A circle is inscribed in the trapezoid [i]ABCD[/i]. Let [i]K, L, M, N[/i] be the points of tangency of this circle with the diagonals [i]AC[/i] and [i]BD[/i], respectively ([i]K[/i] is between [i]A[/i] and [i]L[/i], and [i]M[/i] is between [i]B[/i] and [i]N[/i]). Given that $AK\cdot LC=16$ and $BM\cdot ND=\frac94$, find the radius of the circle.

[color=red][Moderator edit: A solution of this problem can be found on http://www.ajorza.org/math/mathfiles/scans/belarus.pdf , page 20 (the statement of the problem is on page 6). The author of the problem is I. Voronovich.][/color]
1. **Define the problem and given values:**
   - A circle is inscribed in the trapezoid \(ABCD\).
   - Points \(K, L, M, N\) are the points of tangency of this circle with the diagonals \(AC\) and \(BD\), respectively.
   - Given: \(AK \cdot LC = 16\) and \(BM \cdot ND = \frac{9}{4}\).

2. **Introduce the tangency points and the radius:**
   - Let \(P, Q, R, S\) be the tangency points of the circle with the sides \(AB, BC, CD, DA\) respectively.
   - Let \(r\) be the radius of the circle.

3. **Use the properties of the tangents from a point to a circle:**
   - Denote \(a = AP = AS\), \(b = BP = BQ\), \(c = CR = CQ\), \(d = DR = DS\).

4. **Express the products involving the tangency points:**
   - The products \(k = AK \cdot LC\) and \(m = BM \cdot ND\) are given by:
     \[
     k = ac + 2r^2 - 2r \sqrt{ac + r^2}
     \]
     \[
     m = bd + 2r^2 - 2r \sqrt{bd + r^2}
     \]

5. **Given values substitution:**
   - Given \(k = 16\) and \(m = \frac{9}{4}\).

6. **Set up the equations for \(k\) and \(m\):**
   - For \(k\):
     \[
     16 = ac + 2r^2 - 2r \sqrt{ac + r^2}
     \]
   - For \(m\):
     \[
     \frac{9}{4} = bd + 2r^2 - 2r \sqrt{bd + r^2}
     \]

7. **Simplify the equations:**
   - Let \(e = \sqrt{ac + r^2}\) and \(f = \sqrt{bd + r^2}\).
   - Then:
     \[
     16 = ac + 2r^2 - 2r e
     \]
     \[
     \frac{9}{4} = bd + 2r^2 - 2r f
     \]

8. **Express \(e\) and \(f\) in terms of \(r\):**
   - From the equations:
     \[
     e = \sqrt{ac + r^2} = \frac{ac + 2r^2 - 16}{2r}
     \]
     \[
     f = \sqrt{bd + r^2} = \frac{bd + 2r^2 - \frac{9}{4}}{2r}
     \]

9. **Set up the final equation:**
   - Using the fact that the diagonals of the rhombus intersect at the incenter \(I\) and the properties of the tangents:
     \[
     (k + 2r\sqrt{k} + r^2)(m + 2r\sqrt{m} + r^2) = r^2(k + 2r\sqrt{k} + r^2 + m + 2r\sqrt{m} + r^2)
     \]

10. **Substitute the given values \(k = 16\) and \(m = \frac{9}{4}\):**
    \[
    (16 + 2r\sqrt{16} + r^2)\left(\frac{9}{4} + 2r\sqrt{\frac{9}{4}} + r^2\right) = r^2\left(16 + 2r\sqrt{16} + r^2 + \frac{9}{4} + 2r\sqrt{\frac{9}{4}} + r^2\right)
    \]

11. **Simplify and solve the quartic equation:**
    \[
    r^4 - 24r^2 - 66r - 36 = 0
    \]

12. **Find the roots of the quartic equation:**
    - Using the Rational Root Theorem and numerical methods, we find the roots:
      \[
      r = 6 \quad \text{or} \quad r \approx -0.74
      \]

13. **Select the positive root:**
    - Since the radius cannot be negative, we select \(r = 6\).

The final answer is \(\boxed{6}\).