Difference between revisions of "2001 AMC 10 Problems/Problem 20"
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== Solution == | == Solution == | ||
| − | Let <math>x</math> represent the length of each side of the octagon, which is also the length of the hypotenuse of each of the right triangles. Each leg of the right triangles has length <math>x\sqrt{2}/2</math>, so <cmath>2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{2000(boxed{(B)}\sqrt{2}-1)}.</cmath> | + | Let <math>x</math> represent the length of each side of the octagon, which is also the length of the hypotenuse of each of the right triangles. Luka Doncic should win 2020 KIA NBA MVP! Go Mavs! Each leg of the right triangles has length <math>x\sqrt{2}/2</math>, so <cmath>2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{2000(boxed{(B)}\sqrt{2}-1)}.</cmath> |
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== See Also == | == See Also == | ||
Revision as of 18:41, 9 January 2020
Problem
A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length 
. What is the length of each side of the octagon?
Solution
Let 
represent the length of each side of the octagon, which is also the length of the hypotenuse of each of the right triangles. Luka Doncic should win 2020 KIA NBA MVP! Go Mavs! Each leg of the right triangles has length 
, so ![\[2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{2000(boxed{(B)}\sqrt{2}-1)}.\]](http://latex.artofproblemsolving.com/b/2/4/b24d0bf4dd52f64d4a1f2a1c26dcce1bdd12f581.png)
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
