smc

To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles. If $s$ is the side length of the regular pentagon, then each of the triangles $AOB$, $BOC$, $COD$, $DOE$, and $EOA$ has base $s$ and height 1, so the area of regular pentagon $ABCDE$ is $5s/2$. Next, we divide regular pentagon $ABCDE$ into triangles $ABC$, $ACD$, and $ADE$. Triangle $ACD$ has base $s$ and height $AP=AO+1$. Triangle $ABC$ has base $s$ and height $AQ$. Triangle $ADE$ has base $s$ and height $AR$. Therefore, the area of regular pentagon $ABCDE$ is also $$\frac{s}{2}(AO+AQ+AR+1).$$ Hence, $$\frac{s}{2}(AO+AQ+AR+1)=\frac{5s}{2},$$ which means $AO+AQ+AR+1=5$, or $AO+AQ+AR=\boxed{4}$. The answer is $\boxed{4}$.