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

Latest revision as of 17:20, 27 November 2023

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

@@ Line 1: / Line 1: @@
 ==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(6)
+<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)
-==Solution==
+==Solution 1==
+The point <math>(a,b)</math> is the foot of the perpendicular from <math>(12,10)</math> to the line <math>y=-5x+18</math>. The perpendicular has slope <math>\frac{1}{5}</math>, so its equation is <math>y=10+\frac{1}{5}(x-12)=\frac{1}{5}x+\frac{38}{5}</math>. The <math>x</math>-coordinate at the foot of the perpendicular satisfies the equation <math>\frac{1}{5}x+\frac{38}{5}=-5x+18</math>, so <math>x=2</math> and <math>y=-5\cdot2+18=8</math>. Thus <math>(a,b) = (2,8)</math>, and <math>a+b = \boxed{10}</math>.
-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
+==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).
+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 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4).</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>.
+==Solution 3==
+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 4==
+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^2 - 4x + 8) = 26((x - 2)^2 + 4)</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
 ==See also==
-* [[2007 AMC 12A Problems/Problem 12 | Previous problem]]
+{{AMC12 box|year=2007|ab=A|num-b=12|num-a=14}}
-* [[2007 AMC 12A Problems/Problem 14 | Next problem]]
-* [[2007 AMC 12A Problems]]
+[[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

2007 AMC 12A (Problems • Answer Key • Resources)
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