Difference between revisions of "2009 AIME I Problems/Problem 11"

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<cmath>(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))</cmath>
 
<cmath>(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))</cmath>
  
Since <math>2009</math> is not even, <math>((x_1)-(x_2))</math> must be even, thus the two <math>x</math>'s must be of the same parity. Also note that the maximum value for <math>x</math> is <math>49</math> and the minimum is <math>0</math>. There are then <math>25</math> even and <math>25</math> odd numbers available for use as coordinates and thus there are <math>(_{25}C_2)+(_{25}C_2)=\boxed{600}</math> such triangles.
+
Since <math>2009</math> is not even, <math>((x_1)-(x_2))</math> must be even, thus the two <math>x</math>'s must be of the same parity. Also note that the maximum value for <math>x</math> is <math>49</math> and the minimum is <math>0</math>. There are <math>25</math> even and <math>25</math> odd numbers available for use as coordinates and thus there are <math>(_{25}C_2)+(_{25}C_2)=\boxed{600}</math> such triangles.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2009|n=I|num-b=10|num-a=12}}

Revision as of 10:27, 28 March 2009

Problem

Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$. Find the number of such distinct triangles whose area is a positive integer.

Solution

Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$

We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).

$\det \left({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1 \above y_2}\right).$ (Error compiling LaTeX. ! Package amsmath Error: Old form `\matrix' should be \begin{matrix}.)

Since the triangle has half the area of the parallelogram, we just need the determinant to be even.

The determinant is

\[(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))\]

Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$'s must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$. There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=\boxed{600}$ such triangles.

See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions
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