2007 AMC 12A Problems/Problem 13

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 1

The point $(a,b)$ is the foot of the perpendicular from $(12,10)$ to the line $y=-5x+18$ . The perpendicular has slope $\frac{1}{5}$ , so its equation is $y=10+\frac{1}{5}(x-12)=\frac{1}{5}x+\frac{38}{5}$ . The $x$ -coordinate at the foot of the perpendicular satisfies the equation $\frac{1}{5}x+\frac{38}{5}=-5x+18$ , so $x=2$ and $y=-5\cdot2+18=8$ . Thus $(a,b) = (2,8)$ , and $a+b = \boxed{10}$ .

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 $a+b=2+8 = \boxed{10}$ .

Solution 3

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 4

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

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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