2023 AIME II Problems/Problem 14
Problem
A cube-shaped container has vertices and where and are parallel edges of the cube, and and are diagonals of faces of the cube, as shown. Vertex of the cube is set on a horizontal plane so that the plane of the rectangle is perpendicular to vertex is meters above vertex is meters above and vertex is meters above The cube contains water whose surface is parallel to at a height of meters above The volume of water is cubic meters, where and are relatively prime positive integers. Find
Diagram
Solution 1
Let's first view the cube from a direction perpendicular to , as illustrated above. Let be the cube's side length. Since , we have We know , , , . Plug them into the above equation, we get Solving this we get the cube's side length , and
Let be the water's surface, both and are meters from . Notice that is meters from , this means Similarly,
Now, we realize that the 3D space inside the cube without water is a frustum, with on its smaller base and on its larger base. To find its volume, all we need is to find the areas of both bases and the height, which is . To find the smaller base, let's move our viewpoint onto the plane and view the cube from a direction parallel to , as shown above. The area of the smaller base is simply Similarly, the area of the larger base is
Finally, applying the formula for a frustum's volume,
The water's volume is thus giving .
Solution 2
Denote the distance from point to
Let slope to be Notation is shown in the diagram. Let
Similarly
Let the volume without water be volume of the pyramid be
It is clear that from which
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Solution 3
We introduce a Cartesian coordinate system to the diagram. We put the origin at . We let the -components of , , be positive. We set the -axis in a direction such that is on the plane.
The coordinates of , , are , , .
Because , . Thus,
Because is a diagonal of a face, . Thus,
Because plane is perpendicular to plan , . Thus,
Jointly solving (1), (2), (3), we get one solution , , . Thus, the side length of the cube is .
Denote by and two vertices such that and are two edges, and satisfy the right-hand rule that . Now, we compute the coordinates of and .
Because , we have , , .
Hence,
By solving these equations, we get
In addition, we have .
Thus, , .
Therefore, the volume of the water is
Define , , . Thus,
Define . Thus,
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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