$24 \cdot 1$ The volume of a cone is $A$, the volume of a cylinder is $M$, and the volume of a sphere is $C$. They have the same radius, and the height of the cone and the cylinder is equal to the diameter of the sphere. Then
(A) $A-M+C=0$.
(B) $A+M=C$.
(C) $2 A=M+C$.
(D) $A^{2}-M^{2}+C^{2}=0$.
(E) $2 A+2 M=3 C$.
(49th American High School Mathematics Examination, 1998)
[Solution] According to the problem, if the radius of the base of the cone and cylinder is $\gamma$, then their heights are $2 \gamma$.
$$
\begin{array}{l}
V_{\text {cone }}=\frac{1}{3} \pi \cdot r^{2} \cdot 2 r=\frac{2}{3} \pi r^{3}=A . \\
V_{\text {cylinder }}=\pi r^{2} \cdot 2 r=2 \pi r^{3}=M . \\
V_{\text {sphere }}=\frac{4}{3} \pi r^{3}=C .
\end{array}
$$

From the above, we know that $A-M+C=0$.
Therefore, the answer is $(A)$.