The nine angles of three triangles are n1, n2, ..., n9 degrees, positive integers forming an arithmetic sequence with a positive common difference. If one of the angles is 78 degrees, find the sum of all possible values of n1.

Step-by-step solution

1. 1

   Use the total angle sum to locate the middle term. Three triangles contribute $3\times 180=540$ degrees. For an arithmetic sequence of nine terms, the sum is nine times the middle term:

   $\sum_{i=1}^{9}n_i=9n_5=540\quad\Longrightarrow\quad n_5=60.$

   So whatever $d$ is, the sequence is centred on $60$: $n_i=60+(i-5)d$.

2. 2

   Turn the 78 degree condition into a divisibility condition. If $n_k=78$ then

   $60+(k-5)d=78\quad\Longrightarrow\quad (k-5)d=18.$

   Since $d>0$ we need $k>5$, so $k-5\in\{1,2,3,4\}$ and $d=\dfrac{18}{k-5}$ must be a positive integer. Only $k-5=1,2,3$ divide $18$, giving

   $d=18,\ 9,\ 6.$

3. 3

   Discard the case that makes an angle non-positive. The first term is $n_1=60-4d$:

   $d=18\Rightarrow n_1=-12\ (\text{rejected}),\qquad d=9\Rightarrow n_1=24,\qquad d=6\Rightarrow n_1=36.$

   The largest terms are $n_9=60+4d$, namely $96$ and $84$ — both below $180$, so every angle is a legal triangle angle.

4. 4

   Confirm the nine angles really can form three triangles. Writing $n_i=60+(i-5)d$, the offsets are $-4,-3,\ldots,4$. Splitting them as $\{-4,0,4\}$, $\{-3,1,2\}$ and $\{-2,-1,3\}$ makes each group of offsets sum to zero, so each group of three angles sums to $3\times 60=180$. Such a partition exists for every $d$, so this condition adds no new restriction.

5. 5

   Check both surviving sequences.

   $d=6:\ 36,42,48,54,60,66,72,\mathbf{78},84$
   $d=9:\ 24,33,42,51,60,69,\mathbf{78},87,96$

   Both contain $78$, are strictly increasing, and consist of positive integers.

6. 6

   Add the possible first terms.

   $24+36=60.$

   A brute-force search over all admissible common differences confirms these are the only two cases.