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2009 AIME I Problems/Problem 11

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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).

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 (ProblemsResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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