2018 AMC 10B 第 17 题

The four cut corners are congruent right triangles with legs $x$ (on the sides of length $8$) and $y$ (on the sides of length $6$)

The equal sides give $8 - 2x = 6 - 2y = \sqrt{x^2 + y^2};$ the first equality gives $y = x - 1,$ then substitute and square

Let $s$ be the octagon’s side length and let $x=AP=BQ=\frac{8-s}{2}.$ The right triangles $\triangle APH$ and $\triangle BQC$ have the same hypotenuse $s$ and a leg of length $x,$ so they are congruent; write $PH=QC=y.$ Because $CD=HG=s$ and the vertical sides of the rectangle both have length $6,$ it follows that $DR=GS.$ The right triangles at $R$ and $S$ are then congruent, so $RE=SF.$ Since $RS=8$ and $EF=s,$ each of these equal lengths is $\frac{8-s}{2}=x.$ Thus all four cut corners have legs $x$ and $y.$

The equal octagon sides give $$8-2x=6-2y=\sqrt{x^2+y^2}.$$ The first equality gives $y=x-1.$ Substituting and squaring gives $2x^2-30x+63=0,$ so the root with $x<4$ is $x=\frac{15-3\sqrt{11}}{2}.$

The side length is $8-2x=-7+3\sqrt{11},$ so $k+m+n=-7+3+11=7.$ Thus, B is the correct answer.