1. **State the problem:** Three regular polygons A, B, and C meet at a point, and their interior angles are in the ratio $a:b:c=3:4:5$. We need to show that polygon C has twice the number of sides as polygon B. 2. **Recall the formula for interior angles of a regular polygon:** The interior angle $I$ of a regular polygon with $n$ sides is given by $$I = \frac{(n-2) \times 180}{n}$$ 3. **Express the interior angles of polygons A, B, and C:** Let the number of sides be $n_A$, $n_B$, and $n_C$ respectively. Then, $$a = \frac{(n_A - 2) \times 180}{n_A}, \quad b = \frac{(n_B - 2) \times 180}{n_B}, \quad c = \frac{(n_C - 2) \times 180}{n_C}$$ 4. **Use the given ratio of interior angles:** Since $a:b:c = 3:4:5$, there exists a constant $k$ such that $$a = 3k, \quad b = 4k, \quad c = 5k$$ 5. **Sum of angles at the point:** The three polygons meet at a point, so the sum of their interior angles at that point is $360^\circ$: $$a + b + c = 3k + 4k + 5k = 12k = 360$$ 6. **Solve for $k$:** $$12k = 360 \implies k = 30$$ 7. **Find the actual interior angles:** $$a = 3 \times 30 = 90^\circ, \quad b = 4 \times 30 = 120^\circ, \quad c = 5 \times 30 = 150^\circ$$ 8. **Set up equations for $n_B$ and $n_C$ using the interior angle formula:** $$120 = \frac{(n_B - 2) \times 180}{n_B} \implies 120 n_B = 180 n_B - 360 \implies 180 n_B - 120 n_B = 360 \implies 60 n_B = 360 \implies n_B = 6$$ $$150 = \frac{(n_C - 2) \times 180}{n_C} \implies 150 n_C = 180 n_C - 360 \implies 180 n_C - 150 n_C = 360 \implies 30 n_C = 360 \implies n_C = 12$$ 9. **Conclusion:** Polygon C has $12$ sides and polygon B has $6$ sides, so polygon C has twice the number of sides as polygon B. **Final answer:** $$n_C = 2 n_B$$