Let triangle ABC with $ AB=c$ $AC=b$ $BC=a$  $R$ circumradius, $p$ half peremetr of $ABC$.
I f $\frac{acosA+bcosB+ccosC}{asinA+bsinB+csinC}=\frac{p}{9R}$  then find all  value of $cosA$.
Given the problem, we need to prove that in any triangle \(ABC\), if 
\[
\frac{a \cos A + b \cos B + c \cos C}{a \sin A + b \sin B + c \sin C} = \frac{p}{9R}
\]
then \(a = b = c\).

1. **Expressing \(\sum a \cos A\) and \(\sum a \sin A\):**

   We start by expressing \(\sum a \cos A\) and \(\sum a \sin A\) in terms of known trigonometric identities and properties of the triangle.

   \[
   \sum a \cos A = a \cos A + b \cos B + c \cos C
   \]

   Using the identity \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\), we get:

   \[
   a \cos A = a \cdot \frac{b^2 + c^2 - a^2}{2bc}
   \]

   Similarly, for \(\cos B\) and \(\cos C\):

   \[
   b \cos B = b \cdot \frac{a^2 + c^2 - b^2}{2ac}
   \]
   \[
   c \cos C = c \cdot \frac{a^2 + b^2 - c^2}{2ab}
   \]

   Summing these up:

   \[
   \sum a \cos A = \frac{a(b^2 + c^2 - a^2)}{2bc} + \frac{b(a^2 + c^2 - b^2)}{2ac} + \frac{c(a^2 + b^2 - c^2)}{2ab}
   \]

   Simplifying, we get:

   \[
   \sum a \cos A = \frac{a(b^2 + c^2 - a^2) + b(a^2 + c^2 - b^2) + c(a^2 + b^2 - c^2)}{2abc}
   \]

   \[
   \sum a \cos A = \frac{a b^2 + a c^2 - a^3 + b a^2 + b c^2 - b^3 + c a^2 + c b^2 - c^3}{2abc}
   \]

   \[
   \sum a \cos A = \frac{2(ab^2 + ac^2 + ba^2 + bc^2 + ca^2 + cb^2 - a^3 - b^3 - c^3)}{2abc}
   \]

   \[
   \sum a \cos A = \frac{2(ab^2 + ac^2 + ba^2 + bc^2 + ca^2 + cb^2 - a^3 - b^3 - c^3)}{2abc}
   \]

   \[
   \sum a \cos A = \frac{2S}{R}
   \]

2. **Expressing \(\sum a \sin A\):**

   Using the identity \(\sin A = \frac{a}{2R}\), we get:

   \[
   \sum a \sin A = a \sin A + b \sin B + c \sin C
   \]

   \[
   \sum a \sin A = a \cdot \frac{a}{2R} + b \cdot \frac{b}{2R} + c \cdot \frac{c}{2R}
   \]

   \[
   \sum a \sin A = \frac{a^2 + b^2 + c^2}{2R}
   \]

3. **Substituting into the given equation:**

   Given:
   \[
   \frac{\sum a \cos A}{\sum a \sin A} = \frac{p}{9R}
   \]

   Substituting the expressions we derived:

   \[
   \frac{\frac{2S}{R}}{\frac{a^2 + b^2 + c^2}{2R}} = \frac{p}{9R}
   \]

   Simplifying:

   \[
   \frac{2S}{a^2 + b^2 + c^2} = \frac{p}{9R}
   \]

   Since \(S = \frac{abc}{4R}\) and \(p = \frac{a + b + c}{2}\):

   \[
   \frac{2 \cdot \frac{abc}{4R}}{a^2 + b^2 + c^2} = \frac{\frac{a + b + c}{2}}{9R}
   \]

   Simplifying further:

   \[
   \frac{abc}{2R(a^2 + b^2 + c^2)} = \frac{a + b + c}{18R}
   \]

   Cross-multiplying:

   \[
   18R \cdot abc = 2R(a + b + c)(a^2 + b^2 + c^2)
   \]

   Dividing both sides by \(2R\):

   \[
   9abc = (a + b + c)(a^2 + b^2 + c^2)
   \]

4. **Analyzing the equation:**

   For the equation \(9abc = (a + b + c)(a^2 + b^2 + c^2)\) to hold true, we need to consider the equality case in the AM-GM inequality. This implies that \(a = b = c\).

   Therefore, the only possible solution is when \(a = b = c\).

\(\blacksquare\)

The final answer is \( \boxed{ a = b = c } \).