1957 AHSME Problems/Problem 31
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
Let the side length of the regular octagon be , and let the length of the legs of the isosceles right triangles be . The triangles are triangles, so . Because each side of the square is length and is composed of two legs of the triangles and one side of the octagon, . Substituting into this equation, we can now solve for to get our desired answer: \begin{align*} 2x+s &= 1 \\ 2x+x\sqrt2 &= 1 \\ x(2+\sqrt2) &= 1 \\ x &= \frac{1}{2+\sqrt2} \cdot \frac{2-\sqrt2}{2-\sqrt2} \\ x &= \frac{2-\sqrt2}{4-2} = \frac{2-\sqrt2}{2} \end{align*} Thus, our answer is .
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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