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

(See Also)
(10 intermediate revisions by 5 users not shown)
Line 5: Line 5:
 
<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>
  
==Solution==
+
==Solution 1==
  
Plug <math>\sqrt{\pi}</math> in to the equation.  
+
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>
Line 23: Line 23:
 
<math>y = \pi - 1</math>
 
<math>y = \pi - 1</math>
  
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>
  
 
The answer is <math>\boxed{\textbf{(C) }2}</math>
 
The answer is <math>\boxed{\textbf{(C) }2}</math>
 +
 +
==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>
 +
 +
==Video Solution==
 +
https://youtu.be/gKzliDi3zgk
 +
 +
~savannahsolver
 +
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2015|ab=A|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Introductory Algebra Problems]]

Revision as of 19:22, 25 September 2020

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png