2012 AIME I Problems/Problem 8
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Problem 8
Cube labeled as shown below, has edge length and is cut by a plane passing through vertex and the midpoints and of and respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form where and are relatively prime positive integers. Find
Solution: think outside the box
Define a coordinate system with at the origin and and on the , , and axes respectively. The and It follows that the plane going through and has equation Let be the intersection of this plane and edge and let Now since is on the plane. Also, lies on the extensions of segments and so the solid is a right triangular pyramid. Note also that pyramid is similar to with scale factor and thus the volume of solid which is one of the solids bounded by the cube and the plane, is But the volume of is simply the volume of a pyramid with base and height which is So Note, however, that this volume is less than and thus this solid is the smaller of the two solids. The desired volume is then
Alternative Solution (using calculus): think inside the box
Let be the intersection of the plane with edge then is similar to and the volume is a sum of areas of cross sections of similar triangles running parallel to face Let be the distance from face let be the height parallel to of the cross-sectional triangle at that distance, and be the area of the cross-sectional triangle at that distance. and then , and the volume is Thus the volume of the larger solid is
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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