Let `Delta` denote the area of the `DeltaABCand Delta p` be the area of its pedal triangle. If `Delta=k Deltap,` then k is equal to

Answer

To solve the problem, we need to find the value of \( k \) such that the area of triangle \( ABC \) (denoted as \( \Delta \)) is equal to \( k \) times the area of its pedal triangle \( \Delta_p \).

### Step-by-Step Solution:

1. **Understanding the Pedal Triangle**:
The pedal triangle \( DEF \) is formed by dropping perpendiculars from the vertices \( A, B, C \) of triangle \( ABC \) onto the opposite sides. The vertices of the pedal triangle are the feet of these perpendiculars.

2. **Area of the Pedal Triangle**:
The area \( \Delta_p \) of the pedal triangle can be expressed using the lengths of its sides and the sine of the angles between them. The sides of the pedal triangle can be expressed in terms of the sides of triangle \( ABC \) and the angles at its vertices:
- \( DE = C \cos C \)
- \( EF = A \cos A \)
- \( FD = B \cos B \)

Therefore, the area \( \Delta_p \) can be calculated as:
\[ \Delta_p = \frac{1}{2} \times DE \times EF \times \sin D \]
Substituting the expressions for \( DE \) and \( EF \):
\[ \Delta_p = \frac{1}{2} \times (C \cos C) \times (B \cos B) \times \sin(180^\circ - 2A) \]
Since \( \sin(180^\circ - x) = \sin x \), we have:
\[ \Delta_p = \frac{1}{2} \times C \times B \times \cos C \times \cos B \times \sin(2A) \]
Using the identity \( \sin(2A) = 2 \sin A \cos A \), we can rewrite it as:
\[ \Delta_p = \frac{1}{2} \times C \times B \times \cos C \times \cos B \times 2 \sin A \cos A \]
Simplifying gives:
\[ \Delta_p = C \times B \times \cos C \times \cos B \times \sin A \cos A \]

3. **Area of Triangle ABC**:
The area \( \Delta \) of triangle \( ABC \) can be expressed as:
\[ \Delta = \frac{1}{2} \times B \times C \times \sin A \]

4. **Relating the Two Areas**:
We know from the problem statement that:
\[ \Delta = k \Delta_p \]
Substituting the expressions for \( \Delta \) and \( \Delta_p \):
\[ \frac{1}{2} \times B \times C \times \sin A = k \times (C \times B \times \cos C \times \cos B \times \sin A \cos A) \]

5. **Solving for k**:
Dividing both sides by \( C \times B \times \sin A \) (assuming \( \sin A \neq 0 \)):
\[ \frac{1}{2} = k \times \cos C \times \cos B \times \cos A \]
Rearranging gives:
\[ k = \frac{1}{2 \cos A \cos B \cos C} \]

### Final Result:
Thus, the value of \( k \) is:
\[ k = \frac{1}{2 \cos A \cos B \cos C} \]