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

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<math> \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} </math>
 
<math> \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} </math>
  
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==Solution==
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==Solution 1==
  
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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.
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There are only two solutions to the equation <math>(y-\pi)^2 = 1</math>, 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>
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The answer is <math>\boxed{\textbf{(C) }2}</math>
 
The answer is <math>\boxed{\textbf{(C) }2}</math>
  
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{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}
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==Solution 2==
 +
This solution is very closely related to Solution #1 and just simplifies the problem earlier to make it easier.
 +
 
 +
<math>y^2 + x^4 = 2x^2 y + 1</math> can be written as <math>x^4-2x^2y+y^2=1</math>. Recognizing that this is a binomial square, simplify this to <math>(x^2-y)^2=1</math>. This gives us two equations:
 +
 
 +
<math>x^2-y=1</math> and <math>x^2-y=-1</math>.
 +
 
 +
One of these <math>y</math>'s is <math>a</math> and one is <math>b</math>. Substituting <math>\sqrt{\pi}</math> for <math>x</math>, we get <math>a=\pi+1</math> and <math>b=\pi-1</math>.
 +
 
 +
So, <math>|a-b|=|(\pi+1)-(\pi-1)|=2</math>.
 +
 
 +
The answer is <math>\boxed{\textbf{(C) }2}</math>
 +
 
 +
==Solution 3==
 +
This solution is similar to Solution #1 but uses a different way to find <math>y</math> at the end.
 +
 
 +
Just like Solution #1, we arrive at the conclusion that <math>y^2 - 2\pi y + \pi^2 = 1</math>.
 +
 
 +
Simplifying we get:
 +
 
 +
<math>y^2 - 2\pi y + \pi^2 -1 = 0</math>
 +
 
 +
We now can factor this quadratic. We must find two terms that multiply to <math>\pi^2 -1</math> and add to <math>2\pi</math>.
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These terms are <math>\pi+1</math> and <math>\pi-1</math>.
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Subtracting one from the other, we get <math>2</math>.
 +
 
 +
Thus, the answer is <math>\boxed{\textbf{(C) }2}</math>
 +
 
 +
-DuckDuckGooseGoose
 +
 
 +
==Video Solution==
 +
https://youtu.be/gKzliDi3zgk
 +
 
 +
~savannahsolver
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2015|ab=A|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Introductory Algebra Problems]]

Latest revision as of 14:25, 1 August 2022

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 1

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 $(y-\pi)^2 = 1$, 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 $\boxed{\textbf{(C) }2}$

Solution 2

This solution is very closely related to Solution #1 and just simplifies the problem earlier to make it easier.

$y^2 + x^4 = 2x^2 y + 1$ can be written as $x^4-2x^2y+y^2=1$. Recognizing that this is a binomial square, simplify this to $(x^2-y)^2=1$. This gives us two equations:

$x^2-y=1$ and $x^2-y=-1$.

One of these $y$'s is $a$ and one is $b$. Substituting $\sqrt{\pi}$ for $x$, we get $a=\pi+1$ and $b=\pi-1$.

So, $|a-b|=|(\pi+1)-(\pi-1)|=2$.

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

Solution 3

This solution is similar to Solution #1 but uses a different way to find $y$ at the end.

Just like Solution #1, we arrive at the conclusion that $y^2 - 2\pi y + \pi^2 = 1$.

Simplifying we get:

$y^2 - 2\pi y + \pi^2 -1 = 0$

We now can factor this quadratic. We must find two terms that multiply to $\pi^2 -1$ and add to $2\pi$.

These terms are $\pi+1$ and $\pi-1$.

Subtracting one from the other, we get $2$.

Thus, the answer is $\boxed{\textbf{(C) }2}$

-DuckDuckGooseGoose

Video Solution

https://youtu.be/gKzliDi3zgk

~savannahsolver

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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