it is 11

False, the side containing $10$ would also contain two other numbers, so its sum is at least $10+1+2=13$.

Let $a_1,a_2,\ldots ,a_{10}$ be a permutation of $1, 2, \ldots , 10$ where $a_1,a_2,\ldots , a_5$ are on the vertices and the other $5$ numbers are on the sides. In addition, let this minimal sum be $S$. Note that $5S = 2(a_1+a_2+\cdots +a_5)+a_6+a_7+\cdots +a_{10}$ Obviously to minimize $S$ we want as many small numbers to be $a_1,a_2,\ldots , a_5$ as possible. So let's consider the case $a_k=k$. It gives $S=14$, so we know that $S\ge 14$. I'm too lazy to find a solution for $S=14$ but it probably exists so we're done.

bobthesmartypants wrote: Let $a_1,a_2,\ldots ,a_{10}$ be a permutation of $1, 2, \ldots , 10$ where $a_1,a_2,\ldots , a_5$ are on the vertices and the other $5$ numbers are on the sides. In addition, let this minimal sum be $S$. Note that $5S = 2(a_1+a_2+\cdots +a_5)+a_6+a_7+\cdots +a_{10}$ Obviously to minimize $S$ we want as many small numbers to be $a_1,a_2,\ldots , a_5$ as possible. So let's consider the case $a_k=k$. It gives $S=14$, so we know that $S\ge 14$. I'm too lazy to find a solution for $S=14$ but it probably exists so we're done. I'm not too lazy to make my computer do it. Up to symmetry, there are only three solutions: one with $S = 14$ and two with $S = 16$. (The images of these under $a_i \rightarrow 11 - a_i$ yield solutions for $S = 17$ and $S = 19$ .) The $S = 14$ solution is 1 - 9 - 4 - 8 - 2 - 7 - 5 - 6 - 3 - 10 - 1. The two $S = 16$ solutions are 1 - 8 - 7 - 6 - 3 - 4 - 9 - 2 - 5 - 10 - 1 and 1 - 8 - 7 - 6 - 3 - 9 - 4 - 2 - 10 - 5 - 1.

latchberry wrote: bobthesmartypants wrote: Let $a_1,a_2,\ldots ,a_{10}$ be a permutation of $1, 2, \ldots , 10$ where $a_1,a_2,\ldots , a_5$ are on the vertices and the other $5$ numbers are on the sides. In addition, let this minimal sum be $S$. Note that $5S = 2(a_1+a_2+\cdots +a_5)+a_6+a_7+\cdots +a_{10}$ Obviously to minimize $S$ we want as many small numbers to be $a_1,a_2,\ldots , a_5$ as possible. So let's consider the case $a_k=k$. It gives $S=14$, so we know that $S\ge 14$. I'm too lazy to find a solution for $S=14$ but it probably exists so we're done. I'm not too lazy to make my computer do it. Up to symmetry, there are only three solutions: one with $S = 14$ and two with $S = 16$. (The images of these under $a_i \rightarrow 11 - a_i$ yield solutions for $S = 17$ and $S = 19$ .) The $S = 14$ solution is 1 - 9 - 4 - 8 - 2 - 7 - 5 - 6 - 3 - 10 - 1. The two $S = 16$ solutions are 1 - 8 - 7 - 6 - 3 - 4 - 9 - 2 - 5 - 10 - 1 and 1 - 8 - 7 - 6 - 3 - 9 - 4 - 2 - 10 - 5 - 1. You are asked to provide the smallest value so its 14

latchberry wrote: bobthesmartypants wrote: Let $a_1,a_2,\ldots ,a_{10}$ be a permutation of $1, 2, \ldots , 10$ where $a_1,a_2,\ldots , a_5$ are on the vertices and the other $5$ numbers are on the sides. In addition, let this minimal sum be $S$. Note that $5S = 2(a_1+a_2+\cdots +a_5)+a_6+a_7+\cdots +a_{10}$ Obviously to minimize $S$ we want as many small numbers to be $a_1,a_2,\ldots , a_5$ as possible. So let's consider the case $a_k=k$. It gives $S=14$, so we know that $S\ge 14$. I'm too lazy to find a solution for $S=14$ but it probably exists so we're done. I'm not too lazy to make my computer do it. Up to symmetry, there are only three solutions: one with $S = 14$ and two with $S = 16$. (The images of these under $a_i \rightarrow 11 - a_i$ yield solutions for $S = 17$ and $S = 19$ .) The $S = 14$ solution is 1 - 9 - 4 - 8 - 2 - 7 - 5 - 6 - 3 - 10 - 1. The two $S = 16$ solutions are 1 - 8 - 7 - 6 - 3 - 4 - 9 - 2 - 5 - 10 - 1 and 1 - 8 - 7 - 6 - 3 - 9 - 4 - 2 - 10 - 5 - 1. Who cares about $16$?

I am so dizzy when looking at 3 dimensional combinatorics questions zzzzzzzzzdizzy!