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

(Solution)
(Solution)
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Let the two points be point P and Q
 
Let the two points be point P and Q
  
and <math>P=(x_1,y_1),Q=(x_2,y_2)
+
and <math>P=(x_1,y_1),Q=(x_2,y_2)</math>
  
We cna calculate the area of the parallelogram span with the determinant of matrix PQ, P above Q, since triangle is half of the area of the parallelogram. We just need the determinant to be even
+
We can calculate the area of the parallelogram span with the determinant of matrix PQ, P above Q, since triangle is half of the area of the parallelogram. We just need the determinant to be even
  
 
The deteminant is
 
The deteminant is
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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>
 
<cmath>=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))</cmath>
  
since 2009 is not even, </math>((x_1)-(x_2))<math> must be even
+
since 2009 is not even, <math>((x_1)-(x_2))</math> must be even
  
 
Thus the two x's have to be both odd or even.
 
Thus the two x's have to be both odd or even.
  
Also note that the maximum value for x is </math>49<math> and minimum is </math>0<math>.
+
Also note that the maximum value for x is <math>49</math> and minimum is <math>0</math>.
  
There are </math>25<math> even and </math>25<math> odd number
+
There are <math>25</math> even and <math>25</math> odd number
  
 
Thus, there are  
 
Thus, there are  
  
</math>(_{25}C_2)+(_{25}C_2)=\boxed{600}$of such triangle
+
<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 22:28, 20 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

Solution 1 (This solution requires linear algeber knowledgw)

Let the two points be point P and Q

and $P=(x_1,y_1),Q=(x_2,y_2)$

We can calculate the area of the parallelogram span with the determinant of matrix PQ, P above Q, since triangle is half of the area of the parallelogram. We just need the determinant to be even

The deteminant 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 have to be both odd or even.

Also note that the maximum value for x is $49$ and minimum is $0$.

There are $25$ even and $25$ odd number

Thus, there are

$(_{25}C_2)+(_{25}C_2)=\boxed{600}$of such triangle

See also

2009 AIME I (ProblemsAnswer KeyResources)
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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