Question

In rectangle $PQRS,$ $PQ = 8$ and $QR = 6.$ Points $A$ and $B$ lie on $PQ,$ points $C$ and $D$ lie on $QR,$ points $E$ and $F$ lie on $RS,$ and points $G$ and $H$ lie on $SP$ so that $AP = BQ < 4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k + m\sqrt{n},$ where $k,$ $m,$ and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k + m + n?$

$1$

$7$

$21$

$92$

$106$

Answer: B

Concepts:Pythagorean Theorem quadratic

Difficulty rating: 1890

Solution:

Let $s$ be the octagon's side length and let $x=AP=BQ=(8-s)/2.$ 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 $(8-s)/2=x.$ 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 $x=(15-3\sqrt{11})/2.$

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

2018 AMC 10B Problem 17

Answer: B

Difficulty rating: 1890

Solution:

Problem 17 in Other Years