Difference between revisions of "2023 AIME I Problems/Problem 13"
(Created page with "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 large...") |
m (→Solution 1 (3-d Vector Analysis)) |
||
Line 4: | Line 4: | ||
such as the one shown below. | such as the one shown below. | ||
− | ==Solution (3-d Vector Analysis)== | + | ==Solution 1 (3-d Vector Analysis)== |
Denote <math>\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}</math>. | Denote <math>\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}</math>. |
Revision as of 14:26, 8 February 2023
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.
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)