Determine whether there exist ten sets $A_1$, $A_2$, $\dots$, $A_{10}$ such that (i) each set is of the form $\{a,b,c\}$, where $a \in \{1,2,3\}$, $b \in \{4,5,6\}$, $c \in \{7,8,9\}$, (ii) no two sets are the same, (iii) if the ten sets are arranged in a circle $(A_1, A_2, \dots, A_{10})$, then any two adjacent sets have no common element, but any two non-adjacent sets intersect. (Note: $A_{10}$ is adjacent to $A_1$.)
Problem
Source: Taiwan 2014 TST1 Quiz 1, P2
Tags: combinatorics proposed, combinatorics
sqing
19.09.2014 09:37
A solution: $A_{1}=\{1,4,7\},A_{2}=\{2,5,8\},A_{3}=\{1,4,9\},A_{4}=\{2,5,7\},A_{5}=\{1,6,8\},$ $A_{6}=\{2,4,7\},A_{7}=\{1,5,9\},A_{8}=\{2,4,8\},A_{9}=\{1,5,7\},A_{10}=\{2,6,9\}.$
Radar
22.09.2014 16:59
Kindof like http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1225210&sid=7fa2e80a22f2109f7c29149645e8eeef#p1225210
MathPanda1
29.10.2014 22:40
Another solution: $ A_{1}=\{1,4,7\},A_{2}=\{2,5,8\},A_{3}=\{1,6,9\},A_{4}=\{2,4,8\},A_{5}=\{1,5,7\}, A_{6}=\{2,4,9\},A_{7}=\{1,5,8\},A_{8}=\{2,6,7\},A_{9}=\{1,4,8\},A_{10}=\{2,5,9\} $
primesarespecial
03.10.2021 14:23
This problem is a troll.It will get u fricked up if u try to use all numbers. Here's another example $(1,4,7),(2,5,8),(1,6,7),(2,4,9),(1,6,8),(2,4,7),(1,5,9),(2,6,7),(1,4,8),(2,6,9)$
mathleticguyyy
05.04.2022 16:14
1212121212 4564546546 7878978789 virgin problem solvers vs chad brute forcers
YaoAOPS
11.09.2023 17:08
Yes, take the following sets of \begin{align*} A_1 &= \{0,3,6\}, A_2 = \{1,5,8\}, A_3 = \{0,4,6\}, A_4 = \{1,3,7\}, A_5 = \{0,4,8\} \\ A_6 &= \{1,3,6\}, A_7 = \{0,5,7\}, A_8 = \{1,4,6\}, A_9 = \{0,3,8\}, A_{10} = \{1,4,7\}. \end{align*}
InterLoop
01.05.2024 23:01
feels like you've seen it before + love the bash
Yes.
$$(1, 4, 7), (2, 5, 8), (1, 4, 9), (2, 5, 7), (1, 6, 8), (2, 4, 7), (1, 5, 9), (2, 4, 8), (1, 5, 7), (2, 6, 9)$$