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Difference between revisions of "2008 AMC 12B Problems/Problem 17"
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Because <math>m^2-n^2</math> is the length of the altitude of triangle <math>ABC</math> from <math>AB</math>, and <math>2m</math> is the length of <math>AB</math>, the area of <math>\triangle ABC=m(m^2-n^2)=2008</math>. Since <math>m^2-n^2=1</math>, <math>m=2008</math>.  
 
Because <math>m^2-n^2</math> is the length of the altitude of triangle <math>ABC</math> from <math>AB</math>, and <math>2m</math> is the length of <math>AB</math>, the area of <math>\triangle ABC=m(m^2-n^2)=2008</math>. Since <math>m^2-n^2=1</math>, <math>m=2008</math>.  
−
Substituting, <math>2008^2-n^2=1</math> <math>\Rightarrow</math> <math>n^2=2008^2-1=4032063</math>, whose digits sum to <math>18 \Rightarrow \textbf{(C)}</math>.
+
Substituting, <math>2008^2-n^2=1</math> <math>\Rightarrow</math> <math>n^2=2008^2-1=(2000+8)^2-1=4000000+32000+64-1=4032063</math>, whose digits sum to <math>18 \Rightarrow \textbf{(C)}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|num-b=16|num-a=18|ab=B}}
 
{{AMC12 box|year=2008|num-b=16|num-a=18|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:59, 2 February 2015

Problem

Let $A$, $B$ and $C$ be three distinct points on the graph of $y=x^2$ such that line $AB$ is parallel to the $x$-axis and $\triangle ABC$ is a right triangle with area $2008$. What is the sum of the digits of the $y$-coordinate of $C$?

$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$

Solution

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 $BC$. Let C be the point $(n, n^2)$. 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, $2008^2-n^2=1$ $\Rightarrow$ $n^2=2008^2-1=(2000+8)^2-1=4000000+32000+64-1=4032063$, whose digits sum to $18 \Rightarrow \textbf{(C)}$.

See Also

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

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