Difference between revisions of "2011 AMC 10B Problems/Problem 16"
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<cmath> \frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}</cmath> | <cmath> \frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}</cmath> | ||
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+ | Explanation: | ||
+ | The area of the octagon consists of the area of the triangles, the rectangles, and the square in the middle. Assume the octagon has side length <math>1</math>. The triangles are right isosceles triangles with the hypotenuse 1, so their side length is <math>\frac{\sqrt{2}}{2}</math> and the area of one triangle is <math>\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{1}{4}</math>. The area of all 4 triangles is then just <math>1</math>. The rectangles share one side with the octagon, and another with the triangle. The octagon has side length 1, and the triangle has side length <math>\frac{\sqrt{2}}{2}</math> as found earlier. So the area of one rectangle is <math>\frac{\sqrt{2}}{2}</math>. The area of all 4 is <math>4\cdot\frac{\sqrt{2}}{2}=2\sqrt{2}</math>. Finally, the area of the square in the middle is <math>1\cdot1=1</math>. The total area is <math>1+1+2\sqrt{2}=2+2\sqrt{2}</math>. We want the area of the square over the area of the octagon, which is <math>\frac{1}{2+2\sqrt{2}}</math>. Rationalize by multiplying both numerator and denominator by <math>2-2\sqrt{2}</math>: <math>\frac{1}{2+2\sqrt{2}}\cdot\frac{2-2\sqrt{2}}{2-2\sqrt{2}}=\frac{2-2\sqrt{2}}{\left(2+2\sqrt{2}\right)\left(2-2\sqrt{2}\right)}</math>. By the difference of squares, the denominator reduces to <math>-4</math> and the fraction is <math>\frac{2-2\sqrt{2}}{-4}=\frac{\sqrt{2}-1}{2}</math> which is <math>\boxed{\textbf{(A) } \frac{\sqrt{2}-1}{2}}</math>. | ||
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+ | ~Explanation by JH. L | ||
== Solution 2== | == Solution 2== | ||
− | + | The area of a regular octagon is <math>2(1+\sqrt{2})a^2</math> where <math>a</math> is the side. Hence the answer is obvious now. | |
== See Also== | == See Also== |
Latest revision as of 17:54, 2 September 2024
Contents
[hide]Problem
A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?
Solution
If the side lengths of the dart board and the side lengths of the center square are all then the side length of the legs of the triangles are .
Use Geometric probability by putting the area of the desired region over the area of the entire region.
Explanation:
The area of the octagon consists of the area of the triangles, the rectangles, and the square in the middle. Assume the octagon has side length . The triangles are right isosceles triangles with the hypotenuse 1, so their side length is and the area of one triangle is . The area of all 4 triangles is then just . The rectangles share one side with the octagon, and another with the triangle. The octagon has side length 1, and the triangle has side length as found earlier. So the area of one rectangle is . The area of all 4 is . Finally, the area of the square in the middle is . The total area is . We want the area of the square over the area of the octagon, which is . Rationalize by multiplying both numerator and denominator by : . By the difference of squares, the denominator reduces to and the fraction is which is .
~Explanation by JH. L
Solution 2
The area of a regular octagon is where is the side. Hence the answer is obvious now.
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 | ||
All AMC 10 Problems and Solutions |
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