Difference between revisions of "2012 AMC 12A Problems/Problem 12"
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<math> \textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5} </math> | <math> \textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5} </math> | ||
− | == Solution == | + | == Solution 1 == |
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<cmath>s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}</cmath> | <cmath>s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}</cmath> | ||
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+ | == Solution 2 == | ||
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+ | Using the diagram above, we look at the top-right vertex of the square. Let us call this point <math>(x,y)</math>. Then, we that since the square is symmetrical over the y-axis, that the y value is equal to <math>2x+1</math>, since we can multiply the x value(which is half of <math>s</math>) by two to get <math>s</math>, and we add one since the square lies one unit above the origin. Now, all we must do is find the intersection of the larger circle, <math>x^2 + y^2 = 4</math>, and the line <math>y=2x+1</math>. Substituting the second equation into the first, we get: | ||
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+ | <math>5x^2 +4x -3 = 0</math> | ||
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+ | Using the quadratic formula, we arrive with <math>x=\frac{-4 \pm 2\sqrt{19}}{10}</math>. However, recall that the x value is only one half of the side length. Multiplying this value by <math>2</math>, then, and using only the positive root(since the top right vertex of the square has a positive x value), we get: | ||
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+ | <math>\frac{-4 + 2\sqrt{19}}{5} \Rightarrow \boxed{\textbf{(D)}}</math> | ||
== See Also == | == See Also == |
Revision as of 22:58, 7 February 2015
Contents
Problem
A square region is externally tangent to the circle with equation at the point on the side . Vertices and are on the circle with equation . What is the side length of this square?
Solution 1
The circles have radii of and . Draw the triangle shown in the figure above and write expressions in terms of (length of the side of the square) for the sides of the triangle. Because is the radius of the larger circle, which is equal to , we can write the Pythagorean Theorem.
Use the quadratic formula.
Solution 2
Using the diagram above, we look at the top-right vertex of the square. Let us call this point . Then, we that since the square is symmetrical over the y-axis, that the y value is equal to , since we can multiply the x value(which is half of ) by two to get , and we add one since the square lies one unit above the origin. Now, all we must do is find the intersection of the larger circle, , and the line . Substituting the second equation into the first, we get:
Using the quadratic formula, we arrive with . However, recall that the x value is only one half of the side length. Multiplying this value by , then, and using only the positive root(since the top right vertex of the square has a positive x value), we get:
See Also
2012 AMC 12A (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.