Difference between revisions of "2017 AMC 10A Problems/Problem 12"
m (→Solution) |
Pinotation (talk | contribs) (→Solution 2 (Graphing \(S\))) |
||
| (3 intermediate revisions by the same user not shown) | |||
| 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> | ||
| − | ==Solution== | + | ==Solution 1== |
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>. | ||
| Line 13: | Line 13: | ||
Since <math>S</math> is made up of three rays with common endpoint <math>(1, 7)</math>, the answer is <math>\boxed{\textbf{(E) }\text{three rays with a common endpoint}}</math> | Since <math>S</math> is made up of three rays with common endpoint <math>(1, 7)</math>, the answer is <math>\boxed{\textbf{(E) }\text{three rays with a common endpoint}}</math> | ||
| + | |||
| + | ==Solution 2 (Graphing \(S\))== | ||
| + | |||
| + | Similar to above, we make three equations. | ||
| + | |||
| + | \( 3 = x + 2 \) | ||
| + | |||
| + | \( 3 = y - 4 \) | ||
| + | |||
| + | \( x + 3 = y - 4 \) | ||
| + | |||
| + | and solve them in terms of a linear equation | ||
| + | |||
| + | \( x = 1 \) | ||
| + | |||
| + | \( y = 7 \) | ||
| + | |||
| + | \( x + 6 = y \) | ||
| + | |||
| + | We proceed to graph each of the three equations. | ||
| + | |||
| + | [[File:Amc10a 2017 q12.png | center | 650]] | ||
| + | |||
| + | We now see that no third value is less than the common point. Therefore, we want the overlapping region of \( y \ge 7 \) and \( x \ge 1 \). | ||
| + | |||
| + | There are three lines in this region that start at a point \((1, 7)\). A line that extends forever starting at any point is called a ray. Because we have three of these lines, particularly \( x = 1 \), \( y = 7 \), and \( x + 6 = y \), we have <math>\boxed{\textbf{(E) }\text{three rays with a common endpoint}}</math>. | ||
| + | |||
| + | ~Pinotation | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 21:11, 2 October 2025
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 1
If the two equal values are 
and 
, then 
. Also, 
because 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 
. We also know 
, so 
.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 
Solution 2 (Graphing \(S\))
Similar to above, we make three equations.
\( 3 = x + 2 \)
\( 3 = y - 4 \)
\( x + 3 = y - 4 \)
and solve them in terms of a linear equation
\( x = 1 \)
\( y = 7 \)
\( x + 6 = y \)
We proceed to graph each of the three equations.
We now see that no third value is less than the common point. Therefore, we want the overlapping region of \( y \ge 7 \) and \( x \ge 1 \).
There are three lines in this region that start at a point \((1, 7)\). A line that extends forever starting at any point is called a ray. Because we have three of these lines, particularly \( x = 1 \), \( y = 7 \), and \( x + 6 = y \), we have .
~Pinotation
Video Solution
https://youtu.be/s4vnGlwwHHw?t=190
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
