SAT1000 - P894

Let $A$ be the top vertex of the equilateral triangle, $B$ be the bottom left, and $C$ , bottom right. Let the common difference of the arithmetic progression be $d$ and put the values on the vertices as shown. Row 1 ( $A$ being on the row $0$ ) starts with $a+d$ and ends with $a+2d$ . Row 2 starts with $a+2d$ and ends with $a+4d$ , Row 3 starts with $a+3d$ and ends with $a+6d$ , and so on. We note that row $k$ starts with $a+kd$ and ends with $a+2kd$ . We also note that, for an $n$ -row equilateral triangle, $b=a+nd$ and $c = b + nd = a + 2nd$ . Therefore, $a+b+c = 3a + 3nd = 1 \implies a + nd = \frac 13$ .

For an $n$ -row case, there is a total of $T_n a$ , where $T_{n+1}$ is the $n$ th triangular number. On the $k$ th row there are $k+1$ vertices and the number of $d$ starts with $kd$ and ends with $2kd$ . The sum of all the vertex numbers is given by:

$\begin{aligned} f(n) & = T_{n+1} a + \sum_{k=1}^n \frac {(k+1)(k+2k)d}2 \\ & = \frac {(n+1)(n+2)}2 + \frac {3d}2 \sum_{k=1}^n k(k+1) \\ & = \frac {(n+1)(n+2)}2 + \frac {3d}2 \sum_{k=1}^n (k^2 + k) \\ & = \frac {(n+1)(n+2)}2 + \frac {3d}2 \left(\frac {n(n+1)(2n+1)}6 + \frac {n(n+1)}2 \right) \\ & = \frac {(n+1)(n+2)}2 + \frac {dn(n+1)}4 (2n+1+3) \\ & = \frac {(n+1)(n+2)}2 + \frac {n(n+1)(n+2)d}2 \\ & = \frac {(n+1)(n+2)\blue{(a+nd)}}2 & \small \blue{\text{Note that }a+nd = \frac 13} \\ & = \frac {(n+1)(n+2)}6 \end{aligned}$

Therefore, $f(2020) = \dfrac {2021\times 2022}6 = \boxed{681077}$ .

All the vertices is some linear combination of $a,b,c$ (i.e. the number on vertex V can be expressed as $p_va+q_vb+r_vc$ , where $p_v,q_v,r_v$ are some constants.)

However, $a,b,c$ is symmetric in the sum, so the sum is definitely in the form of $ka+kb+kc$ , where $k$ is a constant, so we only need to find $k$ . This means for every $a,b,c$ satisfying $a+b+c=1$ , $f(2020)$ doesn't vary with the choose of $a,b,c$ .

Hence assume that every number on the vertices is $\frac13$ and you will ge the answer.