2023 AIME I Problems/Problem 13
Contents
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.
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 parallelepiped, the three diagonals involving the origin are of length and the volume is .
Consequently, the answer is , giving .
~EVIN-
Solution 3 (No trig, no linear algebra)
Observe that both parallelepipeds have two vertices (one on each base) that have three congruent angles meeting at them. Denote the parallelepiped with three acute angles meeting , and the one with three obtuse angles meeting .
The area of a parallelepiped is simply the base area times the height, but because both parallelepipeds have the same base, what we want is just the ratio of the heights.
Denote the point with three acute angles meeting at it in as , and its neighbors , , and . Similarly, denote the point with three obtuse angles meeting at it in as , and its neighbors , , and .
We have the following equations:
However, and are both half the area of a rhombus with diagonals and , so our ratio is really
Because the diagonals of all of the faces are and , each edge of the parallelepipeds is by the Pythagorean theorem.
We have , and . When we drop a perpendicular to the centroid of from (let's call this point ), we have . Thus,
The area of base is . Hence,
We can apply a similar approach to .
, and . When we drop a perpendicular to the centroid of from (let's call this point ), we have . Thus,
The area of base is . Hence,
Finally,
Our answer is .
~mathboy100
Solution 4 (Pythagorean theorem)
Since the two parallelepipeds have the same base, all we need to do is to find their respective heights.
As illustrated in the above diagram, drop a perpendicular from onto the base at . Denote the center of the base by . By symmetry, must be on . Now we need to find .
Apply Pythagorean theorem to we have
Apply Pythagorean theorem to and then we have
Combining the above two, we have
Since , , , , plug them into the above equation and solve for the only unknown variable , we get
Thus the height
For the other parallelepiped, using the same approach and drop a perpendicular from onto the base at . Similarly applying Pythagorean theorem to , and we have
Plugging known values into the above equation and solve for the only unknown variable , we get
Thus the height
The ratio between the two is therefore , giving .
~sgdzw
Solution 5 (Visual)
Let us inscribe a tetrahedron in given parallelepiped so that its edges coincide with the diagonals of the faces of the parallelepiped. Note that the three edges outgoing from the vertex have the same length , and the three edges at the base have a different length
The volume of the tetrahedron is three times less than the volume of the parallelepiped.
In second parallelepiped and change the positions.
Required ratio is
Claim
Let be the regular pyramid,
The area of
Height
Volume
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~MathProblemSolvingSkills.com
Animated Video Solution
~Star League (https://starleague.us)
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.