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Difference between revisions of "2007 AMC 12A Problems/Problem 13"
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==Solution==
 
==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
 
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>.
  
 
==See also==
 
==See also==

Revision as of 14:57, 20 May 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 $a+b=2+8 = \boxed{10}$.

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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