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

(Solution)
(Solution)
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== Solution ==
 
== Solution ==
Let the two points be <math>P</math> and <math>Q</math>; <math>P=(x_1,y_1)</math> and <math>Q=(x_2,y_2)</math>
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Let the two points <math>P</math> and <math>Q</math> be defined with coordinates; <math>P=(x_1,y_1)</math> and <math>Q=(x_2,y_2)</math>
  
 
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).  
 
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).  

Revision as of 17:47, 22 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. Unknown error_msg)

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 then $25$ even and $25$ odd numbers 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