2020 AMC 8 Problems/Problem 16

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Solution 1: We notice that 3 lines pass through B, and 2 lines pass through all other points. In addition, we are given that $A+B+C+D+E+F=1+2+3+4+5+6=21$. This means that \[2A+3B+2C+2D+2E+2F=47\] \[2(A+B+C+D+E+F)+B=47\] \[2(21)+B=47\] \[B=5\]

~samrocksnature