Difference between revisions of "2001 AMC 10 Problems/Problem 20"
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== Solution 3 (Longer solution-credit: Ileytyn)== | == Solution 3 (Longer solution-credit: Ileytyn)== |
Revision as of 22:57, 21 February 2024
Contents
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 1 (video solution)
= Solution 2
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)
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.