A regular icosahedron is a regular solid of $20$ faces, each in the form of an equilateral triangle, with $12$ vertices, so that each vertex is in $5$ edges.
Twelve indistinguishable candies are glued to the vertices of a regular icosahedron (one at each vertex), and four of these twelve candies are special. André and Lucas want to together create a strategy for the following game:
• First, André is told which are the four special sweets and he must remove exactly four sweets that are not special from the icosahedron and leave the solid on a table, leaving afterwards without communicating with Lucas.
• Later, Sponchi, who wants to prevent Lucas from discovering the special sweets, can pick up the icosahedron from the table and rotate it however he wants.
• After Sponchi makes his move, he leaves the room, Lucas enters and he must determine the four special candies out of the eight that remain in the icosahedron.
Determine if there is a strategy for which Lucas can always properly discover the four special sweets.
Note the diameter of the graph is $3$, and the shortest path between $A_i$ and $A_j$ is $3$ if and only if $j=i+6$. ($A_{12+i}=A_i$ for $i\ge 1$). Now suppose the four special candies are at $X=\{A_{i_1},A_{i_2},A_{i_3},A_{i_4}\}$ We say that a set of antipodal points good if they are both elements of $X$. Obviously there are at most $2$ good pairs. We handle the cases when there are $0,1,2$ such pairs.
Case 1. There are no good pairs.
Andre chooses the antipodal points of each $x\in X$.
Case 2. There is exactly one good pair.
Without loss of generality, assume that $A_{i_3}$ and ${A_{i_4}}$ are antipodes. Clearly the distance between $A_{i_1},A_{i_2}$ is at most $2$. If the two points are adjacent, (WLOG $A_2,A_3$), Andre chooses $A_8,A_9$ and
$$A_1,A_{7} \Longleftrightarrow A_5, A_{11}$$$$A_6,A_{12}\Longleftrightarrow A_4, A_{10}$$
Case 3. There are two good pairs. ($A_{i_1}\leftrightarrow A_{i_2}, A_{i_3}\leftrightarrow A_{i_4}$)
Notice that $d(A_{i_1},A_{i_3})+d(A_{i_3},A_{i_2})\le d(A_{i_1},A_{i_2})=3$, so there exist two adjacent points $B,C$ in $X$, hence we can assume $X=\{A_2,A_3,A_8,A_9\}$. Andre chooses $A_1, A_5, A_7, A_{11}=\{A_i | \text{ there exists } B,C\in X \text{ such that } \{A_i,B,C\}\text{ is a face of the icosahedron}\}$.
Now it suffices to prove that Lucas can find the four special sweets regardless of Sponchi's move. Define $Y$ the set of four points Andre chose. Observe that the number of good pairs in $X$ is equal to the number of good pairs in $Y$.
If there are no good pairs in $Y$, Lucas chooses the antipodal points in $Y$.
If there is one good pair, Lucas chooses the antipodes of the other two points($P,Q$), and finds the other two points as well. (It's easy to do so now that Lucas knows $P,Q$.)
If there are two good pairs, there exists $P,Q,R,S\in Y$ such that $d(P,Q)=d(R,S)=2$. There exists $K,L,M,N$ such that $PKL,QKL, RMN,SMN$ are the faces of the icosahedron. One can easily prove that $K,L,M,N$ are the special candies.