Difference between revisions of "1957 AHSME Problems/Problem 31"
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== Solution == | == Solution == | ||
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| + | <asy> | ||
| + | |||
| + | real x = 8*(2-sqrt(2))/2; | ||
| + | |||
| + | // Square | ||
| + | draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0)); | ||
| + | |||
| + | // Corners | ||
| + | draw((0,x)--(x,0)); | ||
| + | draw((8-x,0)--(8,x)); | ||
| + | draw((8,8-x)--(8-x,8)); | ||
| + | draw((x,8)--(0,8-x)); | ||
| + | |||
| + | </asy> | ||
| + | |||
<math>\boxed{\textbf{(B) }\frac{2-\sqrt{2}}2}</math>. | <math>\boxed{\textbf{(B) }\frac{2-\sqrt{2}}2}</math>. | ||
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== See Also == | == See Also == | ||
Revision as of 17:18, 25 July 2024
Problem
A regular octagon is to be formed by cutting equal isosceles right triangles from the corners of a square. If the square has sides of one unit, the leg of each of the triangles has length:
Solution
![[asy] real x = 8*(2-sqrt(2))/2; // Square draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0)); // Corners draw((0,x)--(x,0)); draw((8-x,0)--(8,x)); draw((8,8-x)--(8-x,8)); draw((x,8)--(0,8-x)); [/asy]](http://latex.artofproblemsolving.com/e/3/6/e3683cda4f3d70ffd3c8ba09e07b22d4c27955cc.png)
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See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
