Answer: Pigeonhole on an Icosahedron

Key Idea / Intuition

The icosahedron has 12 vertices, and every vertex is surrounded by exactly 5 faces. If we assume for contradiction that no two face-sharing faces have the same label, then at every vertex the 5 surrounding faces all carry distinct nonneg integers — meaning they are at least $0,1,2,3,4$ , summing to at least $10$ . Summing over all 12 vertices overcounts each face exactly 3 times (since every face has 3 vertices). This gives a lower bound on the total sum that exceeds 39 — contradiction.

Formal Proof / Solution

Setup.
A regular icosahedron has:

20 faces, each an equilateral triangle,
12 vertices,
each vertex shared by exactly 5 faces,
each face having exactly 3 vertices.

Denote the label on face $i$ by $a_i \geq 0$ , with $\sum_{i=1}^{20} a_i = 39$ .

Assume for contradiction that no two faces sharing a vertex carry the same label.

Local constraint at each vertex.
Fix any vertex $v$ . The 5 faces meeting at $v$ form a "fan," and any two consecutive faces in this fan share an edge (hence share $v$ itself). In fact every pair among these 5 faces shares the vertex $v$ , so by assumption they must all have distinct labels.

Since they are distinct nonneg integers, the 5 labels at vertex $v$ are at least $0, 1, 2, 3, 4$ in some order, giving:

$\sum_{\text{faces } f \ni v} a_f \;\geq\; 0 + 1 + 2 + 3 + 4 = 10.$

Global count.
Sum this inequality over all 12 vertices:

$\sum_{v} \sum_{\text{faces } f \ni v} a_f \;\geq\; 12 \times 10 = 120.$

The left side counts each face $f$ once per vertex of $f$ , and since every face is a triangle it has exactly 3 vertices:

$\sum_{v} \sum_{f \ni v} a_f = \sum_{f} 3\, a_f = 3 \sum_{f} a_f = 3 \times 39 = 117.$

Contradiction.
We have derived $117 \geq 120$ , which is false.

Therefore the assumption was wrong: there exist two faces sharing a vertex with the same label. $\blacksquare$

Why the bound is tight.
The number 39 is carefully chosen: $3 \times 39 = 117 < 120 = 12 \times 10$ . If the total were 40, the argument would fail (117 would become 120, matching the bound with no contradiction). This shows the problem is tight — a beautiful instance of pigeonhole / double counting.