¶ 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:
Answer (751):
Let the other vertices of the cube be , and where , as shown. Let and be the projections of and , respectively, onto , and let , and be where the surface of the water meets , and , respectively, as shown. Because the plane of the rectangle is perpendicular to , it follows that points and also lie in that same plane and thus points , and are colinear.
Let the cube have side length . Then , and . Because points , and are coplanar and , it follows that and thus . Then the Pythagorean Theorem applied to right triangle implies that , so and . Because and are a distance 7 from the horizontal plane while is a distance 8 from the plane, it follows that . Similarly, because and are distances 2 and 10 from the plane, .
Note that and the two triangles are in parallel planes. Lines and intersect at a point . Then is a cross section of pyramid , showing that lines , and all intersect at . Thus and are similar triangular pyramids with being 3 times as large as . It follows that the region of the cube without water has volume of the volume of the . That pyramid has height and base area , so its volume is
The volume of the region of the cube without water is . Finally, the volume of the cube is , so the required volume of water is . The requested sum is .
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions