2018 AMC 10B Problem 17

Problem 17 of 25IntermediateAlgebra Geometry

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?

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Solution

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

x=\frac{15-3\sqrt{11}}{2}.

The side length is

8-2x=-7+3\sqrt{11},

k+m+n=-7+3+11=7.

Thus, B is the correct answer.

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