The diameter of the nine point circle of the triangle with vertices `(3, 4), (5 cos theta, 5 sin theta)` and `(5 sin theta, -5 cos theta)`, where `theta in R`, is

Answer

To find the diameter of the nine-point circle of the triangle with vertices \( A(3, 4) \), \( B(5 \cos \theta, 5 \sin \theta) \), and \( C(5 \sin \theta, -5 \cos \theta) \), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices are given as:
- \( A(3, 4) \)
- \( B(5 \cos \theta, 5 \sin \theta) \)
- \( C(5 \sin \theta, -5 \cos \theta) \)

### Step 2: Find the circumradius \( R \)
The circumradius \( R \) can be found using the distances from the circumcenter (which we will take as the origin \( O(0, 0) \)) to each vertex.

#### Step 2.1: Calculate \( OA \)
Using the distance formula:
\[ OA = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

#### Step 2.2: Calculate \( OB \)
Using the distance formula:
\[ OB = \sqrt{(5 \cos \theta - 0)^2 + (5 \sin \theta - 0)^2} = \sqrt{(5 \cos \theta)^2 + (5 \sin \theta)^2} \]
\[ = \sqrt{25 \cos^2 \theta + 25 \sin^2 \theta} = \sqrt{25(\cos^2 \theta + \sin^2 \theta)} = \sqrt{25 \cdot 1} = 5 \]

#### Step 2.3: Calculate \( OC \)
Using the distance formula:
\[ OC = \sqrt{(5 \sin \theta - 0)^2 + (-5 \cos \theta - 0)^2} = \sqrt{(5 \sin \theta)^2 + (-5 \cos \theta)^2} \]
\[ = \sqrt{25 \sin^2 \theta + 25 \cos^2 \theta} = \sqrt{25(\sin^2 \theta + \cos^2 \theta)} = \sqrt{25 \cdot 1} = 5 \]

### Step 3: Conclude the circumradius
Since \( OA = OB = OC = 5 \), we have:
\[ R = 5 \]

### Step 4: Find the radius of the nine-point circle \( R_n \)
The radius of the nine-point circle is given by the formula:
\[ R_n = \frac{1}{2} R \]
Substituting the value of \( R \):
\[ R_n = \frac{1}{2} \times 5 = \frac{5}{2} \]

### Step 5: Find the diameter of the nine-point circle
The diameter \( D \) of the nine-point circle is:
\[ D = 2 R_n = 2 \times \frac{5}{2} = 5 \]

### Final Answer
Thus, the diameter of the nine-point circle is \( 5 \).

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