Pythagorean theorem on <math>\triangle MPY</math> and calling <math>MP</math> (which is actually the height of the pyramid) <math>x</math>, we get <math>x^2+\frac4{\sqrt{13}}^2=2^2</math>. Therefore, <math>x=\sqrt{4-\frac{16}{13}}=\frac6{\sqrt{13}}</math>. | ||
+ | |||
+ | Now since we have the base and the height of the pyramid, we can find its volume. <math>\dfrac{\sqrt{13}\times\dfrac6{\sqrt{13}}}3=2</math>, so the answer is <math>\boxed{\text{(E)}}</math>. | ||
+ | |||
+ | Written by: SS4 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:05, 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
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
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 |
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