Difference between revisions of "1957 AHSME Problems/Problem 31"
(diagram) |
(→Solution) |
||
| Line 24: | Line 24: | ||
</asy> | </asy> | ||
| − | <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 18:27, 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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
