Question

Consider the square region $S = [0, n] \times [0, n]$ and, let $S_n$ be the set of points $(x, y) \in S$, where $n, x, y$ are integers. Let $L_n$ be the set of all lines passing through at least two distinct points from $S_n$. Suppose we choose a random line $l$ from $L_n$. Let $P_n$ be the probability that $l$ is tangent to the circle $x^2 + y^2 = n^2 \left(1 + \left(1 - \frac{1}{\sqrt{n}}\right)^2 \right)$. Then $\lim_{n \to \infty} P_n$ is

• A. 0
• B. 1
• C. $1/\pi$
• D. $1/\sqrt{2}$

Answer 1

Answer: (A)

0

Answer 2

Let $S_n = \{(x, y) \in \mathbb{Z}^2 \mid 0 \le x \le n, 0 \le y \le n\}$. The number of points in $S_n$ is $(n+1)^2$. $L_n$ is the set of all lines passing through at least two distinct points from $S_n$. We are choosing a random line $l$ from $L_n$. The total number of lines $|L_n|$ for large $n$ is dominated by lines passing through exactly two points. The number of pairs of points is $\binom{(n+1)^2}{2} \approx \frac{n^4}{2}$. A line passing through $(x_1, y_1)$ and $(x_2, y_2)$ with $x_1 \ne x_2$ has slope $m = \frac{y_2 - y_1}{x_2 - x_1}$. Let $p = y_2 - y_1$ and $q = x_2 - x_1$. $|p| \le n$, $|q| \le n$, $q \ne 0$. The slope is $p/q$. We can write this in reduced form $p'/q'$ where $\gcd(|p'|, |q'|)=1$. The equation of such a line can be written as $p'x - q'y = c$ (if $q' \ne 0$) or $x = c$ (if $q'=0$, then $p'=\pm 1$). If $q'=0$, then $x_1=x_2$. The points are $(x_1, y_1)$ and $(x_1, y_2)$. $0 \le x_1 \le n$, $0 \le y_1, y_2 \le n$. These are vertical lines $x=k$ for $k \in \{0, 1, \dots, n\}$. There are $n+1$ such lines. If $p'=0$, then $y_1=y_2$. The points are $(x_1, y_1)$ and $(x_2, y_1)$. $0 \le y_1 \le n$, $0 \le x_1, x_2 \le n$. These are horizontal lines $y=k$ for $k \in \{0, 1, \dots, n\}$. There are $n+1$ such lines. If $p' \ne 0$ and $q' \ne 0$: The slope is $p'/q'$. We can assume $q' > 0$ and $p' \in \mathbb{Z}$. $\gcd(|p'|, q')=1$. The range of slopes $p/q$ is $p \in [-n, n]$, $q \in [-n, n] \setminus \{0\}$. In reduced form $p'/q'$, we have $|p'| \le n$ and $1 \le q' \le n$. The number of lines $ax+by=c$ passing through $S_n$ with $\gcd(|a|,|b|)=1$ can be estimated for large $n$. The number of lines is dominated by those with small $|a|$ and $|b|$. For $a=p', b=-q'$, the line is $p'x - q'y = c$. The range of $c$ for $(x,y) \in S_n$ is $[p' \cdot 0 - q' \cdot n, p' \cdot n - q' \cdot 0]$ if $p'>0, q'>0$, which is $[-q'n, p'n]$. The number of integer values of $c$ is $p'n+q'n+1$. The total number of lines $|L_n|$ is approximately $\sum_{q'=1}^n \sum_{|p'| \le n, \gcd(|p'|,q')=1} (p'n+q'n)$. This sum is roughly $O(n^3 \sum_{q=1}^n \phi(q)/q) = O(n^4)$. A more precise estimate for the number of lines passing through at least two points in an $n \times n$ grid is $O(n^4)$.

The circle is $x^2 + y^2 = R_n^2$, where $R_n^2 = n^2 \left(1 + \left(1 - \frac{1}{\sqrt{n}}\right)^2 \right) = n^2 \left(1 + 1 - \frac{2}{\sqrt{n}} + \frac{1}{n}\right) = n^2 \left(2 - \frac{2}{\sqrt{n}} + \frac{1}{n}\right) = 2n^2 - 2n^{3/2} + n$. The radius is $R_n = n \sqrt{2 - \frac{2}{\sqrt{n}} + \frac{1}{n}}$. For large $n$, $R_n \approx n\sqrt{2}$. A line $ax+by=c$ is tangent to the circle $x^2+y^2=R_n^2$ if the distance from the origin $(0,0)$ to the line is $R_n$. The distance is $|c|/\sqrt{a^2+b^2}$. So, the condition for tangency is $c^2 = (a^2+b^2) R_n^2$.

For vertical lines $x=k$, $a=1, b=0$. $c=k$. $k^2 = (1^2+0^2) R_n^2 = R_n^2$. $k = \pm R_n$. Since $k \in [0, n]$, $k=R_n$. $R_n \approx n\sqrt{2} > n$ for large $n$. So there are no vertical lines tangent to the circle for large $n$. For horizontal lines $y=k$, $a=0, b=1$. $c=k$. $k^2 = (0^2+1^2) R_n^2 = R_n^2$. $k = \pm R_n$. Since $k \in [0, n]$, $k=R_n$. Again, no horizontal lines tangent to the circle for large $n$.

For lines $p'x - q'y = c$ with $p'>0, q'>0, \gcd(p', q')=1$. $a=p', b=-q'$. $c^2 = (p'^2+q'^2) R_n^2$. $c^2 = (p'^2+q'^2) n^2 (2 - 2/\sqrt{n} + 1/n)$. $c$ must be an integer in $[-q'n, p'n]$. $|c| \le \max(p'n, q'n)$. $c^2 \le (\max(p', q') n)^2 = (\max(p', q'))^2 n^2$. $(p'^2+q'^2) n^2 (2 - 2/\sqrt{n} + 1/n) \le (\max(p', q'))^2 n^2$. $(p'^2+q'^2) (2 - 2/\sqrt{n} + 1/n) \le (\max(p', q'))^2$. As $n \to \infty$, $2 - 2/\sqrt{n} + 1/n \to 2$. $(p'^2+q'^2) \cdot 2 \le (\max(p', q'))^2$. If $p' \ge q'$, $2(p'^2+q'^2) \le p'^2 \implies p'^2 + 2q'^2 \le 0$. This is only possible if $p'=q'=0$, which is not allowed. If $q' \ge p'$, $2(p'^2+q'^2) \le q'^2 \implies 2p'^2 + q'^2 \le 0$. This is only possible if $p'=q'=0$. This suggests that for large $n$, there are no lines with fixed $p', q'$ that are tangent to the circle.

However, we must consider lines with $p', q'$ that might grow with $n$. The condition $c^2 = (p'^2+q'^2) R_n^2$ with $c \in [-q'n, p'n]$ implies $|c| \le \max(p'n, q'n)$. $|c| = \sqrt{p'^2+q'^2} R_n = \sqrt{p'^2+q'^2} n \sqrt{2 - 2/\sqrt{n} + 1/n}$. $\sqrt{p'^2+q'^2} n \sqrt{2 - 2/\sqrt{n} + 1/n} \le \max(p', q') n$. $\sqrt{p'^2+q'^2} \sqrt{2 - 2/\sqrt{n} + 1/n} \le \max(p', q')$. As $n \to \infty$, $\sqrt{p'^2+q'^2} \sqrt{2} \le \max(p', q')$, which is impossible for $p', q' > 0$.

Let's re-examine the circle equation. The radius $R_n$ is $n \sqrt{1+(1-1/\sqrt{n})^2}$. $R_n^2 = n^2(1 + 1 - 2/\sqrt{n} + 1/n) = 2n^2 - 2n^{3/2} + n$. The tangent condition is $c^2 = (p'^2+q'^2) R_n^2$. $c$ must be an integer and $|c| \le \max(p'n, q'n)$. Let's consider the lines $x+y=c$. $p'=1, q'=1$. $a=1, b=1$. $\gcd(1,1)=1$. $c^2 = (1^2+1^2) R_n^2 = 2 R_n^2 = 2 (2n^2 - 2n^{3/2} + n) = 4n^2 - 4n^{3/2} + 2n$. $c = \pm \sqrt{4n^2 - 4n^{3/2} + 2n}$. For large $n$, $c \approx \pm 2n$. For $x+y=c$, $x,y \in [0,n]$, $c \in [0, 2n]$. $c = \sqrt{4n^2 - 4n^{3/2} + 2n} = 2n \sqrt{1 - 1/\sqrt{n} + 1/(2n)} \approx 2n (1 - 1/(2\sqrt{n})) = 2n - \sqrt{n}$. For large $n$, $2n-\sqrt{n}$ is close to an integer. For example, if $n=m^2$, $2m^2-m$. $c = 2n - \sqrt{n} (2 - 1/\sqrt{n})$. If $2n-\sqrt{n}$ is an integer, then $x+y=2n-\sqrt{n}$ is a line passing through $S_n$ if there exist $x, y \in [0, n]$ with $x+y=2n-\sqrt{n}$. $x=n, y=n-\sqrt{n}$ is not in $S_n$ if $\sqrt{n}$ is not integer. If $n$ is a perfect square, $n=k^2$, $R_n^2 = k^4(1+(1-1/k)^2) = k^4(1+1-2/k+1/k^2) = 2k^4-2k^3+k^2$. $c^2 = 2 R_n^2 = 4k^4-4k^3+2k^2$. $c = \sqrt{4k^4-4k^3+2k^2}$. The line $x+y=c$ has $c \in [0, 2n]$. Consider lines $x+y=2n-k$ for small integer $k \ge 0$. $x+y=2n-k$ passes through $S_n$ if $x,y \in [0,n]$. $x=n, y=n-k$. This point is in $S_n$ if $0 \le n-k \le n$, which means $0 \le k \le n$. $c=2n-k$. $(2n-k)^2 = 4n^2-4nk+k^2$. We want $(2n-k)^2 = 4n^2 - 4n^{3/2} + 2n$. $4n^2-4nk+k^2 = 4n^2 - 4n^{3/2} + 2n$. $-4nk+k^2 = -4n^{3/2} + 2n$. For large $n$, $4nk \approx 4n^{3/2}$, so $k \approx \sqrt{n}$. Let $k = \sqrt{n}$. Then $-4n\sqrt{n} + n = -4n^{3/2} + 2n$. This implies $n=0$, not possible. Let $k = \sqrt{n} + \delta$. $-4n(\sqrt{n}+\delta) + (\sqrt{n}+\delta)^2 = -4n^{3/2} + 2n$. $-4n^{3/2}-4n\delta + n + 2\delta\sqrt{n} + \delta^2 = -4n^{3/2} + 2n$. $-4n\delta + n + 2\delta\sqrt{n} + \delta^2 = 2n$. $-4n\delta + 2\delta\sqrt{n} + \delta^2 = n$. $\delta(-4n+2\sqrt{n}) + \delta^2 = n$. $\delta \approx -n/(4n) = -1/4$. So $k \approx \sqrt{n}-1/4$. The closest integer $c$ to $R_n\sqrt{2}$ is $2n-\lfloor\sqrt{n}\rfloor$ or $2n-\lceil\sqrt{n}\rceil$. Let $k_0 = \lfloor\sqrt{n}\rfloor$. $c_0 = 2n-k_0$. $(2n-k_0)^2 = 4n^2 - 4n k_0 + k_0^2$. $c_0^2 - R_n^2 = 4n^2 - 4n k_0 + k_0^2 - (4n^2 - 4n^{3/2} + 2n) = -4nk_0 + k_0^2 + 4n^{3/2} - 2n$. $k_0 \approx \sqrt{n}$. $k_0^2 \approx n$. $-4n\sqrt{n} + n + 4n^{3/2} - 2n = -n$. $c_0^2 = R_n^2 - n$. $c_0 = \sqrt{R_n^2-n}$. Distance from origin $|c_0|/\sqrt{2} = \sqrt{R_n^2-n}/\sqrt{2}$. We want this to be $R_n$. $\sqrt{R_n^2-n}/\sqrt{2} = R_n \implies R_n^2-n = 2R_n^2 \implies R_n^2=-n$, impossible.

The lines tangent to the circle for large $n$ are those whose distance to the origin is $R_n$. $|c|/\sqrt{p'^2+q'^2} = R_n$. The number of lines through $S_n$ with $a=p', b=-q'$ is $p'n+q'n+1$. The total number of lines $|L_n|$ is approximately $\frac{6}{\pi^2} n^4$. The number of tangent lines: $c^2 = (p'^2+q'^2)R_n^2$. $c$ must be an integer. $c^2 \approx 2n^2(p'^2+q'^2)$. $c \approx n\sqrt{2(p'^2+q'^2)}$. $c$ must be in $[-q'n, p'n]$. $n\sqrt{2(p'^2+q'^2)} \le \max(p'n, q'n)$. $\sqrt{2(p'^2+q'^2)} \le \max(p', q')$. Impossible for $p',q'>0$.

This suggests that the number of tangent lines is small compared to the total number of lines. Consider lines $px-qy=c$ where $p,q$ are small integers. $|c| = R_n \sqrt{p^2+q^2} = n \sqrt{1+(1-1/\sqrt{n})^2} \sqrt{p^2+q^2} \approx n \sqrt{2(p^2+q^2)}$. $c$ is an integer. For $c$ to be an integer, $n^2 \cdot 2(p^2+q^2)$ must be close to a perfect square. $c \in [-qn, pn]$. Number of lines with parameters $(p,q)$ is $pn+qn+1$. Total number of lines is $O(n^4)$.

Consider lines $x=k$ and $y=k$. For $k \in \{0, \dots, n\}$. $2(n+1)$ lines. $R_n \approx n\sqrt{2}$. $k=R_n \approx n\sqrt{2} > n$. No tangent lines among these for large $n$.

Consider lines $x+y=k$. $k \in \{0, \dots, 2n\}$. $2n+1$ lines. $a=1, b=1$. $k^2 = (1^2+1^2)R_n^2 = 2R_n^2 = 4n^2 - 4n^{3/2} + 2n$. $k = \sqrt{4n^2 - 4n^{3/2} + 2n} \approx 2n - \sqrt{n}$. For large $n$, $2n-\sqrt{n}$ is not an integer. The closest integers are $\lfloor 2n-\sqrt{n} \rfloor$ and $\lceil 2n-\sqrt{n} \rceil$. Let $k_1 = \lfloor 2n-\sqrt{n} \rfloor$, $k_2 = \lceil 2n-\sqrt{n} \rceil$. These lines $x+y=k_1$ and $x+y=k_2$ pass through $S_n$ as $k_1, k_2 \in [0, 2n]$. The distance from origin is $k/\sqrt{2}$. For tangency, $k=\sqrt{2}R_n$. $k^2 = 2R_n^2$. $k^2 = 4n^2 - 4n^{3/2} + 2n$. The values of $k$ for which $x+y=k$ is tangent must satisfy $k^2=4n^2 - 4n^{3/2} + 2n$. As $n \to \infty$, $k/\sqrt{2} \to n\sqrt{2}$. $k \to 2n$. The number of lines tangent to the circle is very small compared to the total number of lines. For large $n$, $R_n \approx n\sqrt{2}$. The circle $x^2+y^2=2n^2$. The lines $x=n, y=n, x=-n, y=-n$ are tangent to $x^2+y^2=n^2$. The circle $x^2+y^2=2n^2$ is tangent to $x+y=\pm 2n$ and $x-y=\pm 2n$ (not in $S_n$). $x+y=2n$ passes through $(n,n)$. $x+y=0$ passes through $(0,0)$.

The number of lines in $L_n$ is roughly $C n^4$ for some constant $C$. The number of lines $px-qy=c$ tangent to $x^2+y^2=R_n^2$ is $c^2=(p^2+q^2)R_n^2$. $c^2 \approx 2n^2(p^2+q^2)$. The number of integer solutions $(p,q,c)$ to $c^2 \approx 2n^2(p^2+q^2)$ with $0<|p|,|q| \le n, \gcd(|p|,|q|)=1, |c| \le \max(|p|,|q|)n$. The number of pairs $(p,q)$ with $1 \le p,q \le n, \gcd(p,q)=1$ is $\sum_{q=1}^n \phi(q) \approx \frac{6}{\pi^2} \frac{n^2}{2}$. For each such $(p,q)$, $c^2 = (p^2+q^2)R_n^2$. $c = \pm R_n \sqrt{p^2+q^2}$. For $c$ to be an integer, $(p^2+q^2)R_n^2$ must be a perfect square. $(p^2+q^2) n^2 (1+(1-1/\sqrt{n})^2)$ must be a perfect square. For large $n$, this is approximately $2n^2(p^2+q^2)$. For this to be a square, $2(p^2+q^2)$ must be a square. Let $2(p^2+q^2)=k^2$. $k$ must be integer. This is related to sum of two squares. $p^2+q^2$ must be of the form $m^2$ or $2m^2$. $2(p^2+q^2)$ is a square if $p^2+q^2$ is of the form $2m^2$. Solutions to $p^2+q^2=2m^2$ with $\gcd(p,q)=1$: $p=q=1 \implies p^2+q^2=2$. $2(p^2+q^2)=4=2^2$. So $(p,q)=(1,1)$ gives $c^2 = 2R_n^2$. $c = \pm R_n\sqrt{2}$. $c = \pm n\sqrt{2(1+(1-1/\sqrt{n})^2)} = \pm n\sqrt{4-4/\sqrt{n}+2/n}$. $c = \pm (2n - \sqrt{n} + O(1))$. The closest integer values for $c$ are $2n-\lfloor\sqrt{n}\rfloor$ and $2n-\lceil\sqrt{n}\rceil$ for $p=q=1$. These two lines $x+y=c$ are candidates for tangent lines. The number of lines is $O(n^4)$. The number of tangent lines is $O(1)$ for large $n$. $P_n = \frac{\text{Number of tangent lines}}{\text{Total number of lines}}$. The number of tangent lines is finite for large $n$. The total number of lines tends to infinity as $O(n^4)$. Thus, $\lim_{n \to \infty} P_n = \lim_{n \to \infty} \frac{O(1)}{O(n^4)} = 0$.

The final answer is $\boxed{0}$.