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2018 AMC 10B Problem 17

Problem 17 of 25IntermediateAlgebraGeometry

In rectangle PQRS,PQRS, PQ=8PQ = 8 and QR=6.QR = 6. Points AA and BB lie on PQ,PQ, points CC and DD lie on QR,QR, points EE and FF lie on RS,RS, and points GG and HH lie on SPSP so that AP=BQ<4AP = BQ < 4 and the convex octagon ABCDEFGHABCDEFGH is equilateral. The length of a side of this octagon can be expressed in the form k+mn,k + m\sqrt{n}, where k,k, m,m, and nn are integers and nn is not divisible by the square of any prime. What is k+m+n?k + m + n?

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Solution

Let ss be the octagon’s side length and let x=AP=BQ=8−s2.x=AP=BQ=\frac{8-s}{2}. The right triangles △APH\triangle APH and △BQC\triangle BQC have the same hypotenuse ss and a leg of length x,x, so they are congruent; write PH=QC=y.PH=QC=y. Because CD=HG=sCD=HG=s and the vertical sides of the rectangle both have length 6,6, it follows that DR=GS.DR=GS. The right triangles at RR and SS are then congruent, so RE=SF.RE=SF. Since RS=8RS=8 and EF=s,EF=s, each of these equal lengths is 8−s2=x.\frac{8-s}{2}=x. Thus all four cut corners have legs xx and y.y. The equal octagon sides give 8−2x=6−2y=x2+y2.8-2x=6-2y=\sqrt{x^2+y^2}. The first equality gives y=x−1.y=x-1. Substituting and squaring gives 2x2−30x+63=0,2x^2-30x+63=0, so the root with x<4x<4 is x=15−3112.x=\frac{15-3\sqrt{11}}{2}. The side length is 8−2x=−7+311,8-2x=-7+3\sqrt{11}, so k+m+n=−7+3+11=7.k+m+n=-7+3+11=7. Thus, B is the correct answer.
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Tagged: Pythagorean Theorem · quadratic

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