Let $ABCD$ be a convex quadrilateral. $O$ is the intersection of $AC$ and $BD$. $AO=3$ ,$BO=4$, $CO=5$, $DO=6$. $X$ and $Y$ are points in segment $AB$ and $CD$ respectively, such that $X,O,Y$ are collinear. The minimum of $\frac{XB}{XA}+\frac{YC}{YD}$ can be written as $\frac{a\sqrt{c}}{b}$ , where $\frac{a}{b}$ is in lowest term and $c$ is not divisible by any square number greater then $1$. What is the value of $10a+b+c$?
1. We start by expressing the ratios \(\frac{XB}{XA}\) and \(\frac{YC}{YD}\) in terms of the areas of triangles:
   \[
   \frac{XB}{XA} = \frac{[OXB]}{[OXA]} = \frac{OB \times \sin \angle BOX}{OA \times \sin \angle AOX}
   \]
   Similarly,
   \[
   \frac{YC}{YD} = \frac{[OYC]}{[OYD]} = \frac{OC \times \sin \angle YOC}{OD \times \sin \angle YOD}
   \]

2. Since \(X, O, Y\) are collinear, we have:
   \[
   \angle AOX = \angle COY \quad \text{and} \quad \angle BOX = \angle DOY
   \]

3. Therefore, we can write:
   \[
   \frac{XB}{XA} + \frac{YC}{YD} = \frac{OB \times \sin \angle BOX}{OA \times \sin \angle AOX} + \frac{OC \times \sin \angle YOC}{OD \times \sin \angle YOD}
   \]
   Substituting the given lengths:
   \[
   = \frac{4}{3} \cdot \frac{\sin \angle BOX}{\sin \angle AOX} + \frac{5}{6} \cdot \frac{\sin \angle AOX}{\sin \angle BOX}
   \]

4. Let \(x = \frac{\sin \angle BOX}{\sin \angle AOX}\). Then:
   \[
   \frac{XB}{XA} + \frac{YC}{YD} = \frac{4}{3}x + \frac{5}{6}x^{-1}
   \]

5. To find the minimum value of \(\frac{4}{3}x + \frac{5}{6}x^{-1}\), we use the AM-GM inequality:
   \[
   \frac{4}{3}x + \frac{5}{6}x^{-1} \geq 2 \sqrt{\frac{4}{3}x \cdot \frac{5}{6}x^{-1}} = 2 \sqrt{\frac{4}{3} \cdot \frac{5}{6}} = 2 \sqrt{\frac{20}{18}} = \frac{2\sqrt{10}}{3}
   \]

6. Equality in the AM-GM inequality holds when:
   \[
   \frac{4}{3}x = \frac{5}{6}x^{-1} \implies x^2 = \frac{5}{8} \implies x = \sqrt{\frac{5}{8}}
   \]

7. We need to verify that there exist angles \(\angle AOX\) and \(\angle BOX\) such that \(x = \sqrt{\frac{5}{8}}\). This is possible by setting:
   \[
   \angle AOX = 90^\circ \quad \text{and} \quad \angle BOX = \arcsin \sqrt{\frac{5}{8}} \approx 52.24^\circ
   \]

Thus, the minimum value of \(\frac{XB}{XA} + \frac{YC}{YD}\) is \(\frac{2\sqrt{10}}{3}\).

The final answer is \( \boxed{ 10a + b + c = 10 \cdot 2 + 3 + 10 = 23 } \).