Difference between revisions of "2017 AMC 12A Problems/Problem 9"
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Similar to the process above, we assume that the two equal values are <math>3</math> and <math>y-4</math>. Solving the equation <math>3=y-4</math> then <math>y=7</math>. Also, <math>x+2\leqslant 3</math> because 3 is the common value. Solving for <math>x</math>, we get <math>x<1</math>. Therefore the portion of the line <math>y=7</math> where <math>x<1</math> is also part of <math>S</math>. This is another ray with the same endpoint as the above ray: <math>(1, 7)</math>. | Similar to the process above, we assume that the two equal values are <math>3</math> and <math>y-4</math>. Solving the equation <math>3=y-4</math> then <math>y=7</math>. Also, <math>x+2\leqslant 3</math> because 3 is the common value. Solving for <math>x</math>, we get <math>x<1</math>. Therefore the portion of the line <math>y=7</math> where <math>x<1</math> is also part of <math>S</math>. This is another ray with the same endpoint as the above ray: <math>(1, 7)</math>. | ||
− | If <math>x+2</math> and <math>y-4</math> are the two equal values, then <math>x+2=y-4</math>. Solving the equation for <math>y</math>, we get <math>y=x+6</math>. Also <math>3\leqslant y-4</math> because <math>y-4</math> is one way to express the common value (<math> | + | If <math>x+2</math> and <math>y-4</math> are the two equal values, then <math>x+2=y-4</math>. Solving the equation for <math>y</math>, we get <math>y=x+6</math>. Also <math>3\leqslant y-4</math> because <math>y-4</math> is one way to express the common value (using <math>x-2</math> as the common value works as well). Solving for <math>y</math>, we get <math>y>7</math>. Therefore the portion of the line <math>y=x+6</math> where <math>y>7</math> is part of <math>S</math> like the other two rays. The lowest possible value that can be achieved is also <math>(1, 7)</math>. |
− | Since < | + | Since <math>S</math> is made up of three rays with common endpoint <math>(1, 7)</math>, the answer is <math>\boxed{E}</math> |
Solution by TheMathematicsTiger7 | Solution by TheMathematicsTiger7 |
Revision as of 20:42, 15 July 2017
Problem
Let be the set of points in the coordinate plane such that two of the three quantities , , and are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of ?
Solution
If the two equal values are and , then . Also, because 3 is the common value. Solving for , we get . Therefore the portion of the line where is part of . This is a ray with an endpoint of .
Similar to the process above, we assume that the two equal values are and . Solving the equation then . Also, because 3 is the common value. Solving for , we get . Therefore the portion of the line where is also part of . This is another ray with the same endpoint as the above ray: .
If and are the two equal values, then . Solving the equation for , we get . Also because is one way to express the common value (using as the common value works as well). Solving for , we get . Therefore the portion of the line where is part of like the other two rays. The lowest possible value that can be achieved is also .
Since is made up of three rays with common endpoint , the answer is
Solution by TheMathematicsTiger7
See Also
2017 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 |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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.