# 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)