2012 AIME I Problems/Problem 8
Contents
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
![[asy]import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair A = (0,0), B = (3.8,0), C = (5.876,1.564), D = (2.076,1.564), E = (0,3.8), F = (3.8,3.8), G = (5.876,5.364), H = (2.076,5.364), M = (1.9,0), N = (5.876,3.465); pair[] dotted = {A,B,C,D,E,F,G,H,M,N}; D(A--B--C--G--H--E--A); D(E--F--B); D(F--G); pathpen=dashed; D(A--D--H); D(D--C); dot(dotted); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NW); label("$E$",E,W); label("$F$",F,SE); label("$G$",G,NE); label("$H$",H,NW); label("$M$",M,S); label("$N$",N,NE); [/asy]](http://latex.artofproblemsolving.com/a/f/1/af154db9d25256bd90239e9f4e37253b234659e0.png)
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
Alternative Solution think inside the box like a total nerd
If you memorized the formula for a frustum, then this problem is very trivial.
The formula for a frustum is:
where
is the area of the base and
is the height from the chopped of 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 area is Thus the answer is:
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