Problem:
A cube-shaped container has vertices A,B,C, and D, where AB and CD are parallel edges of the cube, and AC and BD are diagonals of faces of the cube, as shown. Vertex A of the cube is set on a horizontal plane P so that the plane of the rectangle ABDC is perpendicular to P, vertex B is 2 meters above P, vertex C is 8 meters above P, and vertex D is 10 meters above P. The cube contains water whose surface is parallel to P at a height of 7 meters above P. The volume of water is nm cubic meters, where m and n are relatively prime positive intgers. Find m+n.
Solution:
Let's first view the cube from a direction perpendicular to ABDC, as illustrated above. Let x be the cube's side length. Since △CHA∼△AGB, we have
CHCA=AGAB.
We know AB=x, AG=x2−22, AC=2x, and CH=8. Plug them into the above equation, we get
82x=x2−22x.
Solving this we get the cube's side length x=6, and AC=62.
Let PQ be the water's surface, both P and Q are 7 meters from P. Notice that C is 8 meters from P, this means
CP=81CA=432.
Similarly,
DQ=83CA=492.
Now, we realize that the 3D space inside the cube without water is a frustum, with P on its smaller base and Q on its larger base. To find its volume, all we need is to find the areas of both bases and the height, which is x=6. To find the smaller base, let's move our viewpoint onto the plane ABDC and view the cube from a direction parallel to ABDC, as shown above. The area of the smaller base is simply
S1=CP2=(432)2=89.
Similarly, the area of the larger base is
S2=DQ2=(492)2=881.
Finally, applying the formula for a frustum's volume,
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