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

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== Solution ==
 
== Solution ==
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'''Solution 1 (This solution requires linear alg. knowledge)'''
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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 be point P and Q
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We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).
  
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and <math>P=(x_1,y_1),Q=(x_2,y_2)</math>
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<math>\det \left({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1 \above y_2}\right).</math>
  
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We can calculate the area of the parallelogram span with the determinant of matrix
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Since the triangle has half the area of the parallelogram, we just need the determinant to be even.
  
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<math>\det ({\matrix {P \above Q}})=\det </math><math>({\matrix {x_1 \above x_2} \right \matrix {y_1  \above y_2})</math>
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The determinant is
  
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since triangle is half of the area of the parallelogram. We just need the determinant to be even
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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>
  
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The deteminant is
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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 and thus there are <math>(_{25}C_2)+(_{25}C_2)=\boxed{600}</math> such triangles.
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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))</cmath>
 
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<cmath>=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))</cmath>
 
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since 2009 is not even, <math>((x_1)-(x_2))</math> must be even
 
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Thus the two x's have to be both odd or even.
 
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Also note that the maximum value for x is <math>49</math> and minimum is <math>0</math>.
 
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There are <math>25</math> even and <math>25</math> odd number
 
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Thus, there are  
 
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<math>(_{25}C_2)+(_{25}C_2)=\boxed{600}</math>of such triangle
 
  
 
== 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 18:46, 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 be $P$ and $Q$; $P=(x_1,y_1)$ 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 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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