Difference between revisions of "2023 AIME I Problems/Problem 13"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~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 <math>u</math>, <math>v</math>, and <math>w</math>. For one of the parallelepipeds, the three diagonals involving the origin have length <math>\sqrt {21}</math>. Hence, <math>(u+v)\cdot (u+v)=u\cdot u+v\cdot v+2u\cdot v=21</math> and <math>(u-v)\cdot (u-v)=u\cdot u+v\cdot v-2u\cdot v=31</math>. Since all of <math>u</math>, <math>v</math>, and <math>w</math> have equal length, <math>u\cdot u=13</math>, <math>v\cdot v=13</math>, and <math>u\cdot v=-2.5</math>. Symmetrically, <math>w\cdot w=13</math>, <math>u\cdot w=-2.5</math>, and <math>v\cdot w=-2.5</math>. Hence the volume of the parallelepiped is given by <math>\sqrt{\operatorname{det} | ||
+ | |||
+ | For the other parallelpiped, the three diagonals involving the origin are of length <math>\sqrt{31}</math> and the volume is <math>\sqrt{\operatorname{det} | ||
+ | |||
+ | Consequently, the answer is <math>\sqrt\frac{10.5^2\cdot 18}{15.5^2\cdot 8}=\frac{63}{62}</math>, giving <math>\boxed{125}</math>. | ||
+ | |||
+ | ~EVIN- | ||
{{AIME box|year=2023|n=I|num-b=12|num-a=14}} | {{AIME box|year=2023|n=I|num-b=12|num-a=14}} |
Revision as of 16:09, 8 February 2023
Problem 13
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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