In the diagram, four circles of radius $1$ with centers $P$, $Q$, $R$, and $S$ are tangent to one another and to the sides of $\triangle ABC$, as shown.
Find the circumradius of triangle $AQS$.
[asy]defaultpen(.98+fontsize(9)); pair A=10*dir(50), B=10*dir(170), C=10*dir(350), P=8*dir(20), R=8*dir(80), Q=8*dir(140), S=8*dir(200); pair M=N=(0,50/sqrt(3)); pair Y=X=(A+N)/2, Z=W=dir(50+65)*Y; draw(X--Q--B--Z);draw(Z--R--C--W);draw(W--S--A--X);draw(Q--R--S--cycle);draw(A--B--C--cycle);draw(C--Y,dashed);draw(X--B,dashed);draw(R--M--Q,dashed); pair[] ps={A,B,C,M,N,P,Q,R,S,X,Y,W,Z}; dot(ps); label("$A$",A,NE);label("$B$",B,W);label("$C$",C,SE);label("$M$",M,NW);label("$N$",N,NE); label("$P$",P,NW);label("$Q$",Q,W);label("$R$",R,S);label("$S$",S,SW); label("$X$",X,S);label("$Y$",Y,W);label("$Z$",Z,N);label("$W$",W,SW); [/asy]
As shown, let $M$ be the foot of the perpendicular from $A$ to $BC$, and let $N$ be the midpoint of $BC$.
By ASA, it is apparent that $RMS\cong NAQ.$ Therefore, $RQ=AN$. Since $PN=\textfrac{3}{2}$ (Problems for the Intermediate Olympiad 4.42), we have $NP = NQ = \textfrac{1}{2}$. Thus $1 = NA^{2} = RQ^{2} = AQ^{2}+AR^{2}-2 \cdot AQ \cdot AR \cdot \cos \left(\angle QAR \right)$.
By the Law of Cosines on triangle $AQR$, $17 = 1+1-\frac{1}{2} \cdot QR \cdot \cos \left(\angle QAR \right) = QR-\frac{QR}{2} \cdot \cos \left(\angle QAR \right) \Longrightarrow QR = 6$.
Therefore, $\cos \left(\angle QAR \right) = \frac{3}{4}$.
Applying the Law of Sines to triangle $AQS$, $\frac{AS}{\sin \left(\angle AQS \right)} = \frac{1}{\sin \left(\angle QAS \right)}$
Since $\angle QAS + \angle QAR = 180^{\circ}$, $\sin \left(\angle QAS \right) = \sin \left(\angle QAR \right) = \sqrt{1-\cos^{2} \left(\angle QAR \right)} = \sqrt{\frac{7}{16}}$.
$\angle AQS = \angle AQR + \angle RQS = 180^{\circ}-2 \cdot \cos^{-1} \left(\frac{3}{4} \right) - \sin^{-1} \left(\frac{\sqrt{7}}{4} \right)$ (draw triangle $SQR$ and notice what the angles have to add up to).
By the Law of Sines on triangle $AQS$, $\frac{AQ}{\sin \left(\angle AQS \right)} = \frac{AS}{\sin AQS} = 2R$.
Therefore, $2R = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{7}}{2 \cdot \sqrt{2}} = \boxed{\textbf{(E)} \frac{\sqrt{21}}{4}}$.
To find the circumradius of triangle $AQS$, we need to find the length of the side $AQ$ and the distance between $A$ and $Q$.
Let's start by finding the length of $AQ$. Since the circles are tangent to sides $AB$ and $AC$ at $B$ and $C$ respectively, the length of $AQ$ is equal to $AB + AC$.
Since the circles are tangent to sides $AB$ and $AC$, we have $BP = BQ = 1$ and $CR = CS = 1$.
Let's denote the point where the circles centered at $P$ and $S$ intersect as $T$. Since $PQ = 1$ and $CS = 1$, we have $QS = 2$.
Since $APTQ$ is a rectangle, we have $PT = AQ = 2$.
Similarly, if we denote the point where the circles centered at $Q$ and $R$ intersect as $U$, we have $QU = UR = 1$.
Since $BUQ$ is an equilateral triangle with side length $1$, we have $BQ = 1$ and $AB = 2$.
Similarly, since $CVR$ is an equilateral triangle with side length $1$, we have $CR = 1$ and $AC = 2$.
Therefore, $AQ = AB + AC = 2 + 2 = 4$.
Now, let's find the distance between $A$ and $Q$. Since $AQ = 4$ and $PT = AQ = 2$, we have $AT = AQ - PT = 4 - 2 = 2$.
In a right triangle, the distance between the circumcenter and the midpoint of the hypotenuse is equal to half the length of the hypotenuse.
Therefore, the distance between $A$ and the circumcenter of triangle $AQS$ is equal to half the length of $AQ$, which is $2/2 = 1$.
Finally, the circumradius of triangle $AQS$ is the distance between $A$ and the circumcenter of triangle $AQS$, which is equal to $1$.
So, the circumradius of triangle $AQS$ is $\boxed{1}$.