Let $ABC$ be a triangle with $AB=13$, $BC=14$, and $CA=15$. Points $P$, $Q$, and $R$ are chosen on segments $BC$, $CA$, and $AB$, respectively, such that triangles $AQR$, $BPR$, $CPQ$ have the same perimeter, which is $\frac{4}{5}$ of the perimeter of $PQR$. What is the perimeter of $PQR$?


[i]2021 CCA Math Bonanza Individual Round #2[/i]
1. Denote \( p(X) \) as the perimeter of figure \( X \). We are given that the perimeters of triangles \( AQR \), \( BPR \), and \( CPQ \) are equal and each is \(\frac{4}{5}\) of the perimeter of triangle \( PQR \).

2. Let \( p(PQR) \) be the perimeter of triangle \( PQR \). Then, the perimeter of each of the triangles \( AQR \), \( BPR \), and \( CPQ \) is \(\frac{4}{5} p(PQR) \).

3. The sum of the perimeters of triangles \( AQR \), \( BPR \), and \( CPQ \) is:
   \[
   p(AQR) + p(BPR) + p(CPQ) = 3 \left( \frac{4}{5} p(PQR) \right)
   \]

4. According to the problem, the sum of the perimeters of triangles \( AQR \), \( BPR \), and \( CPQ \) minus the perimeter of triangle \( PQR \) equals the perimeter of triangle \( ABC \), which is 42 (since \( AB + BC + CA = 13 + 14 + 15 = 42 \)).

5. Therefore, we have:
   \[
   3 \left( \frac{4}{5} p(PQR) \right) - p(PQR) = 42
   \]

6. Simplify the left-hand side:
   \[
   3 \left( \frac{4}{5} p(PQR) \right) = \frac{12}{5} p(PQR)
   \]
   \[
   \frac{12}{5} p(PQR) - p(PQR) = 42
   \]

7. Combine the terms:
   \[
   \frac{12}{5} p(PQR) - \frac{5}{5} p(PQR) = 42
   \]
   \[
   \frac{7}{5} p(PQR) = 42
   \]

8. Solve for \( p(PQR) \):
   \[
   p(PQR) = 42 \times \frac{5}{7}
   \]
   \[
   p(PQR) = 30
   \]

The final answer is \( \boxed{30} \).