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> | ||
+ | |||
+ | ==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 now factor this quadratic. We must find two terms that multiply to <math>\pi^2 -1</math> and add to <math>-2\pi</math>. | ||
+ | |||
+ | These terms are <math>-\pi+1</math> and <math>-\pi-1</math>. | ||
+ | |||
+ | Subtracting one from the other, we get <math>2</math>. | ||
+ | |||
+ | Thus, the answer is <math>\boxed{\textbf{(C) }2}</math> | ||
==Video Solution== | ==Video Solution== |
Revision as of 21:44, 3 November 2021
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
Solution 3
This solution is similar to Solution #1 but uses a different way to find at the end.
Just like Solution #1, we arrive at the conclusion that .
Simplifying we get:
We now can now factor this quadratic. We must find two terms that multiply to and add to .
These terms are and .
Subtracting one from the other, we get .
Thus, 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.