Given a right triangle ABC right angled at C and whose legs are given `1+4log_(p^2)(2p),1+2^(log_2(log_2(p))` and hypotenuse is given to be `1+log_2(4p)`. The area of `trianleABC` and circle circumscribing it are `Delta_1` and`Delta_2` respectively.

Options

A. `(1)/(2)`
B. `(1)/(sqrt(2))`
C. `(sqrt(3))/(2)`
D. 1

Answer

To solve the problem, we need to find the area of triangle ABC and the value of the expression involving the area and the circumradius. Let's break this down step by step.

### Step 1: Define the sides of the triangle
Given the legs of the triangle:
- \( AC = 1 + 4 \log_{p^2}(2p) \)
- \( BC = 1 + 2^{\log_2(\log_2(p))} \)

And the hypotenuse:
- \( AB = 1 + \log_2(4p) \)

### Step 2: Simplify the sides
**For side AC:**
Using the property of logarithms:
\[ \log_{p^2}(2p) = \frac{\log_2(2p)}{\log_2(p^2)} = \frac{\log_2(2) + \log_2(p)}{2 \log_2(p)} = \frac{1 + \log_2(p)}{2 \log_2(p)} \]
Thus,
\[ AC = 1 + 4 \cdot \frac{1 + \log_2(p)}{2 \log_2(p)} = 1 + \frac{4 + 4\log_2(p)}{2\log_2(p)} = 1 + \frac{2 + 2\log_2(p)}{\log_2(p)} = 3 + \frac{2}{\log_2(p)} \]

**For side BC:**
Using the property \( 2^{\log_2(x)} = x \):
\[ BC = 1 + \log_2(p) \]

**For hypotenuse AB:**
Using the property of logarithms:
\[ AB = 1 + \log_2(4) + \log_2(p) = 1 + 2 + \log_2(p) = 3 + \log_2(p) \]

### Step 3: Apply Pythagorean theorem
Using the Pythagorean theorem:
\[ AC^2 + BC^2 = AB^2 \]
Substituting the values:
\[ \left(3 + \frac{2}{\log_2(p)}\right)^2 + (1 + \log_2(p))^2 = (3 + \log_2(p))^2 \]

### Step 4: Expand and simplify
Expanding both sides:
\[ \left(3 + \frac{2}{\log_2(p)}\right)^2 = 9 + 2 \cdot 3 \cdot \frac{2}{\log_2(p)} + \left(\frac{2}{\log_2(p)}\right)^2 = 9 + \frac{12}{\log_2(p)} + \frac{4}{(\log_2(p))^2} \]
\[ (1 + \log_2(p))^2 = 1 + 2\log_2(p) + (\log_2(p))^2 \]
\[ (3 + \log_2(p))^2 = 9 + 6\log_2(p) + (\log_2(p))^2 \]

Combining:
\[ 9 + \frac{12}{\log_2(p)} + \frac{4}{(\log_2(p))^2} + 1 + 2\log_2(p) = 9 + 6\log_2(p) + (\log_2(p))^2 \]

### Step 5: Solve for \( p \)
This leads to a quadratic equation in terms of \( \log_2(p) \). Solving this will give us the value of \( p \).

### Step 6: Calculate the area of triangle ABC
The area \( \Delta_1 \) of triangle ABC is given by:
\[ \Delta_1 = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times \left(3 + \frac{2}{\log_2(p)}\right) \times (1 + \log_2(p)) \]

### Step 7: Find the circumradius \( R \)
The circumradius \( R \) of triangle ABC can be calculated using:
\[ R = \frac{AB}{2 \sin C} \]
Since \( C \) is the right angle, \( \sin C = 1 \).

### Step 8: Final expression
Finally, we need to evaluate:
\[ \sin(\phi) \cdot 2P^2 \Delta_1 + 2/6 \]
Substituting the values of \( P \) and \( \Delta_1 \) will give the final answer.