1957 AHSME Problems/Problem 31: Difference between revisions
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<math>\boxed{\textbf{(B) }\frac{2-\sqrt{2}}2}</math>. | Let the side length of the regular octagon be <math>s</math>, and let the length of the legs of the isosceles right triangles be <math>x</math>. The triangles are [[45-45-90|<math>45-45-90</math>]] triangles, so <math>s=x\sqrt2</math>. Because each side of the square is length <math>1</math> and is composed of two legs of the triangles and one side of the octagon, <math>2x+s=1</math>. Substituting <math>s=x\sqrt2</math> into this equation, we can now solve for <math>x</math> to get our 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 <math>\boxed{\textbf{(B) }\frac{2-\sqrt{2}}2}</math>. | |||
== See Also == | == See Also == | ||
Revision as of 17:27, 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)
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 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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| All AHSME Problems and Solutions | ||
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