Let a,b,c be the sides BC, CA, AB of `DeltaABC` on xy plane. If abscissa and ordinate of vertices of the triangle are integers and R is the circumradius, then 2R can be equal to

Options

A. `(8)/(9)abc`
B. abc
C. `(9)/(8)abc`
D. `(abc)/(2)`

Explanation

To solve the problem, we need to determine the possible values of \(2R\) (where \(R\) is the circumradius) for a triangle \(ABC\) with integer coordinates for its vertices. We will derive the relationship between the circumradius and the area of the triangle, and then analyze the options given.

### Step-by-Step Solution:

1. **Understanding the Triangle Vertices**:
The vertices of triangle \(ABC\) are given to have integer coordinates. Let's denote the vertices as \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\), where \(x_i\) and \(y_i\) are integers.

2. **Area of Triangle**:
The area \(A\) of triangle \(ABC\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Since the coordinates are integers, the area \(A\) will also be a rational number.

3. **Circumradius Formula**:
The circumradius \(R\) of triangle \(ABC\) is given by the formula:
\[ R = \frac{abc}{4A} \]
where \(a\), \(b\), and \(c\) are the lengths of the sides opposite to vertices \(A\), \(B\), and \(C\) respectively.

4. **Finding \(2R\)**:
To find \(2R\), we multiply the circumradius by 2:
\[ 2R = \frac{abc}{2A} \]

5. **Minimum Area**:
The minimum area of a triangle with integer vertices is \(\frac{1}{2}\) square units. Therefore, we can say:
\[ A \geq \frac{1}{2} \]

6. **Inequality for \(2R\)**:
Substituting the minimum area into the formula for \(2R\):
\[ 2R = \frac{abc}{2A} \leq \frac{abc}{2 \times \frac{1}{2}} = abc \]
This means:
\[ 2R \leq abc \]

7. **Analyzing the Options**:
We need to check which of the given options satisfy the inequality \(2R \leq abc\):
- **Option 1**: \(2R \leq \frac{8}{9}abc\) - This is true since \(\frac{8}{9}abc < abc\).
- **Option 2**: \(2R \leq \frac{9}{8}abc\) - This is false since \(\frac{9}{8}abc > abc\).
- **Option 3**: \(2R \leq \frac{abc}{2}\) - This is false since \(\frac{abc}{2} < abc\).
- **Option 4**: \(2R \leq abc\) - This is true.

Thus, the valid options are **Option 1** and **Option 4**.

### Final Answer:
The values that \(2R\) can equal are \(2R \leq \frac{8}{9}abc\) and \(2R \leq abc\).