It seems like there's some information missing from your question. However, I'll try to provide a general answer.

If you're asking about the number of triangles with integral sides AB and BC measuring 2001 units and 1002 units respectively, we need to consider the triangle inequality theorem. The theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

In this case, let's denote the third side as AC. For the triangle to exist:

AB + BC > AC, so 2001 + 1002 > AC, which means AC < 3003.

BC + AC > AB, so 1002 + AC > 2001, which means AC > 999.

AB + AC > BC, so 2001 + AC > 1002, which means AC > 1.

Combining these inequalities, we find that AC must be an integer between 1000 and 3002, inclusive. Therefore, there are 3002 - 1000 + 1 = 2003 such triangles.

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It seems like there's some information missing from your question. However, I'll try to provide a general answer.

If you're asking about the number of triangles with integral sides AB and BC measuring 2001 units and 1002 units respectively, we need to consider the triangle inequality theorem. The theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

In this case, let's denote the third side as AC. For the triangle to exist:

1. AB + BC > AC, so 2001 + 1002 > AC, which means AC < 3003.
2. BC + AC > AB, so 1002 + AC > 2001, which means AC > 999.
3. AB + AC > BC, so 2001 + AC > 1002, which means AC > 1.

Combining these inequalities, we find that AC must be an integer between 1000 and 3002, inclusive. Therefore, there are 3002 - 1000 + 1 = 2003 such triangles.

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It seems like there's some information missing from your question. However, I'll try to provide a general answer.

If you're asking about the number of triangles with integral sides AB and BC measuring 2001 units and 1002 units respectively, we need to consider the triangle inequality theorem. The theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

In this case, let's denote the third side as AC. For the triangle to exist:

1. AB + BC > AC, so 2001 + 1002 > AC, which means AC < 3003.
2. BC + AC > AB, so 1002 + AC > 2001, which means AC > 999.
3. AB + AC > BC, so 2001 + AC > 1002, which means AC > 1.

Combining these inequalities, we find that AC must be an integer between 1000 and 3002, inclusive. Therefore, there are 3002 - 1000 + 1 = 2003 such triangles.

has integral sides AB, BC measuring 2001 units and 1002 units. The number of such triangles is..

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