Difference between revisions of "2017 AMC 12B Problems/Problem 8"
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<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> | ||
| − | ==Solution== | + | ==Solution 1: Cross Multiplying== |
Let <math>a</math> be the short side of the rectangle, and <math>b</math> be the long side of the rectangle. The diagonal, therefore, is <math>\sqrt{a^2 + b^2}</math>. We can get the equation <math>\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}</math>. Cross-multiplying, we get <math>a\sqrt{a^2 + b^2} = b^2</math>. Squaring both sides of the equation, we get <math>a^2 (a^2 + b^2) = b^4</math>, which simplifies to <math>a^4 + a^2b^2 - b^4 = 0</math>. Solving for a quadratic in <math>a^2</math>, using the quadratic formula we get <math>a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}</math> which gives us <math>\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}</math>. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is <math>\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}</math>. | Let <math>a</math> be the short side of the rectangle, and <math>b</math> be the long side of the rectangle. The diagonal, therefore, is <math>\sqrt{a^2 + b^2}</math>. We can get the equation <math>\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}</math>. Cross-multiplying, we get <math>a\sqrt{a^2 + b^2} = b^2</math>. Squaring both sides of the equation, we get <math>a^2 (a^2 + b^2) = b^4</math>, which simplifies to <math>a^4 + a^2b^2 - b^4 = 0</math>. Solving for a quadratic in <math>a^2</math>, using the quadratic formula we get <math>a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}</math> which gives us <math>\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}</math>. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is <math>\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}</math>. | ||
Solution by: vedadehhc | Solution by: vedadehhc | ||
| + | |||
| + | ==Solution 2: Substitution== | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=7|num-a=9}} | {{AMC12 box|year=2017|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:26, 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 1: Cross Multiplying
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
Solution 2: Substitution
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
