Difference between revisions of "2017 AMC 12B Problems/Problem 8"

(Problem 8)
(Solution)
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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==
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==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 14:26, 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?

$\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}$

Solution 1: Cross Multiplying

Let $a$ be the short side of the rectangle, and $b$ be the long side of the rectangle. The diagonal, therefore, is $\sqrt{a^2 + b^2}$. We can get the equation $\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}$. Cross-multiplying, we get $a\sqrt{a^2 + b^2} = b^2$. Squaring both sides of the equation, we get $a^2 (a^2 + b^2) = b^4$, which simplifies to $a^4 + a^2b^2 - b^4 = 0$. Solving for a quadratic in $a^2$, using the quadratic formula we get $a^2 = \frac{-b^2 \pm \sqrt{5b^4}}{2}$ which gives us $\frac{a^2}{b^2} = \frac{-1 \pm \sqrt{5}}{2}$. We know that the square of the ratio must be positive (the square of any real number is positive), so the solution is $\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}$.

Solution by: vedadehhc

Solution 2: Substitution

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
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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