Difference between revisions of "2007 AMC 12A Problems/Problem 13"
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− | If the mouse is at <math>(x, y) = (x, 18 - 5x)</math>, then the square of the distance from the mouse to the cheese is | + | If the mouse is at <math>(x, y) = (x, 18 - 5x)</math>, then the square of the distance from the mouse to the cheese is\[ |
(x - 12)^2 + (8 - 5x)^2 = | (x - 12)^2 + (8 - 5x)^2 = | ||
26(x^2 - 4x + 8) = 26((x - 2)^2 + 4). | 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4). | ||
− | \] | + | \]The value of this expression is smallest when <math>x = 2</math>, so the mouse is closest to the cheese at the point <math>(2, 8)</math>, and <math>a+b=2+8 = \boxed{10}</math>. |
-Paixiao | -Paixiao | ||
Revision as of 12:18, 6 July 2021
Contents
Problem
A piece of cheese is located at in a coordinate plane. A mouse is at and is running up the line . At the point the mouse starts getting farther from the cheese rather than closer to it. What is ?
Solution
We are trying to find the point where distance between the mouse and is minimized. This point is where the line that passes through and is perpendicular to intersects . By basic knowledge of perpendicular lines, this line is . This line intersects at . So . - MegaLucario1001
Solution 2
If the mouse is at , then the square of the distance from the mouse to the cheese is\[ (x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4). \]The value of this expression is smallest when , so the mouse is closest to the cheese at the point , and . -Paixiao
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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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