If `P_1, P_2, P_3` be the perpendiculars from the vertices of a triangle to the opposite sidės then are the following statements true or false ? If not true, write the corrrect answer.
`1/p_1 + 1/p+1/p_3=1/R`

Answer

To determine whether the statement \( \frac{1}{P_1} + \frac{1}{P_2} + \frac{1}{P_3} = \frac{1}{R} \) is true or false, we will analyze the relationship between the perpendiculars from the vertices of a triangle to the opposite sides and the circumradius \( R \).

### Step-by-Step Solution:

1. **Identify the Perpendiculars**:
Let \( P_1, P_2, P_3 \) be the lengths of the perpendiculars dropped from vertices \( A, B, C \) of triangle \( ABC \) to the opposite sides \( BC, CA, AB \) respectively.

2. **Area of the Triangle**:
The area \( \Delta \) of triangle \( ABC \) can be expressed using the base and height:
- Using side \( BC \) as the base:
\[ \Delta = \frac{1}{2} \times a \times P_1 \]
- Using side \( CA \) as the base:
\[ \Delta = \frac{1}{2} \times b \times P_2 \]
- Using side \( AB \) as the base:
\[ \Delta = \frac{1}{2} \times c \times P_3 \]

3. **Expressing Perpendiculars**:
From these area expressions, we can derive the formulas for the perpendiculars:
- \( P_1 = \frac{2\Delta}{a} \)
- \( P_2 = \frac{2\Delta}{b} \)
- \( P_3 = \frac{2\Delta}{c} \)

4. **Finding the Reciprocals**:
Now we find the reciprocals of the perpendiculars:
\[ \frac{1}{P_1} = \frac{a}{2\Delta}, \quad \frac{1}{P_2} = \frac{b}{2\Delta}, \quad \frac{1}{P_3} = \frac{c}{2\Delta} \]

5. **Summing the Reciprocals**:
Adding these reciprocals together:
\[ \frac{1}{P_1} + \frac{1}{P_2} + \frac{1}{P_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} + \frac{c}{2\Delta} = \frac{a + b + c}{2\Delta} \]

6. **Relating to the Circumradius**:
The circumradius \( R \) of triangle \( ABC \) is given by the formula:
\[ R = \frac{abc}{4\Delta} \]
Therefore, we can express \( \Delta \) in terms of \( R \):
\[ \Delta = \frac{abc}{4R} \]

7. **Substituting Back**:
Substitute \( \Delta \) back into the sum of reciprocals:
\[ \frac{1}{P_1} + \frac{1}{P_2} + \frac{1}{P_3} = \frac{a + b + c}{2 \cdot \frac{abc}{4R}} = \frac{2R(a + b + c)}{abc} \]

8. **Conclusion**:
The final expression does not simplify to \( \frac{1}{R} \). Instead, it shows that:
\[ \frac{1}{P_1} + \frac{1}{P_2} + \frac{1}{P_3} = \frac{2R(a + b + c)}{abc} \]
Thus, the statement \( \frac{1}{P_1} + \frac{1}{P_2} + \frac{1}{P_3} = \frac{1}{R} \) is **false**.

### Correct Answer:
The correct relationship is:
\[ \frac{1}{P_1} + \frac{1}{P_2} + \frac{1}{P_3} = \frac{2R(a + b + c)}{abc} \]