Difference between revisions of "2018 AMC 10B Problems/Problem 17"
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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 Pythagoras, we find that: | 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 Pythagoras, 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> | <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> | + | 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 |
==See Also== | ==See Also== |
Revision as of 17:08, 18 February 2018
Contents
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 Pythagoras, we find that:
Since , we can say that . We can discard the negative solution, so ~ blitzkrieg21
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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