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<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<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<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>, | + | 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 <math>S</math> like the other two rays. The lowest possible value that can be achieved is also <math>(1, 7)</math>. |
+ | |||
+ | Since <math>S</math> is made up of three rays with common endpoint <math>(1, 7)</math>, the answer is <math>\boxed{E}</math> | ||
==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 18:31, 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
. 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 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 |
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