2023 AIME II Problems/Problem 14
Contents
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
Diagram
Solution (3D Vector Analysis, Analytic Geometry + Calculus)
We introduce a Cartesian coordinate system to the diagram.
We put the origin at . We let the
-components of
,
,
be positive.
We set the
-axis in a direction such that
is on the
plane.
The coordinates of ,
,
are
,
,
.
Because ,
.
Thus,
Because is a diagonal of a face,
.
Thus,
Because plane is perpendicular to plan
,
.
Thus,
Jointly solving (1), (2), (3), we get one solution ,
,
.
Thus, the side length of the cube is
.
Denote by and
two vertices such that
and
are two edges, and satisfy the right-hand rule that
.
Now, we compute the coordinates of
and
.
Because , we have
,
,
.
Hence,
By solving these equations, we get \[ y_P^2 + y_Q^2 = 36 . ]\
In addition, we have .
Thus,
,
.
Therefore, the volume of the water is
Define ,
,
.
Thus,
Define .
Thus,
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (3D Geometry)
Let's first view the cube from a direction perpendicular to , as illustrated above. Let
be the cube's side length. Since
, we have
We know
,
,
,
. Plug them into the above equation, we get
Solving this we get the cube's side length
, and
Let be the water's surface, both
and
are
meters from
. Notice that
is
meters from
, this means
Similarly,
Now, we realize that the 3D space inside the cube without water is a frustum, with on its smaller base and
on its larger base. To find its volume, all we need is to find the areas of both bases and the height, which is
. To find the smaller base, let's move our viewpoint onto the plane
and view the cube from a direction parallel to
, as shown above. The area of the smaller base is simply
Similarly, the area of the larger base is
Finally, applying the formula for a frustum's volume,
The water's volume is thus
,
giving
.
~sgdzw
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.