Difference between revisions of "2017 AMC 10A Problems/Problem 12"
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<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
Contents
[hide]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
Diagram
~BakedPotato66
Solution
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 . 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
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 |
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