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2018 AMC 10B Problems/Problem 17

Revision as of 12:52, 27 January 2019 by Rejas (talk | contribs) (→‎Solution 2)

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: 

\[\left(\frac{8-x}{2}\right)^2+\overline{CQ}^2=x^2\implies \overline{CQ}=\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}\]

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 $k+m+n=-7+3+11=\boxed{\textbf{(B) }7}$ ~ blitzkrieg21

Solution 3

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 $(\frac{6 - x}{2})^2 + (\frac{8 - x}{2})^2 = x^2$. Solving this, we get one positive solution, $x=-7+3\sqrt{11}$, so $k+m+n=-7+3+11=\boxed{\textbf{(B) }7}$

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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