Difference between revisions of "2017 AMC 10A Problems/Problem 12"
Eatingstuff (talk | contribs) m (→See Also) |
Eatingstuff (talk | contribs) m (→See Also) |
||
Line 15: | Line 15: | ||
==See Also== | ==See Also== | ||
− | {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13 | + | {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}} |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:55, 17 February 2018
Problem
Let be a 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 this common value. Which of the following is a correct description for
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. 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.