Difference between revisions of "2017 AMC 12B Problems/Problem 8"
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− | ==Problem | + | ==Problem== |
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? | 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> | ||
− | ==Solution 1: Cross | + | ==Solution 1: Cross-Multiplication== |
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>. | ||
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− | Let the short side of the rectangle be <math>a</math> and let the long side of the rectangle be <math>b</math>. Then, the diagonal, according to the Pythagorean Theorem, is <math>sqrt{a+b | + | Let the short side of the rectangle be <math>a</math> and let the long side of the rectangle be <math>b</math>. Then, the diagonal, according to the Pythagorean Theorem, is <math>\sqrt{a^2+b^2}</math>. Therefore, we can write the equation: |
<math>\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}</math>. | <math>\frac{a}{b} = \frac{b}{\sqrt{a^2 + b^2}}</math>. | ||
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<math>k=\frac{-1\pm\sqrt{1^2-4(1)(-1)}}{2}=\frac{-1\pm\sqrt{5}}{2}</math>. | <math>k=\frac{-1\pm\sqrt{1^2-4(1)(-1)}}{2}=\frac{-1\pm\sqrt{5}}{2}</math>. | ||
− | Since the ratio of lengths and diagonals of a rectangle cannot be negative, and <math>sqrt{5}>1</math>, the <math>\pm</math> symbol can only take on the <math>+</math>. Therefore, | + | Since the ratio of lengths and diagonals of a rectangle cannot be negative, and <math>\sqrt{5}>1</math>, the <math>\pm</math> symbol can only take on the <math>+</math>. Therefore, |
<math>k=\frac{-1+\sqrt{5}}{2}=\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}</math>. | <math>k=\frac{-1+\sqrt{5}}{2}=\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | WLOG, assume the short side of the rectangle is 1 and the long side is <math>x</math>. Then, we wish to find <math>\frac{1}{x^2}</math> when we are given <cmath>\frac{1}{x}=\frac{x}{\sqrt{x^2+1}}\implies x^4-x^2-1=0.</cmath> Let <math>y=x^2</math> and note that <math>y^2-y-1=0</math> has solutions <math>y=\frac{1\pm\sqrt{5}}{2}</math>. Since we want the positive answer, then <math>x^2=y=\frac{1+\sqrt{5}}{2}</math> and <math>\frac{1}{x^2}=\boxed{\textbf{(C)} \frac{\sqrt{5}-1}{2}}</math>. | ||
+ | |||
+ | Solution by Jzhang21 | ||
==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}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 12:54, 15 February 2021
Contents
Problem
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-Multiplication
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
Solution by HydroQuantum
Let the short side of the rectangle be and let the long side of the rectangle be . Then, the diagonal, according to the Pythagorean Theorem, is . Therefore, we can write the equation:
.
We are trying to find the square of the ratio of to . Let's let our answer, , be . Then, squaring the above equation,
.
Thus, .
Multiplying each side of the equation by ,
.
Adding each side by ,
.
Solving for using the Quadratic Formula,
.
Since the ratio of lengths and diagonals of a rectangle cannot be negative, and , the symbol can only take on the . Therefore,
.
Solution 3
WLOG, assume the short side of the rectangle is 1 and the long side is . Then, we wish to find when we are given Let and note that has solutions . Since we want the positive answer, then and .
Solution by Jzhang21
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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