Difference between revisions of "2007 AMC 12A Problems/Problem 13"

(Solution 2)
(Solution 2)
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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<math>\[
 
(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>.
+
\]</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 11:17, 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$\[ (x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4). \]$ (Error compiling LaTeX. Unknown error_msg)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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