Difference between revisions of "2007 AMC 12A Problems/Problem 13"
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==Problem== | ==Problem== | ||
− | A piece of cheese is located at (12,10) in a coordinate plane. A mouse is at (4,-2) and is running up the line y=-5x+18. At the point (a,b) the mouse starts getting farther from the cheese rather than closer to it. What is a+b? | + | A piece of cheese is located at <math>(12,10)</math> in a [[coordinate plane]]. A mouse is at <math>(4,-2)</math> and is running up the [[line]] <math>y=-5x+18</math>. At the point <math>(a,b)</math> the mouse starts getting farther from the cheese rather than closer to it. What is <math>a+b</math>? |
− | A( | + | <math>\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22</math> |
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+ | ==Solution== | ||
+ | We are trying to find the point where distance between the mouse and <math>(12, 10)</math> is minimized. This point is where the line that passes through <math>(12, 10)</math> and is perpendicular to <math>y=-5x+18</math> intersects <math>y=-5x+18</math>. By basic knowledge of perpendicular lines, this line is <math>y=\frac{x}{5}+\frac{38}{5}</math>. This line intersects <math>y=-5x+18</math> at <math>(2,8)</math>. So <math>a+b=\boxed{10}</math>. - MegaLucario1001 | ||
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− | + | ==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\[ | |
+ | (x - 12)^2 + (8 - 5x)^2 = | ||
+ | 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4). | ||
+ | \]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 | ||
==See also== | ==See also== | ||
− | + | {{AMC12 box|year=2007|ab=A|num-b=12|num-a=14}} | |
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− | + | [[Category:Introductory Geometry Problems]] | |
+ | {{MAA Notice}} |
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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