Let $ABC$ be a triangle with area $5$ and $BC = 10.$ Let $E$ and $F$ be the midpoints of sides $AC$ and $AB$ respectively, and let $BE$ and $CF$ intersect at $G.$ Suppose that quadrilateral $AEGF$ can be inscribed in a circle. Determine the value of $AB^2+AC^2.$

[i]Proposed by Ray Li[/i]
1. Given that \( \triangle ABC \) has an area of 5 and \( BC = 10 \). Let \( E \) and \( F \) be the midpoints of sides \( AC \) and \( AB \) respectively, and let \( BE \) and \( CF \) intersect at \( G \). Since \( G \) is the centroid of \( \triangle ABC \), it divides each median in the ratio 2:1.

2. Since \( E \) and \( F \) are midpoints, \( AE = \frac{AC}{2} \) and \( AF = \frac{AB}{2} \). The centroid \( G \) divides \( BE \) and \( CF \) such that \( BG = \frac{2}{3}BE \) and \( CG = \frac{2}{3}CF \).

3. Given that quadrilateral \( AEGF \) can be inscribed in a circle, we apply the Power of a Point theorem from \( C \) to the circumcircle of quadrilateral \( AEGF \). This gives us:
   \[
   CE \cdot (CE + CA) = 2x(2x + x)
   \]
   Simplifying, we get:
   \[
   2 \cdot CE^2 = 6x^2 \implies CE = x\sqrt{3} = AE
   \]
   Similarly, for \( BF \):
   \[
   BF = y\sqrt{3} = AF
   \]

4. We need to find \( AB^2 + AC^2 \). Since \( AE = x\sqrt{3} \) and \( AF = y\sqrt{3} \), we have:
   \[
   AB^2 + AC^2 = (2x\sqrt{3})^2 + (2y\sqrt{3})^2 = 12x^2 + 12y^2
   \]

5. Let \( \angle EAF = \theta \), so that \( \angle EGF = \pi - \theta \) (since \( AEGF \) is cyclic). Because \( \triangle AEF \sim \triangle ACB \) with similarity ratio \( \frac{1}{2} \), we have:
   \[
   EF = \frac{1}{2} BC = \frac{1}{2} \cdot 10 = 5
   \]

6. Applying the Law of Cosines to \( \triangle AEF \):
   \[
   25 = 3x^2 + 3y^2 - 6xy \cos \theta
   \]

7. Applying the Law of Cosines to \( \triangle GEF \):
   \[
   25 = x^2 + y^2 - 2xy \cos(\pi - \theta) = x^2 + y^2 + 2xy \cos \theta
   \]

8. Multiplying the second equation by 3 and adding to the first equation:
   \[
   25 + 3(25) = (3x^2 + 3y^2) + 3(x^2 + y^2)
   \]
   Simplifying, we get:
   \[
   100 = 6(x^2 + y^2) \implies x^2 + y^2 = \frac{100}{6} = \frac{50}{3}
   \]

9. Therefore:
   \[
   12(x^2 + y^2) = 12 \cdot \frac{50}{3} = 200
   \]

The final answer is \( \boxed{200} \).