Difference between revisions of "2017 AMC 12A Problems/Problem 9"

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==Solution==
 
==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 <math>S</math>. This is a ray with an endpoint of <math>(1, 7)</math>.
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If the two equal values are <math>3</math> and <math>x+2</math>, then <math>x=1</math>. Also, <math>y-4\leqslant 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 <math>S</math>. 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 <math>S</math>. This is another ray with the same endpoint as the above ray: <math>(1, 7)</math>.
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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>.
  
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>.
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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>Using x-2 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 <math>S</math> is made up of three rays with common endpoint <math>(1, 7)</math>, the answer is <math>\boxed{E}</math>
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Since </math>S<math> is made up of three rays with common endpoint </math>(1, 7)<math>, the answer is </math>\boxed{E}$
  
 
Solution by TheMathematicsTiger7
 
Solution by TheMathematicsTiger7

Revision as of 20:42, 15 July 2017

Problem

Let $S$ be the set of points $(x,y)$ in the coordinate plane such that two of the three quantities $3$, $x+2$, and $y-4$ 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 $S$?

$\textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point}$

Solution

If the two equal values are $3$ and $x+2$, then $x=1$. Also, $y-4\leqslant 3$ because 3 is the common value. Solving for $y$, we get $y<7$. Therefore the portion of the line $x=1$ where $y<7$ is part of $S$. This is a ray with an endpoint of $(1, 7)$.

Similar to the process above, we assume that the two equal values are $3$ and $y-4$. Solving the equation $3=y-4$ then $y=7$. Also, $x+2\leqslant 3$ because 3 is the common value. Solving for $x$, we get $x<1$. Therefore the portion of the line $y=7$ where $x<1$ is also part of $S$. This is another ray with the same endpoint as the above ray: $(1, 7)$.

If $x+2$ and $y-4$ are the two equal values, then $x+2=y-4$. Solving the equation for $y$, we get $y=x+6$. Also $3\leqslant y-4$ because $y-4$ is one way to express the common value ($Using x-2 as the common value works as well). Solving for$y$, we get$y>7$. Therefore the portion of the line$y=x+6$where$y>7$is part of$S$like the other two rays. The lowest possible value that can be achieved is also$(1, 7)$.

Since$ (Error compiling LaTeX. ! Missing $ inserted.)S$is made up of three rays with common endpoint$(1, 7)$, the answer is$\boxed{E}$

Solution by TheMathematicsTiger7

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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