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

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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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==Solution 3==
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This solution is similar to Solution #1 but uses a different way to find <math>y</math> at the end.
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Just like Solution #1, we arrive at the conclusion that <math>y^2 - 2\pi y + \pi^2 = 1</math>.
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Simplifying we get:
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<math>y^2 - 2\pi y + \pi^2 -1 = 0</math>
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We now can now 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>.
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Thus, the answer is <math>\boxed{\textbf{(C) }2}</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 21:44, 3 November 2021

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 now 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 $\boxed{\textbf{(C) }2}$

Video Solution

https://youtu.be/gKzliDi3zgk

~savannahsolver

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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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