Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m
 
(2 intermediate revisions by 2 users not shown)
Line 5: Line 5:
 
<math>\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22</math>
 
<math>\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22</math>
  
−
==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>.
 +
 
 +
==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 <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
 
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 2==
+
==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\[
+
If the mouse is at <math>(x, y) = (x, 18 - 5x)</math>, then the square of the distance from the mouse to the cheese is
−
(x - 12)^2 + (8 - 5x)^2 =
+
<math>(x - 12)^2 + (8 - 5x)^2 =
−
26(x^2 - 4x + 8) = 26((x - 2)^2 + 4).
+
26(x^2 - 4x + 8) = 26((x - 2)^2 + 4)</math>.
−
\]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>.
+
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
  

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_13&oldid=206161"