Author: Tinku Tara

Question Number 219869 by universe last updated on 02/May/25 Answered by MrGaster last updated on 03/May/25 Commented by MrGaster last updated on 03/May/25 Original text:\[ \begin{aligned} &\text{Solution:}\\ &\text{Let the equation of the circle be}C:x^2+y^2=n^2\left(2-\frac{2}{\sqrt{n}}+\frac{1}{n}\right).\text{As}n\to\infty,\text{the radius}R\sim n\sqrt{2}.\\ &\text{For a lattice line}l:ax+by+c=0,\text{its distance to the origin is}\frac{|c|}{\sqrt{a^2+b^2}}=R,\text{i.e.,}|c|=R\sqrt{a^2+b^2}.\\ &\text{Since}R\sim n\sqrt{2},\text{for sufficiently large}n,\text{we have}|c|\sim n\sqrt{2(a^2+b^2)}.\\ &\text{Consider that the line}l\text{must pass through at least two lattice points}(x_1,y_1),(x_2,y_2)\in S_n.\text{Their parameters satisfy:}\\ &\text{(1)The slope}m=\frac{y_2-y_1}{x_2-x_1}\text{is a rational number,and the intercept}c=y_1-mx_1\text{is an integer combination.}\\ &\text{(2)The intercept range is limited by}0\leq y_1,y_2\leq n,\text{so the upper bound of}|c|\text{is}O(n).\\ &\text{As}n\to\infty,\sqrt{a^2+b^2}\sim\sqrt{m^2+1}\text{(let}m=\frac{b}{a}\text{)},\text{then}|c|\sim n\sqrt{2(m^2+1)}.\\ &\text{However,the allowed range of}|c|\text{is}O(n),\text{so we need}\sqrt{2(m^2+1)}\leq 1,\text{i.e.,}m^2\leq\frac{1}{2}-1,\text{which has no real solutions.}\\ &\text{Therefore,the number of tangent lines}N{\text{tangent}}=o(N{\text{total}}).\\ &\text{The total number of lines}N{\text{total}}\sim\frac{3}{\pi^2}n^4\text{(based on the asymptotic estimate of lattice lines)},\text{and the number of tangent lines}N{\text{tangent}}\sim O(n^2)\text{(only a limited number of directions satisfy the intercept condition)}.\\ &\text{Thus,the probability}P_n=\frac{N{\text{tangent}}}{N{\text{total}}}\sim\frac{O(n^2)}{n^4}\to 0.\\ &\text{The final limit is:}\\ &\boxed{A} \end{aligned} \] Terms…