Difference between revisions of "2007 AMC 12A Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
| − | 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 = | + | <math>(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)</math>. |
| − | + | 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 23:18, 1 December 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
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
