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Difference between revisions of "2018 AMC 10B Problems/Problem 17"

(Created page with "In rectangle <math>PQRS</math>, <math>PQ=8</math> and <math>QR=6</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and...")
 
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== Problem ==
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In rectangle <math>PQRS</math>, <math>PQ=8</math> and <math>QR=6</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and <math>D</math> lie on <math>\overline{QR}</math>, points <math>E</math> and <math>F</math> lie on <math>\overline{RS}</math>, and points <math>G</math> and <math>H</math> lie on <math>\overline{SP}</math> so that <math>AP=BQ<4</math> and the convex octagon <math>ABCDEFGH</math> is equilateral. The length of a side of this octagon can be expressed in the form <math>k+m\sqrt{n}</math>, where <math>k</math>, <math>m</math>, and <math>n</math> are integers and <math>n</math> is not divisible by the square of any prime. What is <math>k+m+n</math>?
 
In rectangle <math>PQRS</math>, <math>PQ=8</math> and <math>QR=6</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and <math>D</math> lie on <math>\overline{QR}</math>, points <math>E</math> and <math>F</math> lie on <math>\overline{RS}</math>, and points <math>G</math> and <math>H</math> lie on <math>\overline{SP}</math> so that <math>AP=BQ<4</math> and the convex octagon <math>ABCDEFGH</math> is equilateral. The length of a side of this octagon can be expressed in the form <math>k+m\sqrt{n}</math>, where <math>k</math>, <math>m</math>, and <math>n</math> are integers and <math>n</math> is not divisible by the square of any prime. What is <math>k+m+n</math>?
  
 
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106</math>
 
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106</math>
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== Solution ==
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Let <math>AP=BQ=x</math>. Then <math>AB=8-2x</math>.
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Now notice that since <math>CD=8-2x</math> we have <math>QC=DR=x-1</math>.
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Thus by the Pythagorean Theorem we have <math>x^2+(x-1)^2=(8-2x)^2</math> which becomes <math>2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}</math>.
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Our answer is <math>8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}</math>. (Mudkipswims42)
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==See Also==
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{{AMC10 box|year=2018|ab=B|num-b=16|num-a=18}}
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{{MAA Notice}}

Revision as of 14:03, 16 February 2018

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

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)

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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