jmc

Join $PQ$, $PR$, $PS$, $RQ$, and $RS$. Since the circles with centers $Q$, $R$ and $S$ are all tangent to $BC$, then $QR$ and $RS$ are each parallel to $BC$ (as the centers $Q$, $R$ and $S$ are each 1 unit above $BC$). This tells us that $QS$ passes through $R$.

Similarly, since $P$ and $S$ are each one unit from $AC$, then $PS$ is parallel to $AC$. Also, since $P$ and $Q$ are each one unit from $AB$, then $PQ$ is parallel to $AB$. Therefore, the sides of $△PQS$ are parallel to the corresponding sides of $△ABC$.

When the centers of tangent circles are joined, the line segments formed pass through the associated point of tangency, and so have lengths equal to the sum of the radii of those circles. Therefore, $QR=RS=PR=PS=1+1=2$.



Since $PR=PS=RS$, we know $△PRS$ is equilateral, so $\angle PSR=\angle PRS=60^{\circ }$. Since $\angle PRS=60^{\circ }$ and $QRS$ is a straight line, we have $\angle QRP=180^{\circ }-60^{\circ }=120^{\circ }$.

Since $QR=RP$, we know $△QRP$ is isosceles, so $$\angle PQR=\frac{1}{2}(180^{\circ }-120^{\circ })=30^{\circ }.$$Since $\angle PQS=30^{\circ }$ and $\angle PSQ=60^{\circ }$, we have $$\angle QPS=180^{\circ }-30^{\circ }-60^{\circ }=90^{\circ },$$so $△PQS$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle.

The angles of $△ABC$ are equal to the corresponding angles of $△PQS$, so $△ABC$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle. This means that if we can determine one of the side lengths of $△ABC$, we can then determine the lengths of the other two sides using the side ratios in a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle.

Consider side $AC$. Since the circle with center $P$ is tangent to sides $AB$ and $AC$, the line through $A$ and $P$ bisects $\angle BAC$. Thus, $\angle PAC=45^{\circ }$. Similarly, the line through $C$ and $S$ bisects $\angle ACB$. Thus, $\angle SCA=30^{\circ }$. We extract trapezoid $APSC$ from the diagram, obtaining



Drop perpendiculars from $P$ and $S$ to $X$ and $Z$, respectively, on side $AC$. Since $PS$ is parallel to $AC$, and $PX$ and $SZ$ are perpendicular to $AC$, we know that $PXZS$ is a rectangle, so $XZ=PS=2$.

Since $△AXP$ is right-angled at $X$, has $PX=1$ (the radius of the circle), and $\angle PAX=45^{\circ }$, we have $AX=PX=1$. Since $△CZS$ is right-angled at $Z$, has $SZ=1$ (the radius of the circle), and $\angle SCZ=30^{\circ }$, we have $CZ=\sqrt{3}SZ=\sqrt{3}$ (since $△SZC$ is also a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle). Thus, $AC=1+2+\sqrt{3}=3+\sqrt{3}$.

Since $△ABC$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle, with $\angle ACB=60^{\circ }$ and $\angle CAB=90^{\circ }$, we have $BC=2AC=6+2\sqrt{3}$, and $$AB=\sqrt{3}AC=\sqrt{3}(3+\sqrt{3})=3\sqrt{3}+3.$$Therefore, the side lengths of $△ABC$ are $AC=3+\sqrt{3}$, $AB=3\sqrt{3}+3$, and $BC=6+2\sqrt{3}$. Thus, the perimeter is $$3+\sqrt{3}+3\sqrt{3}+3+6+2\sqrt{3}=\boxed{12+6\sqrt{3}}.$$