2012 AIME I Problems/Problem 8
Contents
- 1 Problem
- 2 Solution: think outside the box (pun intended)
- 3 Solution 2
- 4 Solution 2 but slightly better
- 5 Alternative Solution (using calculus) : think inside the box
- 6 Alternative Solution : think inside the box
- 7 Alternative Solution : think inside the box with pyramids
- 8 Solution outside visual
- 9 Solution 7
- 10 Video Solution by Richard Rusczyk
- 11 Video Solution
- 12 See also
Problem
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 (pun intended)
Define a coordinate system with at the origin and and on the , , and axes respectively. Then 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
- Another way to finish after using coordinates: Take the volume of DMBCNQ as the sum of the volumes of DMBQ and DBCNQ. Then, we have the sum of the volumes of a tetrahedron and a pyramid with a trapezoidal base. Their volumes, respectively, are and , giving the same answer as above. ~mathtiger6
Solution 2
Define a coordinate system with , , . The plane formed by , , and is . It intersects the base of the unit cube at . The z-coordinate of the plane never exceeds the height of the unit cube for . Therefore, the volume of one of the two regions formed by the plane is Since , our answer is .
Solution 2 but slightly better
Same coordinate system, but we will use domains instead of using triple integrals. By the way, the method to obtain the equation of the plane is the cross product (). We can multiply this vector by to make things look cleaner and get . We then get the desired plane, , or . We use a double integral with a Type I domain. Observing the diagram, the domain is where . The integral is then which becomes which becomes which then becomes and finally So our answer is
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
Alternative Solution : think inside the box
The volume of a frustum is where is the area of the base and is the height from the chopped off apex to the base.
We can easily see that from symmetry, the area of the smaller front base is and the area of the larger back base is
Now to find the height of the apex.
Extend the and (call the intersection of the plane with G) to meet at . Now from similar triangles and we can easily find the total height of the triangle to be
Now straight from our formula, the volume is Thus the answer is:
Alternative Solution : think inside the box with pyramids
We will solve for the area of the smaller region, and then subtract it from 1.
Let be the point where plane intersects . Then can be split into triangular pyramid and quadrilateral pyramid .
Pyramid has base with area . The height is , so the volume of is .
Similarly, pyramid has base with area . The height is , so the volume of is .
Adding up the volumes of and , we find that the volume of is . Therefore the volume of the larger solid is
This is basically mathtiger6's solution, but you don't need coordinates or thinking outside the box.
Solution outside visual
Solution 7
We use a coordinate system with , , , . Then the plane going through , , and has equation . We set up a double integral in this coordinate system. Consider the region to be integrated over in the -plane. From to , the upper bound of the region is . From to , the upper bound of the region is . In both cases, the lower bound of the region is . Thus, we have the double integral . We find that this sum evaluates to . However, this is the volume of the smaller region, so the larger region's volume is that of the cube minus that of the smaller region. Since the cube has side length , its volume is , so the volume of the larger region is . Thus, our answer is .
~ cxsmi
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/344
~ dolphin7
Video Solution
https://www.youtube.com/watch?v=LWUN_ZymNnw ~Shreyas S
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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All AIME Problems and Solutions |
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