Difference between revisions of "2011 AMC 12B Problems/Problem 12"
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== Solution == | == Solution == | ||
− | Let's assume that the side length of the octagon is <math>x</math>. The area of the center square is just <math>x^2</math>. The triangles are all <math>45-45-90</math> triangles, with a side length ratio of <math>1:1:\sqrt{2}</math>. The area of each of the <math>4</math> identical triangles is <math>\left(\dfrac{x}{\sqrt{2}}\right)^2\times\dfrac{1}{2}=\dfrac{x^2}{4}</math>, so the total area of all of the triangles is also <math>x^2</math>. Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is <math>x</math> and the other side length is <math>\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}</math>, so the area of all of the rectangles is <math>2x^2\sqrt{2}</math>. The ratio of the area of the square to the area of the octagon is <math>\dfrac{x^2}{2x^2+2x^2\sqrt{2}}</math>. Cancelling <math>x^2</math> from the fraction, the ratio becomes <math>\dfrac{1}{2\sqrt2+2}</math>. Multiplying the numerator and the denominator each by <math>2\sqrt{2}-2</math> will cancel out the radical, so the fraction is now <math>\dfrac{1}{2\sqrt2+2}\times\dfrac{2\sqrt{2}-2}{2\sqrt{2}-2}=\dfrac{2\sqrt{2}-2}{4}=\boxed{\mathrm{( | + | Let's assume that the side length of the octagon is <math>x</math>. The area of the center square is just <math>x^2</math>. The triangles are all <math>45-45-90</math> triangles, with a side length ratio of <math>1:1:\sqrt{2}</math>. The area of each of the <math>4</math> identical triangles is <math>\left(\dfrac{x}{\sqrt{2}}\right)^2\times\dfrac{1}{2}=\dfrac{x^2}{4}</math>, so the total area of all of the triangles is also <math>x^2</math>. Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is <math>x</math> and the other side length is <math>\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}</math>, so the area of all of the rectangles is <math>2x^2\sqrt{2}</math>. The ratio of the area of the square to the area of the octagon is <math>\dfrac{x^2}{2x^2+2x^2\sqrt{2}}</math>. Cancelling <math>x^2</math> from the fraction, the ratio becomes <math>\dfrac{1}{2\sqrt2+2}</math>. Multiplying the numerator and the denominator each by <math>2\sqrt{2}-2</math> will cancel out the radical, so the fraction is now <math>\dfrac{1}{2\sqrt2+2}\times\dfrac{2\sqrt{2}-2}{2\sqrt{2}-2}=\dfrac{2\sqrt{2}-2}{4}=\boxed{\mathrm{(A)}\ \dfrac{\sqrt{2}-1}{2}}</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2011|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2011|ab=B|num-b=11|num-a=13}} |
Revision as of 20:36, 11 August 2011
Problem
A dart board is a regular octagon divided into regions as shown below. 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
Let's assume that the side length of the octagon is . The area of the center square is just . The triangles are all triangles, with a side length ratio of . The area of each of the identical triangles is , so the total area of all of the triangles is also . Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is and the other side length is , so the area of all of the rectangles is . The ratio of the area of the square to the area of the octagon is . Cancelling from the fraction, the ratio becomes . Multiplying the numerator and the denominator each by will cancel out the radical, so the fraction is now
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 12 Problems and Solutions |