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 == | + | == 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) | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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: | ||
+ | <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> | ||
+ | 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 | ||
+ | |||
+ | == Solution 3 == | ||
+ | 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== | ||
+ | https://youtu.be/8sts_hn7cpQ | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Latest revision as of 01:21, 26 September 2020
Problem
In rectangle , and . Points and lie on , points and lie on , points and lie on , and points and lie on so that and the convex octagon is equilateral. The length of a side of this octagon can be expressed in the form , where , , and are integers and is not divisible by the square of any prime. What is ?
Solution 1
Let . Then .
Now notice that since we have .
Thus by the Pythagorean Theorem we have which becomes .
Our answer is . (Mudkipswims42)
Solution 2
Denote the length of the equilateral octagon as . The length of can be expressed as . By the Pythagorean Theorem, we find that: Since , we can say that . We can discard the negative solution, so ~ blitzkrieg21
Solution 3
Let the octagon's side length be . Then and . By the Pythagorean theorem, , so . By expanding the left side and combining the like terms, we get . Solving this using the quadratic formula, , we use , , and , to get one positive solution, , so
Video Solution
~IceMatrix
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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