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?
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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(6)
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<math>\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22</math>
B(10)
 
C(14)
 
D(18)
 
E(22)
 
  
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==Solution==
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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
  
==Solution==
 
  
We are trying to find the point where (12,10) is perpendicular to y=-5x+18. Then the slope of the line that passes through the cheese and (a,b) is 1/5. Therefore, the line is
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==Solution 2==
* <math>y=\frac{1}{5}x+\frac{38}{5}</math> The point where y=-5x+18 and y=.2x+7.6 intersect is (2,8), and 2+8=10(B).
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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\[
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(x - 12)^2 + (8 - 5x)^2 =
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26(x^2 - 4x + 8) = 26((x - 2)^2 + 4).
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\]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>.
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-Paixiao
  
 
==See also==
 
==See also==
* [[2007 AMC 12A Problems/Problem 11 | Previous problem]]
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{{AMC12 box|year=2007|ab=A|num-b=12|num-a=14}}
* [[2007 AMC 12A Problems/Problem 13 | Next problem]]
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* [[2007 AMC 12A Problems]]
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Revision as of 11:18, 6 July 2021

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 $a+b$?

$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$

Solution

We are trying to find the point where distance between the mouse and $(12, 10)$ is minimized. This point is where the line that passes through $(12, 10)$ and is perpendicular to $y=-5x+18$ intersects $y=-5x+18$. By basic knowledge of perpendicular lines, this line is $y=\frac{x}{5}+\frac{38}{5}$. This line intersects $y=-5x+18$ at $(2,8)$. So $a+b=\boxed{10}$. - MegaLucario1001


Solution 2

If the mouse is at $(x, y) = (x, 18 - 5x)$, 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 $x = 2$, so the mouse is closest to the cheese at the point $(2, 8)$, and $a+b=2+8 = \boxed{10}$. -Paixiao

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
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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