AIME 1989 Problem 12

Let $ABCD$ be a tetrahedron with $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, and $CD=13$, as shown in the figure. Let $d$ be the distance between the midpoints of edges $AB$ and $CD$. Find $d^{2}$. [asy] defaultpen(fontsize(10)+0.8); size(175); pair A,B,C,D,M,P,Q; C=origin; B=(8,0); D=IP(CR(C,6.5),CR(B,8)); A=(4,-3); P=midpoint(A--B); Q=midpoint(C--D); draw(B--C--D--B--A--C^^A--D); draw(D--P--C^^P--Q, gray+dashed+0.5); pen p=fontsize(12)+linewidth(3); dot("$A$",A,down,p); dot("$B$",B,right,p); dot("$C$",C,left,p); dot("$D$",D,up,p); dot("$M$",P,dir(-45),p); dot("$N$",Q,0.2*(Q-P),p); label("$27$",B--D,2dir(30),fontsize(10)); label("$7$",A--C,2dir(210),fontsize(10)); label("$18$",A--D,1.5*dir(30),fontsize(10)); label("$36$",(3,0),up,fontsize(10)); [/asy]