3. Given a circle centered at the origin with radius $R$ intersects the sides of $\triangle A B C$, where $A(4,0)$, $B(6,8)$, and $C(2,4)$. Then the range of values for $R$ is $\qquad$
3. $\left[\frac{8 \sqrt{5}}{5}, 10\right]$.

Draw a perpendicular from the origin $O$ to $AC$, with the foot of the perpendicular being $D$, so point $D$ lies on side $AC$. It is easy to see that
$$
l_{A C}: y=-2 x+8 \Rightarrow 2 x+y-8=0 \text {. }
$$

Then $O D=\frac{|2 \times 0+0-8|}{\sqrt{2^{2}+1}}=\frac{8 \sqrt{5}}{5}$.
Also, $O A=4, O B=10, O C=2 \sqrt{5}$, so $O D<O A<O C<O B$.
When and only when $O D \leqslant R \leqslant O B$, the circle intersects with $\triangle A B C$.
Therefore, $R \in\left[\frac{8 \sqrt{5}}{5}, 10\right]$.