6. In an acute triangle $\triangle ABC$ with sides $a, b, c$, the distances from the circumcenter to the sides are $m, n, p$. Then, $m: n: p=(\quad)$.
(A) $\frac{1}{a}: \frac{1}{b}: \frac{1}{c}$
(B) $a: b: c$
(C) $\cos A: \cos B: \cos C$
(I) $\sin A: \sin B: \sin C$
6. (C).

As shown in Figure 1, let the circumradius of $\triangle ABC$ be $R$. Then
$$
\begin{array}{l}
m=\sqrt{R^{2}-\left(\frac{a}{2}\right)^{2}} \\
=\frac{1}{2} \sqrt{4 R^{2}-a^{2}} \\
=\frac{1}{2} \sqrt{\frac{a^{2}}{\sin ^{2} A}-a^{2}} \\
=\frac{a}{2} \cdot \frac{\cos A}{\sin A}=R \cos A .
\end{array}
$$

Similarly, $n=R \cos B, p=$
$R \cos C$.
$m: n: p=\cos A: \cos B: \cos C$. Therefore, the answer is (C).