Angles in sequence (2)

Since $ABCD$ is cyclic, its opposite angles are supplementary. Thus $m\angle B + m\angle D = m\angle A + m\angle C = 180$ . Since $\angle B$ is the smallest angle, $\angle D$ must be the largest. WLOG, we'll say $m\angle A \leq m\angle C$ . Let $b = m\angle B$ be a (positive) integer. Then $m\angle B = b$ $m\angle A = b + n$ $m\angle C = b + 2n$ $m\angle D = b + 3n$

for some positive integer $n\geq 0$ . Since the sum of the interior angles of $ABCD$ must be $360^\circ$ (all cyclic quadrilaterals are convex), $4b + 6n = 360$ . Additionally, for any value of $n$ , $m\angle B + m\angle D = m\angle A + m\angle C = 2b + 3n = 180$ , so the quadrilateral is indeed cyclic. For this equation to have integer solutions, $b$ must be a multiple of 3. So let $b=3k$ ; then $n=60-2k$ . Note that $1\leq k \leq 30$ .

We can verify that both $k=1$ and $k=30$ yield valid solutions: the former gives angle measures $3^\circ, 61^\circ, 119^\circ, 177^\circ$ while the latter gives $90^\circ, 90^\circ, 90^\circ, 90^\circ$ . Thus the sum of all possible values of $b$ is $\sum_{k=1}^{30} 3k = 3\frac{(30)(31)}{2} = 1395.$