Question

对于某些实数有序对 $(a,b)$ ，方程组 $\begin{aligned}ax+by&=1,\\x^2+y^2&=50\end{aligned}$ 至少有一个解，并且每个解都是整数有序对 $(x,y)$ 。这样的有序对 $(a,b)$ 有多少个？

For certain ordered pairs $(a,b)$ of real numbers, the system of equations $\begin{aligned}ax+by&=1,\\x^2+y^2&=50\end{aligned}$ has at least one solution, and each solution is an ordered pair $(x,y)$ of integers. How many such ordered pairs $(a,b)$ are there?

答案

答案：72

知识点：格点弦切线

难度评级：2350

小提示：

列出圆 $x^2+y^2=50$ 上的所有整点

List all integer points on $x^2+y^2=50$

大提示：

既要计算经过两个非对径整点的弦，也要计算每个整点处的切线

Count both chords through two non-antipodal lattice points and tangents at one lattice point

解答：

圆上共有 $12$ 个整点，它们来自 $(\pm1,\pm7)$ 、 $(\pm5,\pm5)$ 和 $(\pm7,\pm1)$ 。一条符合条件的割线由任意两个非对径整点确定，因此共有 $\binom{12}{2}-6=60$ 条；要排除这些对径点对，因为它们所在的直线经过原点，不能写成 $ax+by=1$ 。此外，每个整点处还有一条切线，共 $12$ 条。每条直线都能唯一地规范化为 $ax+by=1$ ，所以总数为 $60+12=72$ 。

The circle has the $12$ lattice points obtained from $(\pm1,\pm7),$ $(\pm5,\pm5),$ and $(\pm7,\pm1).$ A secant satisfying the condition is determined by any two non-antipodal lattice points. This gives $\binom{12}{2}-6=60$ lines; antipodal pairs are excluded because their line passes through the origin and cannot have equation $ax+by=1.$ There are also $12$ tangents, one at each lattice point. Each line has a unique normalization $ax+by=1,$ so the total is $60+12=72.$

1994 AIME 第 7 题

答案：72

难度评级：2350

小提示：

大提示：

解答：

其他年份的第 7 题