Let $ABC$ be a triangle with $AB=17$ and $AC=23$. Let $G$ be the centroid of $ABC$, and let $B_1$ and $C_1$ be on the circumcircle of $ABC$ with $BB_1\parallel AC$ and $CC_1\parallel AB$. Given that $G$ lies on $B_1C_1$, the value of $BC^2$ can be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Determine $100m+n$.

[i]Proposed by Michael Ren[/i]
1. **Given Information and Setup:**
   - Let \( \triangle ABC \) be a triangle with \( AB = 17 \) and \( AC = 23 \).
   - \( G \) is the centroid of \( \triangle ABC \).
   - \( B_1 \) and \( C_1 \) are points on the circumcircle of \( \triangle ABC \) such that \( BB_1 \parallel AC \) and \( CC_1 \parallel AB \).
   - Given that \( G \) lies on \( B_1C_1 \), we need to determine \( BC^2 \) in the form \( \frac{m}{n} \) where \( m \) and \( n \) are relatively prime positive integers, and find \( 100m + n \).

2. **Key Observations:**
   - Since \( BB_1 \parallel AC \) and \( CC_1 \parallel AB \), quadrilaterals \( ACBB_1 \) and \( ABCC_1 \) are isosceles trapezoids.
   - The centroid \( G \) divides each median in the ratio \( 2:1 \).

3. **Using the Hint:**
   - Let \( E \) be the foot of the altitude from \( B \) to \( AC \), and \( F \) be the foot of the altitude from \( C \) to \( AB \).
   - Let \( M \) be the midpoint of \( AC \).
   - Since \( ACBB_1 \) is an isosceles trapezoid, \( BB_1 = 2EM \).
   - Similarly, \( CC_1 = 2FM \).

4. **Collinearity and Symmedian:**
   - Since \( B_1, G, E \) are collinear and \( C_1, G, F \) are collinear, \( E, G, F \) must also be collinear.
   - \( AG \) is the symmedian of \( \triangle AEF \), implying \( \frac{EG}{FG} = \left( \frac{EA}{FA} \right)^2 = \left( \frac{BA}{CA} \right)^2 = \left( \frac{17}{23} \right)^2 \).

5. **Homothety and Similarity:**
   - Consider a homothety centered at \( A \) with a factor of \( \frac{3}{2} \), mapping \( E \) to \( E' \), \( F \) to \( F' \), and \( G \) to the midpoint \( K \) of \( BC \).
   - Triangles \( BF'K \) and \( E'CK \) are similar, so \( \frac{E'K}{BK} = \frac{CK}{F'K} \).

6. **Calculating \( BC^2 \):**
   - Since \( BK = CK \) and \( \frac{E'K}{F'K} = \left( \frac{17}{23} \right)^2 \), we have:
     \[
     \frac{E'K}{BK} \cdot \frac{CK}{F'K} = \left( \frac{17}{23} \right)^2
     \]
   - Solving for \( \cos A \) using the cosine rule:
     \[
     \cos A = \frac{b^2 + c^2 - a^2}{2bc}
     \]
     where \( a = BC \), \( b = AC = 23 \), and \( c = AB = 17 \).

7. **Final Calculation:**
   - Using the cosine rule:
     \[
     BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos A
     \]
     \[
     BC^2 = 17^2 + 23^2 - 2 \cdot 17 \cdot 23 \cdot \cos A
     \]
   - Given \( \cos A = \frac{b^2 + c^2}{3bc} \):
     \[
     BC^2 = 17^2 + 23^2 - \frac{2}{3} (17^2 + 23^2)
     \]
     \[
     BC^2 = \frac{1}{3} (17^2 + 23^2)
     \]
     \[
     BC^2 = \frac{1}{3} (289 + 529) = \frac{818}{3}
     \]

The final answer is \( \boxed{ 100m + n = 100 \times 818 + 3 = 81803 } \).