$17 \cdot 3$ If $\alpha, \beta, \gamma$ are the three interior angles of a triangle, and $x=\alpha+\beta, y=\beta+\gamma, z=\gamma+\alpha$. Then the incorrect conclusion about the number of acute angles among $x, y, z$ is
(A) There may be no acute angles.
(B) There may be one acute angle.
(C) There may be two acute angles.
(D) There is at most one acute angle.
(E) There cannot be two acute angles.
(Chongqing, Sichuan Province, China Junior High School Mathematics Competition, 1984)
[Solution] Given that $\alpha+\beta+\gamma=180^{\circ}$, and
$$
x+y+z=2(\alpha+\beta+\gamma)=360^{\circ} \text {. }
$$

Now assume that two of $x, y, z$ are acute angles, without loss of generality, let them be $x, y$.
That is, $0^{\circ}<x, y<90^{\circ}$. Then $z=360^{\circ}-(x+y)>270^{\circ}$, which implies $z-90^{\circ}>180^{\circ}$, i.e., $\gamma+\alpha>180^{\circ}$. This is impossible (since $\alpha+\beta+\gamma=180^{\circ}$).
Therefore, the answer is $(C)$.