Problem:
Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths 21 and 31. The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is nm, where m and n are relatively prime positive integers. Find m+n. A parallelepiped is a solid with six parallelogram faces such as the one shown below.
Solution:
Answer (125):
Let 2a and 2b be the lengths of the diagonals of the faces of the parallelepipeds. Then each face is a rhombus with side a+b. Because the volume of a parallelepiped is its base area times its height, the desired ratio is the same as the ratio of the heights of the parallelepipeds.
Consider the parallelepiped that has a vertex A where its neighboring vertices B,C, and D satisfy BC=BD=CD=2a. Then AB=AC=AD=a+b. To compute the altitude from D in tetrahedron ABCD, note that
Area(△ABC)⋅hD=3⋅Volume(ABCD)=Area(△BCD)⋅hA,
where hA and hD denote the altitude lengths from A and D, respectively.
Triangle △BCD is equilateral with side 2a. Hence Area(△BCD)=a3. The area of △ABC is equal to half of the area of the rhombus, and is thus equal to ab.
Start with a unit cube in coordinate space with vertices at (0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,1,0), (1,0,1),(0,1,1), and (1,1,1). Let Lk be the linear transformation that maps the vector ⟨1,1,1⟩ to ⟨k,k,k⟩ and acts as the identity map on vectors perpendicular to ⟨1,1,1⟩. Note that for any k,Lk maps the three vectors ⟨1,0,0⟩,⟨0,1,0⟩, and ⟨0,0,1⟩ to three vectors that are equal in length, and because all edges of the cube are parallel to one of these vectors, Lk maps the cube to a parallelepiped whose faces are congruent rhombi, as needed. For any k,Lk maps the six face diagonals in the three directions ⟨1,−1,0⟩,⟨1,0,−1⟩, and ⟨0,1,−1⟩ to diagonals of the same lengths because these vectors are perpendicular to ⟨1,1,1⟩. But Lk maps the other six face diagonals to diagonals of length 32(2k2+1). For example, Lk maps
⟨1,1,0⟩=32⟨1,1,1⟩+31⟨1,1,−2⟩ to 31⟨2k+1,2k+1,2k−2⟩.
The unit cube has face diagonals of length 2. To obtain a parallelepiped with the correct face diagonals, dilate the unit cube by a factor of 221 and then apply the Lk map to the resulting cube that dilates by a factor of 32(2k2+1)=31⋅212. The required value of k is k=216, and the resulting parallelepiped has volume
216⋅(221)3=263
Similarly, to find the volume of the other parallelepiped, dilate the unit cube by a factor of 231 and then apply the Lk map to the resulting cube that dilates by a factor of 32(2k2+1)=21⋅312. The required value of k is k=314, and the resulting parallelepiped has volume
314⋅(231)3=262
The required ratio is then
262263=6263
as above.
Note: The first solution is valid only when b<3a and a<3b. Similarly, in the second solution, the Lk map that maps ⟨1,1,0⟩ to the shortest possible vector is L0. It maps ⟨1,1,0⟩ to 31⟨1,1,−2⟩, which has length 36. Again, the lengths of the diagonals of the rhombus are in the ratio 3. In other words, if the values of a and b are not within a factor of 3 of each other, there is only one resulting parallelepiped, not two.
