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
![[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)
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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