A regular tetrahedron is a pyramid with four faces, each of which is an equilateral triangle.

Let $ABCD$ be a regular tetrahedron and let $P$ be the unique point equidistant from points $A,B,C,D$. Extend $\overrightarrow{AP}$ to hit face $BCD$ at point $Q$. What is the ratio $PQ/AQ$ ?
Let's start with a picture: [asy]
import three;
triple d = (0,0,0);
triple b = (1,0,0);
triple c = (1/2,sqrt(3)/2,0);
triple a = (1/2,sqrt(3)/6,sqrt(6)/3);
triple p = (a+b+c+d)/4;
triple q = (d+b+c)/3;
draw(a--b--c--a); draw(c--d--b,dotted); draw(d--a,dotted); draw(a--q,dashed);
dot(a); dot(b); dot(c); dot(d); dot(p); dot(q);
label("$A$",a,N);
label("$B$",b,WSW);
label("$C$",c,ESE);
label("$D$",d,ENE);
label("$P$",p,W);
label("$Q$",q,W);
[/asy] We can carve $ABCD$ into four (non-regular) tetrahedra that share $P$ as a vertex and have respective bases $ABC$, $ABD$, $ACD$, and $BCD$ (the faces of $ABCD$). For example, this diagram shows one of these four tetrahedra, namely $BCDP$: [asy]
import three;
triple d = (0,0,0);
triple b = (1,0,0);
triple c = (1/2,sqrt(3)/2,0);
triple a = (1/2,sqrt(3)/6,sqrt(6)/3);
triple p = (a+b+c+d)/4;
triple q = (d+b+c)/3;
draw(a--b--c--a); draw(c--d--b,dotted); draw(d--a,dotted); draw(a--q,dashed);
draw(surface(b--p--c--cycle),red,nolight);
draw(surface(d--p--c--cycle),red+white,nolight);
dot(a); dot(b); dot(c); dot(d); dot(p);
label("$A$",a,N);
label("$B$",b,WSW);
label("$C$",c,ESE);
label("$D$",d,ENE);
label("$P$",p,W);
[/asy] The four tetrahedra formed in this way are congruent, so each contains one-quarter the volume of $ABCD$.

The height of tetrahedron $BCDP$ is $PQ$, so the volume of $BCDP$ is $$\frac 13\cdot (\text{area of }\triangle BCD)\cdot PQ.$$The volume of the original tetrahedron, $ABCD$, is $$\frac 13\cdot (\text{area of }\triangle BCD)\cdot AQ.$$Thus $PQ/AQ$ is equal to the ratio of the volume of $BCDP$ to the volume of $ABCD$, which we already know to be $\boxed{\frac{1}{4}}$.