China Mathematical Competition

Let $t = x + \frac{a}{2}$. Then the equation $\sqrt{|t - \frac{a}{2}|} + \sqrt{|t + \frac{a}{2}|} = b$ for $t$ has exactly three different real solutions $t_i = x_i + \frac{a}{2}$ ($i = 1, 2, 3$).

Since $f(t) = \sqrt{|t - \frac{a}{2}|} + \sqrt{|t + \frac{a}{2}|}$ is an even function, the three real solutions of equation $f(t) = b$ are symmetrically distributed about the origin of the number axis, so that there must be $b = f(0) = \sqrt{2a}$. In the following, we will find the real solutions of equation $f(t) = \sqrt{2a}$.

When $|t| \le \frac{a}{2}$, $f(t) = \sqrt{\frac{a}{2} - t} + \sqrt{\frac{a}{2} + t} = \sqrt{a + \sqrt{a^2 - 4t^2}} \le \sqrt{2a}$ and the equal sign holds if and only if $t = 0$; when $|t| > \frac{a}{2}$, $f(t)$ is monotonically increasing, and when $t = \frac{5a}{8}$, $f(t) = \sqrt{2a}$; when $t < -\frac{a}{2}$, $f(t)$ is monotonically decreasing, and when $t = -\frac{5a}{8}$, $f(t) = \sqrt{2a}$.

Thus, equation $f(t) = \sqrt{2a}$ has exactly three real solutions $t_1 = -\frac{5}{8}a$, $t_2 = 0$, $t_3 = \frac{5}{8}a$.

By the given conditions, we can find $b = x_3 = t_3 - \frac{a}{2} = \frac{a}{8}$. Combining $b = \sqrt{2a}$, we get $a = 128$.

Consequently, $a + b = \frac{9a}{8} = 144$. □