Example 1 Given an acute triangle $\triangle A B C$ with three interior angles satisfying $A>B>C$. Let $\alpha$ denote the minimum of $A-B, B-C$, and $90^{\circ}-A$. Then the maximum value of $\alpha$ is $\qquad$
Solution 1: Since $\alpha$ is the minimum of $90^{\circ}-A$, $A-B$, and $B-C$, we have
$$
\begin{array}{l}
90^{\circ}-A \geqslant \alpha, \\
90^{\circ}-B=\left(90^{\circ}-A\right)+(A-B) \geqslant 2 \alpha, \\
90^{\circ}-C=\left(90^{\circ}-A\right)+(A-B)+(B-C) \geqslant 3 \alpha .
\end{array}
$$
(1) + (2) + (3) gives
$$
90^{\circ} \times 3-(A+B+C) \geqslant 6 \alpha.
$$

Thus, $\alpha \leqslant \frac{270^{\circ}-180^{\circ}}{6}=15^{\circ}$.
This shows that $\alpha$ cannot exceed $15^{\circ}$ at most.
Next, we show that $\alpha$ can indeed reach $15^{\circ}$, i.e., equations (1), (2), and (3) can all hold as equalities. By
$$
\left\{\begin{array}{l}
90^{\circ}-A=15^{\circ}, \\
90^{\circ}-B=2 \times 15^{\circ}, \\
90^{\circ}-C=3 \times 15^{\circ},
\end{array}\right.
$$

we get $A=75^{\circ}$, $B=60^{\circ}$, $C=45^{\circ}$.
Therefore, when the three interior angles of $\triangle A B C$ are $75^{\circ}$, $60^{\circ}$, and $45^{\circ}$, $\alpha$ reaches its maximum value of $15^{\circ}$.
Note: The sum (1) + (2) + (3) is essentially
$$
3\left(90^{\circ}-A\right)+2(A-B)+(B-C) \geqslant 6 \alpha,
$$

Thus, we can also have the following solution.
Solution 2: From the given conditions, we have
$$
\begin{aligned}
\alpha & =\frac{3 \alpha+2 \alpha+\alpha}{6} \\
& \leqslant \frac{3\left(90^{\circ}-A\right)+2(A-B)+(B-C)}{6}, \\
& \leqslant \frac{270^{\circ}-(A+B+C)}{6}=15^{\circ}.
\end{aligned}
$$

On the other hand, when equation (4) holds as an equality, i.e.,
$$
90^{\circ}-A=A-B=B-C=15^{\circ},
$$

we have $A=75^{\circ}$, $B=60^{\circ}$, $C=45^{\circ}$, i.e., when the three interior angles of $\triangle A B C$ are $75^{\circ}$, $60^{\circ}$, and $45^{\circ}$, $\alpha$ can reach its maximum value of $15^{\circ}$.