2001 AMC 10 Problems/Problem 20
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 1 (video 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. Each leg of the right triangles has length , so
Solution 3 (Longer solution-credit: Ileytyn)
First, realize that each triangle is congruent, a right triangle and that the two legs are equal. Also, each side of the octagon is equal, because of the definition of regular shapes. Let be the length of a leg of the isosceles right triangle. In terms of , the hypotenuse of the isosceles right triangle, which is also the length of a side of the regular octagon, is . Since the length of each side of the square is 2000, the length of each side of the regular octagon is equal to the length of a side of the square () subtracted by times the length of a leg of the isosceles right triangle ( the total length of the side is , being the length of a side of the regular octagon), which is the same as . As an expression, this is , which we can equate to , ( since the octagon is regular, meaning all of the side's lengths are congruent) giving us the following equation:. By isolating the variable and simplifying the right side, we get the following: . Dividing both sides by , we arrive with , now, to find the length of the side of the octagon, we can plug in and use the equation , being the length of a side of the octagon, to derive the value of a side of the octagon. After plugging in the values, we derive , which is the same as , factoring out a , we derive the following: , by rationalizing the denominator of , we get , after expanding, finally, we get !(not a factorial symbol, just an exclamation point)
|2001 AMC 10 (Problems • Answer Key • Resources)|
|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.