¶ 2023 AIME II 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 intgers. Find .
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Solution:
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 D 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 .
The problems on this page are the property of the MAA's American Mathematics Competitions