Difference between revisions of "2007 AMC 12A Problems/Problem 13"
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<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> | ||
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| − | + | ==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>. | ||
| − | )<math> to < | + | ==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 | ||
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| + | |||
| + | ==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 = | ||
| + | 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>. | ||
| + | -Paixiao | ||
==See also== | ==See also== | ||
Latest revision as of 17:20, 27 November 2023
Problem
A piece of cheese is located at 
in a coordinate plane. A mouse is at 
and is running up the line 
. At the point 
the mouse starts getting farther from the cheese rather than closer to it. What is 
?
Solution 1
The point is the foot of the perpendicular from to the line . The perpendicular has slope 
, so its equation is 
. The 
-coordinate at the foot of the perpendicular satisfies the equation 
, so 
and 
. Thus 
, and 
.
Solution 2
If the mouse is at 
, then the square of the distance from the mouse to the cheese is 
The value of this expression is smallest when 
, so the mouse is closest to the cheese at the point 
, and 
.
Solution 3
We are trying to find the point where distance between the mouse and 
is minimized. This point is where the line that passes through and is perpendicular to intersects . By basic knowledge of perpendicular lines, this line is 
. This line intersects at 
. So 
. - MegaLucario1001
Solution 4
If the mouse is at , then the square of the distance from the mouse to the cheese is
.
The value of this expression is smallest when , so the mouse is closest to the cheese at the point , and .
-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. 
