Let $ABCD$ be a regular tetrahedron. Let $E$, $F$, $G,$ $H$ be the centers of faces $BCD$, $ACD$, $ABD$, $ABC$, respectively. The volume of pyramid $DEFG$ is $18.$ Find the volume of pyramid $EFGH$
[asy] import three; unitsize(1cm); currentprojection=orthographic(1/4,1,1/2); triple A=(0,0,1); triple B=(1,0,0); triple C=(0,1,0); triple D=(-1,-1,-1); triple EE=(2/3,-1/3,-1/3); triple FF=(-1/3,2/3,-1/3); triple GG=(-1/3,-1/3,2/3); triple HH=(0,0,0); three[] dots = {A,B,C,D,EE,FF,GG,HH}; triple[] X = {A,A,A,A,EE,FF,GG,HH}; triple[] Y = {B,B,B,B,FF,GG,EE,HH}; triple[] Z = {C,C,C,C,GG,EE,FF,HH}; for(int i=0;i<dots.length;++i){ label("$"+string(i+1)+"$",dots[i],(1.2,0,0)); } triple[] dotlabels = {X[3],Y[3],Z[3],X[7],Y[7],Z[7]}; string[] dotlabelnames = {"A","B","C","D","E","F"}; for(int i=0;i<dotlabelnames.length;++i){ label("$"+dotlabelnames[i]+"$",dotlabels[i],(-1,0,2)); } draw(surface(EE--GG--FF--cycle),gray(0.8),nolight); draw(surface(EE--FF--HH--cycle),gray(0.7),nolight); pen p=fontsize(10); for(int i=0;i<=3;++i) draw(C--X[i]^^A--Y[i]^^B--Z[i]); draw(EE--GG--FF--cycle^^EE--FF--HH--cycle); [/asy]
Note that $DEFG$ is similar to $ABCD$ with ratio $\frac12$. The ratio of their volumes is $(\frac12)^3=\frac18$, so the volume of $ABCD$ is $8(18)=144$. Using the fact that $ABCD$ is a regular tetrahedron with side length $2$, we find that $[ABC]=12\sqrt{2}$, and so
\[[EFGH]=[ABCD]-4[ABC]=144-4\cdot12\sqrt{2}=\boxed{144-48\sqrt{2}}.\]