Problem: Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , 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.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70);  draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy]

Solution: Since the two parallelepipeds have the same base, all we need to do is to find their respective heights.
[asy] unitsize(2cm); pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2;  label("A",a,S); label("B",b,S); label("C",c,S); label("D",d,N); label("P",p,S); label("O",o,E);  draw(a--b--(b+c)); draw(a--c--(b+c), dotted); draw(shift(d)*(a--b--(b+c)--c--cycle)); draw(a--d); draw(b--(b+d)); draw(c--(c+d), dotted); draw((b+c)--(b+c+d));  draw(d--p, dotted); draw(c--b, dotted); draw(a--(b+c), dotted); draw(p--c, dotted); draw(d--c, dotted); [/asy]
As illustrated in the above diagram, drop a perpendicular from $D$ onto the base at $P$ . Denote the center of the base by $O$ . By symmetry, $P$ must be on $AO$ . Now we need to find $DP$
Apply Pythagorean theorem to $\triangle DPA$ we have \[DP^2 = AD^2 - AP^2.\]
Apply Pythagorean theorem to $\triangle DPC$ and then $\triangle CPO$ we have \[DP^2 = DC^2 - CP^2 = DC^2 - (CO^2 + OP^2) = DC^2 - (CO^2 + (AO-AP)^2) = DC^2 - CO^2 - (AO-AP)^2.\]
Combining the above two, we have \[AD^2 - AP^2 = DC^2 - CO^2 - (AO-AP)^2.\]
Since $AD=\sqrt{13}$ $DC=\sqrt{21}$ $CO=\frac{\sqrt{21}}{2}$ $AO=\frac{\sqrt{31}}{2}$ , plug them into the above equation and solve for the only unknown variable $AP$ , we get $AP=\frac{5}{\sqrt{31}}.$
Thus the height \[DP = \sqrt{AD^2 - AP^2} = \sqrt{13 - \frac{25}{31}} = \sqrt{\frac{378}{31}}.\]
[asy] unitsize(2cm); pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = 0.7*dir(80), e = c+d, p = 0.9*dir(10), o = (b+c)/2;  label("A'",a,S); label("B'",b,S); label("C'",c,W); label("E'",e,N); label("P'",p,S); label("O'",o,W);  draw(a--b--(b+c)); draw(a--c--(b+c), dotted); draw(shift(d)*(a--b--(b+c)--c--cycle)); draw(a--d); draw(b--(b+d)); draw(c--(c+d), dotted); draw((b+c)--(b+c+d));  draw(e--p, dotted); draw(c--b, dotted); draw(a--(b+c), dotted); draw(p--a, dotted); draw(e--a, dotted); [/asy]
For the other parallelepiped, using the same approach and drop a perpendicular from $E'$ onto the base at $P'$ . Similarly applying Pythagorean theorem to $\triangle E'P'C'$ $\triangle E'P'A'$ and $\triangle A'P'O'$ we have \[C'E'^2 - C'P'^2 = A'E'^2 - A'O'^2 - (C'P'-C'O')^2.\]
Plugging known values into the above equation and solve for the only unknown variable $C'P'$ , we get $C'P'=\frac{5}{\sqrt{21}}.$
Thus the height \[E'P' = \sqrt{C'E'^2 - C'P'^2} = \sqrt{13 - \frac{25}{21}} = \sqrt{\frac{248}{21}}.\]
The ratio between the two is therefore \[\frac{DP}{E'P'} = \frac{\sqrt{\frac{378}{31}}}{\sqrt{\frac{248}{21}}} = \sqrt{\frac{2\cdot3^3\cdot7}{31}\cdot\frac{3\cdot7}{2^3\cdot31}} = \frac{3^2\cdot7}{2\cdot31} = \frac{63}{62}\] , giving $\boxed{125}$

Answer:  125