2018 AMC 10B Problems/Problem 17

Problem

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 $k+m+n$ ?

$\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106$

Solution 1

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 $2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}$ .

Our answer is $8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}$ . (Mudkipswims42)

Solution 2

Denote the length of the equilateral octagon as $x$ . The length of $\overline{BQ}$ can be expressed as $\frac{8-x}{2}$ . By Pythagoras, we find that:

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

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2018 AMC 10B Problems/Problem 17

Contents

Problem

Solution 1

Solution 2

See Also