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Home » FÃ¼r $ x = 5 $: $ x \mid y \Rightarrow y \mid 5 $? - Portal da AcÃºstica

FÃ¼r $ x = 5 $: $ x \mid y \Rightarrow y \mid 5 $? - Portal da AcÃºstica

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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📌 But $I$ must be an integer, and $9\sqrt{3} \approx 9 \cdot 1.732 = 15.588$, so $I \approx 15.588 - 8 = 7.588$, not an integer. Contradiction—so our assumption about lattice structure must be reconsidered.

📌 Instead, interpret the problem as a combinatorial lattice: suppose the triangle is discretized into a triangular grid with $6$ units per side, and points are placed only when coordinates satisfy $x + y + z = 6$, $x,y,z \geq 0$ integers—this is a standard triangular lattice. But the triangle is equilateral in Euclidean sense, embedded in a grid.

📌 From combinatorial geometry, the number of lattice points strictly inside an equilateral triangle of side length $n$ (in unit triangles) is approximately $\frac{(n-1)^2}{2}$, but a known formula for a large equilateral triangle of side $n$ with integer vertices and area $\frac{\sqrt{3}}{4}n^2$ is:

📌 Number of interior points (in unit triangular lattice): $ \frac{(n-2)(n-1)}{2} $ for $n \geq 2$. But this gives approximations.

📌 Alternatively, for a large equilateral triangle of side $n$ in a triangular lattice (where each small triangle has side 1), the number of lattice points strictly inside is:

📌 I = \frac{(n-1)(n-2)}{2}

📌 For $n = 6$:

📌 I = \frac{(6-1)(6-2)}{2} = \frac{5 \cdot 4}{2} = 10

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