Difference between revisions of "2023 AIME I Problems/Problem 13"
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Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths <math>\sqrt{21}</math> and <math>\sqrt{31}</math>. | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths <math>\sqrt{21}</math> and <math>\sqrt{31}</math>. | ||
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> | The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> |
Revision as of 00:21, 10 February 2023
Problem
Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths and
.
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is
, where
and
are relatively prime positive integers. Find
. A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
{insert diagram here}
Solution 1 (3-D Vector Analysis)
Denote .
Denote by
the length of each side of a rhombus.
Now, we put the solid to the 3-d coordinate space.
We put the bottom face on the plane.
For this bottom face, we put a vertex with an acute angle
at the origin, denoted as
.
For two edges that are on the bottom face and meet at
, we put one edge on the positive side of the
-axis. The endpoint is denoted as
. Hence,
.
We put the other edge in the first quadrant of the
plane. The endpoint is denoted as
. Hence,
.
For the third edge that has one endpoint , we denote by
its second endpoint.
We denote
.
Without loss of generality, we set
.
Hence,
We have
and
Case 1: or
.
By solving (2) and (3), we get
Plugging these into (1), we get
Case 2: and
, or
and
.
By solving (2) and (3), we get
Plugging these into (1), we get
We notice that . Thus, (4) (resp. (5)) is the parallelepiped with a larger (resp. smaller) height.
Therefore, the ratio of the volume of the larger parallelepiped to the smaller one is
Recall that .
Thus,
.
Plugging this into the equation above, we get
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (no trig)
Let one of the vertices be at the origin and the three adjacent vertices be ,
, and
. For one of the parallelepipeds, the three diagonals involving the origin have length
. Hence,
and
. Since all of
,
, and
have equal length,
,
, and
. Symmetrically,
,
, and
. Hence the volume of the parallelepiped is given by
.
For the other parallelpiped, the three diagonals involving the origin are of length and the volume is
.
Consequently, the answer is , giving
.
~EVIN-
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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