2015 AMC 10A Problems/Problem 12

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

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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