Difference between revisions of "2017 AMC 12B Problems/Problem 8"
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==Problem 8== | ==Problem 8== | ||
− | The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side | + | The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle? |
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math> | <math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math> |
Revision as of 15:16, 17 February 2017
Problem 8
The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?
Solution
Let be the short side of the rectangle, and be the long side of the rectangle. The diagonal, therefore, is . We can get the equation . Cross-multiplying, we get . Squaring both sides of the equation, we get , which simplifies to . Solving for a quadratic in , using the quadratic formula we get which gives us . We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is .
Solution by: vedadehhc
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 |
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