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Difference between revisions of "2015 AMC 10A Problems/Problem 12"

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==Solution==
 
==Solution==
  
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Plug <math>\sqrt{\pi}</math> in to the equation.  
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Since points on the graph make the equation true, substitute <math>\sqrt{\pi}</math> in to the equation and then solve to find <math>a</math> and <math>b</math>.
  
 
<math>y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1</math>
 
<math>y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1</math>
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<math>y = \pi - 1</math>
 
<math>y = \pi - 1</math>
  
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There are only two solutions to the equation, so one of them is the value of <math>a</math> and the other is <math>b</math>. The order does not matter because of the absolute value signs.
+
There are only two solutions to the equation, so one of them is the value of <math>a</math> and the other is <math>b</math>. The order does not matter because of the absolute value sign.
  
 
<math>| (\pi + 1) - (\pi - 1) | = 2</math>
 
<math>| (\pi + 1) - (\pi - 1) | = 2</math>

Revision as of 10:11, 5 February 2015

Problem

Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$. What is $|a-b|$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$

Solution

Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$.

$y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1$

$y^2 + \pi^2 = 2\pi y + 1$

$y^2 - 2\pi y + \pi^2 = 1$

$(y-\pi)^2 = 1$

$y-\pi = \pm 1$

$y = \pi + 1$

$y = \pi - 1$

There are only two solutions to the equation, so one of them is the value of $a$ and the other is $b$. The order does not matter because of the absolute value sign.

$| (\pi + 1) - (\pi - 1) | = 2$

The answer is $\boxed{\textbf{(C) }2}$

See Also

2015 AMC 10A (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 10 Problems and Solutions

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