2022 AMC 12A 第 25 题

A segment from $(0,a)$ to $(b,0)$ tangent to this circle makes $r$ the inradius or the semiperimeter of the right triangle with legs $a,b$

For the inradius case, $(a-2r)(b-2r)=2r^2,$ so positive divisors of $2r^2$ count the oriented segments

The circle centered $(r,r)$ with radius $r$ is tangent to both axes. A segment from $(0,a)$ to $(b,0)$ with $a^2+b^2=c^2$ is tangent to it when $r$ equals either the inradius $\tfrac{a+b-c}{2}$ or the semiperimeter $\tfrac{a+b+c}{2}$ of the right triangle with legs $a,b.$

In the inradius case, put $x=a-2r$ and $y=b-2r.$ Then $xy=2r^2,$ and every positive divisor $x$ determines one oriented segment, with $a=x+2r,$ $b=\frac{2r^2}{x}+2r,$ and $c=x+\frac{2r^2}{x}+2r.$ Thus there are exactly $d(2r^2)$ such segments.

For $r=1,2,3,4,5,$ these counts are $2,4,6,6,6,$ and no semiperimeter case is possible because the smallest integer right triangle has semiperimeter $6.$ At $r=6,$ $d(72)=12,$ and the $3$-$4$-$5$ triangle contributes two more oriented segments. Hence $6$ is the least possible radius and gives exactly $14$ segments.

The two semiperimeter segments have $c=5.$ In the inradius family, $c=x+\frac{72}{x}+12$ is largest at $x=1$ or $72,$ giving $c=85.$ Therefore $c_1=5,$ $c_{14}=85,$ and $\frac{c_{14}}{c_1}=17.$

Thus, the correct answer is E.