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

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<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>
  
== Solution ==
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== Solution 1==
  
 
Let <math>AP=BQ=x</math>. Then <math>AB=8-2x</math>.
 
Let <math>AP=BQ=x</math>. Then <math>AB=8-2x</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)
 
Our answer is <math>8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}</math>. (Mudkipswims42)
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== Solution 2 ==
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Denote the length of the equilateral octagon as <math>x</math>. The length of <math>\overline{BQ}</math> can be expressed as <math>\frac{8-x}{2}</math>. By the Pythagorean Theorem, we find that:
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<cmath>\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}</cmath>
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Since <math>\overline{CQ}=\overline{DR}</math>, we can say that <math>x+2\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}=6\implies x=-7\pm3\sqrt{11}</math>. We can discard the negative solution, so <math>k+m+n=-7+3+11=\boxed{\textbf{(B) }7}</math> ~ blitzkrieg21
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== Solution 3 ==
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Let the octagon's side length be <math>x</math>. Then <math>PH = \frac{6 - x}{2}</math> and <math>PA = \frac{8 - x}{2}</math>. By the Pythagorean theorem, <math>PH^2 + PA^2 = HA^2</math>, so <math>\left(\frac{6 - x}{2} \right)^2 + \left(\frac{8 - x}{2} \right)^2 = x^2</math>. By expanding the left side and  combining the like terms, we get <math>\frac{x^2}{2} - 7x + 25 = x^2 \implies -\frac{x^2}{2} - 7x + 25 = 0</math>. Solving this using the quadratic formula, <math>\frac{-b \pm \sqrt{b^2-4ac}}{2a}</math>, we use <math>a = -\frac{1}{2}</math>, <math>b = -7</math>, and <math>c = 25</math>, to get one positive solution, <math>x=-7+3\sqrt{11}</math>, so <math>k+m+n=-7+3+11=\boxed{\textbf{(B) }7}</math>
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==Video Solution==
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https://youtu.be/8sts_hn7cpQ
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~IceMatrix
  
 
==See Also==
 
==See Also==

Latest revision as of 01:21, 26 September 2020

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 the Pythagorean Theorem, 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 $\left(\frac{6 - x}{2} \right)^2 + \left(\frac{8 - x}{2} \right)^2 = x^2$. By expanding the left side and combining the like terms, we get $\frac{x^2}{2} - 7x + 25 = x^2 \implies -\frac{x^2}{2} - 7x + 25 = 0$. Solving this using the quadratic formula, $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we use $a = -\frac{1}{2}$, $b = -7$, and $c = 25$, to get one positive solution, $x=-7+3\sqrt{11}$, so $k+m+n=-7+3+11=\boxed{\textbf{(B) }7}$

Video Solution

https://youtu.be/8sts_hn7cpQ

~IceMatrix

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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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