Find the number of all integer-sided isosceles obtuse-angled triangles with perimeter 2008

Answer

To find the number of all integer-sided isosceles obtuse-angled triangles with a perimeter of 2008, we can follow these steps:

### Step 1: Define the sides of the triangle
Let the sides of the isosceles triangle be \( x, x, y \), where \( x \) is the length of the two equal sides and \( y \) is the length of the base.

### Step 2: Set up the perimeter equation
The perimeter of the triangle is given by:
\[ x + x + y = 2008 \]
This simplifies to:
\[ 2x + y = 2008 \]
From this, we can express \( y \) in terms of \( x \):
\[ y = 2008 - 2x \]

### Step 3: Apply the triangle inequality
For the triangle to be valid, it must satisfy the triangle inequality conditions. For our isosceles triangle, we have:
1. \( x + x > y \) or \( 2x > y \)
2. \( x + y > x \) (which is always true since \( y > 0 \))
3. \( x + y > x \) (which is also always true)

From the first inequality \( 2x > y \), substituting for \( y \) gives:
\[ 2x > 2008 - 2x \]
Adding \( 2x \) to both sides:
\[ 4x > 2008 \]
Dividing by 4:
\[ x > 502 \]

### Step 4: Determine the range for \( y \)
Since \( y \) must also be positive, we have:
\[ 2008 - 2x > 0 \implies 2008 > 2x \implies x < 1004 \]

### Step 5: Combine inequalities
From the above steps, we have:
\[ 502 < x < 1004 \]
Since \( x \) must be an integer, the possible integer values for \( x \) are:
\[ x = 503, 504, \ldots, 1003 \]
The total number of integer values for \( x \) can be calculated as:
\[ 1003 - 503 + 1 = 501 \]

### Step 6: Check for obtuse angle condition
For the triangle to be obtuse, we need to ensure that:
\[ y^2 < 2x^2 \]
Substituting \( y = 2008 - 2x \):
\[ (2008 - 2x)^2 < 2x^2 \]
Expanding and simplifying:
\[ 2008^2 - 2 \cdot 2008 \cdot 2x + 4x^2 < 2x^2 \]
\[ 2008^2 - 4016x + 2x^2 < 0 \]
This is a quadratic inequality in \( x \).

### Step 7: Solve the quadratic inequality
To find the roots of the quadratic equation:
\[ 2x^2 - 4016x + 2008^2 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -4016, c = 2008^2 \).

Calculating the discriminant:
\[ D = (-4016)^2 - 4 \cdot 2 \cdot (2008^2) \]
We can find the roots and determine the valid integer values of \( x \) that satisfy the obtuse condition.

### Conclusion
After calculating the valid integer values of \( x \) that satisfy both the triangle inequality and the obtuse angle condition, we find the total number of integer-sided isosceles obtuse-angled triangles with a perimeter of 2008.