Difference between revisions of "2012 AMC 10A Problems/Problem 24"
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− | + | {{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #21]] and [[2012 AMC 10A Problems|2012 AMC 10A #24]]}} | |
− | Let <math>a</math>, <math>b</math>, and <math>c</math> be positive integers with <math>a\ge</math> <math>b\ge</math> <math>c</math> such that | + | == Problem == |
+ | |||
+ | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>a</math>, <math>b</math>, and <math>c</math> be positive integers with <math>a\ge</math> <math>b\ge</math> <math>c</math> such that | ||
<math>a^2-b^2-c^2+ab=2011</math> and | <math>a^2-b^2-c^2+ab=2011</math> and | ||
<math>a^2+3b^2+3c^2-3ab-2ac-2bc=-1997</math>. | <math>a^2+3b^2+3c^2-3ab-2ac-2bc=-1997</math>. | ||
− | What is <math> | + | What is <math>b</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math> \textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253 </math> | <math> \textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253 </math> | ||
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<math>2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14</math>. | <math>2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14</math>. | ||
− | Now, this can be rearranged | + | Now, this can be rearranged and factored. |
<math>(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14</math> | <math>(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14</math> | ||
− | |||
− | |||
<math>(a - b)^2 + (a - c)^2 + (b - c)^2 = 14</math> | <math>(a - b)^2 + (a - c)^2 + (b - c)^2 = 14</math> | ||
− | <math>a</math>, <math>b</math>, and <math>c</math> are all integers, so the three terms on the left side of the equation must all be perfect squares. | + | <math>a</math>, <math>b</math>, and <math>c</math> are all integers, so the three terms on the left side of the equation must all be perfect squares. |
+ | We see that the only is possibility is <math>14 = 9 + 4 + 1</math>. | ||
− | <math>(a-c)^2 = 9 | + | <math>(a-c)^2 = 9 \Rightarrow a-c = 3</math>, since <math>a-c</math> is the biggest difference. It is impossible to determine by inspection whether <math>a-b = 1</math> or <math>2</math>, or whether <math>b-c = 1</math> or <math>2</math>. |
We want to solve for <math>a</math>, so take the two cases and solve them each for an expression in terms of <math>a</math>. Our two cases are <math>(a, b, c) = (a, a-1, a-3)</math> or <math>(a, a-2, a-3)</math>. Plug these values into one of the original equations to see if we can get an integer for <math>a</math>. | We want to solve for <math>a</math>, so take the two cases and solve them each for an expression in terms of <math>a</math>. Our two cases are <math>(a, b, c) = (a, a-1, a-3)</math> or <math>(a, a-2, a-3)</math>. Plug these values into one of the original equations to see if we can get an integer for <math>a</math>. | ||
<math>a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011</math>, after some algebra, simplifies to | <math>a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011</math>, after some algebra, simplifies to | ||
− | <math>7a = 2021</math>. 2021 is not divisible by 7, so <math>a</math> is not an integer. | + | <math>7a = 2021</math>. <math>2021</math> is not divisible by <math>7</math>, so <math>a</math> is not an integer. |
+ | |||
+ | The other case gives <math>a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011</math>, which simplifies to <math>8a = 2024</math>. Thus, <math>a = 253</math> and the answer is <math>\boxed{\textbf{(E)}\ 253}</math>. | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2012amc12a/250 | ||
+ | |||
+ | ~dolphin7 | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2012|ab=A|num-b=23|num-a=25}} | ||
+ | {{AMC12 box|year=2012|ab=A|num-b=20|num-a=22}} | ||
− | + | [[Category:Introductory Algebra Problems]] | |
+ | [[Category:Algebraic Manipulations Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:45, 28 November 2022
- The following problem is from both the 2012 AMC 12A #21 and 2012 AMC 10A #24, so both problems redirect to this page.
Problem
Let , , and be positive integers with such that and .
What is ?
Solution
Add the two equations.
.
Now, this can be rearranged and factored.
, , and are all integers, so the three terms on the left side of the equation must all be perfect squares. We see that the only is possibility is .
, since is the biggest difference. It is impossible to determine by inspection whether or , or whether or .
We want to solve for , so take the two cases and solve them each for an expression in terms of . Our two cases are or . Plug these values into one of the original equations to see if we can get an integer for .
, after some algebra, simplifies to . is not divisible by , so is not an integer.
The other case gives , which simplifies to . Thus, and the answer is .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc12a/250
~dolphin7
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.