Difference between revisions of "2017 AMC 12A Problems/Problem 9"
Draziwfozo (talk | contribs) |
|||
| Line 4: | Line 4: | ||
<math> \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} </math> | <math> \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} </math> | ||
| + | |||
| + | ==Solution== | ||
| + | If the two equal values are <math>3</math> and <math>x+2</math>, then <math>x=1</math>. Also, <math>y-4<3</math> because 3 is the common value. Solving for <math>y</math>, we get <math>y<7</math>. Therefore the portion of the line <math>x=1</math> where <math>y<7</math> is part of S. This is a ray with an endpoint of <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<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 S. 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<y-4</math> because <math>y-4</math> is one way to express the common value. 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 S like the other two rays. The lowest possible value that can be achieved is also <math>(1, 7)</math>, which makes the answer E. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2017|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:23, 8 February 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 S. 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 S. 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 S like the other two rays. The lowest possible value that can be achieved is also , which makes the answer E.
See Also
| 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. 
