Cutting up a Triangle

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In \(\triangle ABC\), let points \(P_i (i=1,2,3,4,5)\) be 5 points on side \(\overline{AB}\), in that order, such that
\[
AP_1=P_1P_2=P_2P_3=P_3P_4=P_4P_5=P_5B;
\]
let points \(Q_j (j=1,2,3,4,5,6,7)\) be 7 points on side \(\overline{BC}\), in that order, such that
\[
BQ_1 = Q_1Q_2 = Q_2Q_3 = Q_3Q_4 = Q_4Q_5 = Q_5Q_6 = Q_6Q_7 = Q_7C;
\]
and let \(R_k(k=1,2,3)\) be 3 points on side \(\overline{CA}\), in that order, such that
\[
CR_1 = R_1R_2 = R_2R_3 = R_3A.
\]
Connect each vertex with the points on the opposite side to get the segments \(\overline{CP_i} (1\leq i\leq 5)\), \(\overline{AQ_j}(1\leq j\leq 7)\), and \(\overline{BR_k}(1\leq k\leq 3)\). There are 15 segments in total. These segments divide \(\triangle ABC\) into small non-overlapping regions. How many such regions are there?

(Source: ZIML Monthly Online Math Contest October 2019)

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