Difference between revisions of "2017 AMC 10A Problems/Problem 12"

m (Video Solution)
m (Solution)
Line 5: Line 5:
 
<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 endpoint}</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 endpoint}</math>
  
 +
==Diagram==
 +
[[File:BakedPotato66 (Desmos Graph).png|900px|caption]]
 +
 +
~BakedPotato66
 
==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\le 3</math> because <math>3</math> is the common value. Solving for <math>y</math>, we get <math>y \le 7</math>. Therefore the portion of the line <math>x=1</math> where <math>y \le 7</math> is part of <math>S</math>. This is a ray with an endpoint of <math>(1, 7)</math>.
 
If the two equal values are <math>3</math> and <math>x+2</math>, then <math>x=1</math>. Also, <math>y-4\le 3</math> because <math>3</math> is the common value. Solving for <math>y</math>, we get <math>y \le 7</math>. Therefore the portion of the line <math>x=1</math> where <math>y \le 7</math> is part of <math>S</math>. This is a ray with an endpoint of <math>(1, 7)</math>.

Revision as of 13:20, 8 June 2021

Problem

Let $S$ be a 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 this common value. Which of the following is a correct description for $S?$

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

Diagram

caption

~BakedPotato66

Solution

If the two equal values are $3$ and $x+2$, then $x=1$. Also, $y-4\le 3$ because $3$ is the common value. Solving for $y$, we get $y \le 7$. Therefore the portion of the line $x=1$ where $y \le 7$ 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, $x+2\le 3$ because 3 is the common value. Solving for $x$, we get $x\le1$. Therefore the portion of the line $y=7$ where $x\le 1$ 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 $3\le y-4$ because $y-4$ is one way to express the common value. Solving for $y$, we get $y\ge 7$. Therefore the portion of the line $y=x+6$ where $y\ge 7$ is part of $S$ like the other two rays. The lowest possible value that can be achieved is also $(1, 7)$.

Since $S$ is made up of three rays with common endpoint $(1, 7)$, the answer is $\boxed{\textbf{(E) }\text{three rays with a common endpoint}}$

Video Solution

https://youtu.be/s4vnGlwwHHw?t=190

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png