2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 17 Medium

In rectangle PQRS, PQ=8 and QR=6. Points A and B lie on \overline {PQ}, points C and D lie on \overline {QR}, points E and F lie on \overline {RS}, and points G and H lie on \overline {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? (2018 AMC 10B Problem, Question#17)

  • A.

    1

  • B.

    7

  • C.

    21

  • D.

    92

  • E.

    106

Answer:B

Let AP=BQ+x. Then AB=8-2x.

Now notice that since CD=8-2x we have QC=DR=x-1.

Thus by the Pythagorean Theorem we have x^2+(x-1)^2=(8-2x)^2 which becomes 2x^2-30x+63=0\Rightarrow x=\frac{15-3\sqrt{11}}{2}.

Our answer is 8-(15-3\sqrt{11})=3\sqrt{11}-7\Rightarrow \rm (B) \ 7.

Denote the length of the equilateral octagon as x. The length of \overline{BQ} can be expressed as \frac{8-x}{2}. By the Pythagorean Theorem, we find that: \left(\frac{8-x}{2}\right)^2+\overline{CQ}^2=x^2\Rightarrow\overline{CQ}=\sqrt{x^2-\left(\dfrac{8-x}{2}\right)^2},

Since \overline{CQ}=\overline{DR} , we can say that x+2\sqrt{x^2-\left(\dfrac{8-x}{2}\right)^2}=6\Rightarrow x=-7\pm3\sqrt{11} . We can discard the negative solution, so k+m+n=-7+3+11=\rm (B) \ 7.

Let the octagon's side length be x. Then PH=\frac{6-x}{2} and PA=\frac{8-x}{2}. By the Pythagorean theorem, PH^2+PA^2=HA^2, so \left(\frac{6-x}{2}\right)^2+\left(\frac{8-x}{2}\right)^2=x^2. Solving this, we get one positive solution, x=-7\pm3\sqrt{11}, so k+m+n=-7+3+11=\rm (B) \ 7.

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