Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Coordinate Geometry.

Coordinate Geometry Problem – AIME 2009


Consider the set of all triangles OPQ where O  is the origin and P and Q are distinct points in the plane with non negative integer coordinates (x,y) such that 41x+y=2009 . Find the number of such distinct triangles whose area is a positive integer.

  • is 107
  • is 600
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Geometry

Check the Answer


But try the problem first…

Answer: is 600.

Source
Suggested Reading

AIME, 2009, Question 11

Geometry Revisited by Coxeter

Try with Hints


First hint

 let P and Q be defined with coordinates; P=(\(x_1,y_1)\) and Q(\(x_2,y_2)\). Let the line 41x+y=2009 intersect the x-axis at X and the y-axis at Y . X (49,0) , and Y(0,2009). such that there are 50 points.

here [OPQ]=[OYX]-[OXQ] OY=2009 OX=49 such that [OYX]=\(\frac{1}{2}\)OY.OX=\(\frac{1}{2}\)2009.49 And [OYP]=\(\frac{1}{2}\)\(2009x_1\)  and [OXQ]=\(\frac{1}{2}\)(49)\(y_2\).

Second Hint

2009.49 is odd, area OYX not integer of form k+\(\frac{1}{2}\) where k is an integer

Final Step

41x+y=2009 taking both 25  \(\frac{25!}{2!23!}+\frac{25!}{2!23!}\)=300+300=600.

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