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

Solution: Draw a perpendicular from the origin $O$ to $AC$, with the foot of the perpendicular being $D$. Then point $D$ lies on side $AC$. It is easy to see that
$l_{AC}: y = -2x + 8 \Rightarrow 2x + y - 8 = 0$, so $OD = \frac{|2 \times 0 + 0 - 8|}{\sqrt{2^2 + 1}} = \frac{8 \sqrt{5}}{5}$,

Also, $OA = 4$, $OB = 10$, $OC = 2 \sqrt{5}$, hence $OD < OA < OC < OB$,

When and only when $OD \leqslant R \leqslant OB$, the circle intersects with $\triangle ABC$, so, $R \in \left[\frac{8 \sqrt{5}}{5}, 10\right]$.