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Difference between revisions of "1957 AHSME Problems/Problem 31"

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== Solution ==
 
== Solution ==
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<math>\boxed{\textbf{(B) }\frac{2-\sqrt{2}}2}</math>.
 
  
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<asy>
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 +
real x = 8*(2-sqrt(2))/2;
 +
 +
// Square
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draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0));
 +
 +
// Corners
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draw((0,x)--(x,0));
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draw((8-x,0)--(8,x));
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draw((8,8-x)--(8-x,8));
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draw((x,8)--(0,8-x));
 +
 +
</asy>
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 +
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 desired answer:
 +
\begin{align*}
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2x+s &= 1 \\
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2x+x\sqrt2 &= 1 \\
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x(2+\sqrt2) &= 1 \\
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x &= \frac{1}{2+\sqrt2} \cdot \frac{2-\sqrt2}{2-\sqrt2} \\
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x &= \frac{2-\sqrt2}{4-2} = \frac{2-\sqrt2}{2}
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\end{align*}
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Thus, our answer is <math>\boxed{\textbf{(B) }\frac{2-\sqrt{2}}2}</math>.
  
 
== See Also ==
 
== See Also ==

Latest 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:

$\textbf{(A)}\ \frac{2 + \sqrt{2}}{3} \qquad \textbf{(B)}\ \frac{2 - \sqrt{2}}{2}\qquad \textbf{(C)}\ \frac{1+\sqrt{2}}{2}\qquad\textbf{(D)}\ \frac{1+\sqrt{2}}{3}\qquad\textbf{(E)}\ \frac{2-\sqrt{2}}{3}$


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]

Let the side length of the regular octagon be $s$, and let the length of the legs of the isosceles right triangles be $x$. The triangles are $45-45-90$ triangles, so $s=x\sqrt2$. Because each side of the square is length $1$ and is composed of two legs of the triangles and one side of the octagon, $2x+s=1$. Substituting $s=x\sqrt2$ into this equation, we can now solve for $x$ 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 $\boxed{\textbf{(B) }\frac{2-\sqrt{2}}2}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 30 		Followed by
Problem 32
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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