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> | ||
− | ==Solution== | + | ==Solution 1== |
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>. | 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>. | ||
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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>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: | ||
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The answer is <math>\boxed{\textbf{(C) }2}</math> | The answer is <math>\boxed{\textbf{(C) }2}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/gKzliDi3zgk | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 19:22, 25 September 2020
Problem
Points and are distinct points on the graph of . What is ?
Solution 1
Since points on the graph make the equation true, substitute in to the equation and then solve to find and .
There are only two solutions to the equation, so one of them is the value of and the other is . The order does not matter because of the absolute value sign.
The answer is
Solution 2
This solution is very closely related to Solution #1 and just simplifies the problem earlier to make it easier.
can be written as . Recognizing that this is a binomial square, simplify this to . This gives us two equations:
and .
One of these 's is and one is . Substituting for , we get and .
So, .
The answer is
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.