Difference between revisions of "2009 AIME I Problems/Problem 11"
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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 | + | 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 | ||
| Line 18: | Line 18: | ||
<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, < | + | 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 < | + | Also note that the maximum value for x is <math>49</math> and minimum is <math>0</math>. |
| − | There are < | + | 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}</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 21:28, 20 March 2009
Problem
Consider the set of all triangles 
where 
is the origin and 
and 
are distinct points in the plane with nonnegative integer coordinates 
such that 
. 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 
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
![\[(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))\]](http://latex.artofproblemsolving.com/7/f/6/7f629e2c99a4b43ff565a7d3f11e29bdbe3b9751.png)
![\[=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))\]](http://latex.artofproblemsolving.com/b/b/f/bbfe693cbc8aa9dffc3f4d31668b12b66ba0fba0.png)
since 2009 is not even, 
must be even
Thus the two x's have to be both odd or even.
Also note that the maximum value for x is 
and minimum is 
.
There are 
even and odd number
Thus, there are
of such triangle
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 | ||
