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 "2008 AMC 12B Problems/Problem 17"
Line 3: Line 3:
 
The slope of line <math>BC</math> is <math>\frac{m^2-n^2}{-m-n}=-\frac{(m+n)(m-n)}{m+n}=-(m-n)</math>.  
 
The slope of line <math>BC</math> is <math>\frac{m^2-n^2}{-m-n}=-\frac{(m+n)(m-n)}{m+n}=-(m-n)</math>.  
  
−
Supposing <math>\angle A=90^\circ</math>, <math>AC</math> is perpendicular to <math>AB</math> and, it follows, to the <math>x</math>-axis, making <math>AB</math>  a segment of the line x=m. But that would mean that the coordinates of <math>C</math> <math>(m, m^2)</math>, contradicting the given that points <math>A</math> and <math>C</math> are distinct. So <math>\angle A</math> is not <math>90^\circ</math>. By a similar logic, neither is <math>\angle B</math>.  
+
Supposing <math>\angle A=90^\circ</math>, <math>AC</math> is perpendicular to <math>AB</math> and, it follows, to the <math>x</math>-axis, making <math>AB</math>  a segment of the line x=m. But that would mean that the coordinates of <math>C</math> are <math>(m, m^2)</math>, contradicting the given that points <math>A</math> and <math>C</math> are distinct. So <math>\angle A</math> is not <math>90^\circ</math>. By a similar logic, neither is <math>\angle B</math>.  
  
 
This means that <math>\angle C=90^\circ</math> and <math>AC</math> is perpendicular to <math>BC</math>. So the slope of <math>BC</math> is the negative reciprocal of the slope of <math>AC</math>, yielding <math>m+n=\frac{1}{m-n}</math> <math>\Rightarrow</math> <math>m^2-n^2=1</math>.  
 
This means that <math>\angle C=90^\circ</math> and <math>AC</math> is perpendicular to <math>BC</math>. So the slope of <math>BC</math> is the negative reciprocal of the slope of <math>AC</math>, yielding <math>m+n=\frac{1}{m-n}</math> <math>\Rightarrow</math> <math>m^2-n^2=1</math>.  

Revision as of 01:36, 23 December 2009

Let the coordinates of $A$ be $(m, m^2)$ and the coordinates of $C$ be $(n, n^2)$. Since the line $AB$ is parallel to the $x$-axis, the coordinates of $B$ must be $(-m, m^2)$. Then the slope of line $AC$ is $\frac{m^2-n^2}{m-n}=\frac{(m+n)(m-n)}{m-n}=m+n$. The slope of line $BC$ is $\frac{m^2-n^2}{-m-n}=-\frac{(m+n)(m-n)}{m+n}=-(m-n)$.

Supposing $\angle A=90^\circ$, $AC$ is perpendicular to $AB$ and, it follows, to the $x$-axis, making $AB$ a segment of the line x=m. But that would mean that the coordinates of $C$ are $(m, m^2)$, contradicting the given that points $A$ and $C$ are distinct. So $\angle A$ is not $90^\circ$. By a similar logic, neither is $\angle B$.

This means that $\angle C=90^\circ$ and $AC$ is perpendicular to $BC$. So the slope of $BC$ is the negative reciprocal of the slope of $AC$, yielding $m+n=\frac{1}{m-n}$ $\Rightarrow$ $m^2-n^2=1$.

Because $m^2-n^2$ is the length of the altitude of triangle $ABC$ from $AB$, and $2m$ is the length of $AB$, the area of $\triangle ABC=m(m^2-n^2)=2008$. Since $m^2-n^2=1$, $m=2008$. Substituting, $2008^2-n^2=1$ $\Rightarrow$ $n^2=2008^2-1=4032063$, whose digits sum to $18 \Rightarrow \textbf{(C)}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_17&oldid=33083"