HMMT February

Solution:
Recall that the functions $\sinh(x) = \frac{e^{x} - e^{-x}}{2}$ and $\cosh(x) = \frac{e^{x} + e^{-x}}{2}$ satisfy the relation
$$\sinh(x + y) = \sinh(x)\cosh(y) + \cosh(x)\sinh(y) = \sinh(x)\sqrt{1 + \sinh(y)^{2}} + \sinh(y)\sqrt{1 + \sinh(x)^{2}}.$$
Since $\sinh$ is surjective, we can perform the substitution $a = \sinh(x)$, $b = \sinh(y)$, and $c = \sinh(z)$, which turns the equations into
$$\sinh(x + y) = \frac{2 - \frac{1}{2}}{2},$$
$$\sinh(y + z) = \frac{\frac{3}{2} - \frac{2}{3}}{2},$$
$$\sinh(z + x) = \frac{\frac{5}{2} - \frac{2}{5}}{2}.$$
Thus, $x + y = \log(2)$, $y + z = \log(3 / 2)$, and $z + x = \log(5 / 2)$. Solving these equations gives $x = \log(\sqrt{10 / 3})$, so
$$a = \frac{1}{2}\left(\sqrt{\frac{10}{3}} - \sqrt{\frac{3}{10}}\right) = \left[\frac{7}{2\sqrt{30}}\right]$$

Solution 2:
We can find positive real numbers $x$, $y$, and $z$ such that $a = \frac{x^{2} - 1}{2x}$, $b = \frac{y^{2} - 1}{2y}$, and $c = \frac{z^{2} - 1}{2z}$. Then, the first equation becomes
$$\frac{x^{2} - 1}{2x} \cdot \frac{y^{2} + 1}{2y} + \frac{y^{2} - 1}{2y} \cdot \frac{x^{2} + 1}{2x} = \frac{3}{4},$$
which simplifies to
$$xy - \frac{1}{xy} = \frac{3}{2},$$
from which it follows that $xy = 2$. Similarly, $yz - \frac{1}{yz} = \frac{5}{6}$ and $zx - \frac{1}{zx} = \frac{21}{10}$, so $yz = \frac{3}{2}$ and $zx = \frac{5}{2}$. Thus $x = \sqrt{(2 \cdot \frac{5}{2}) / (\frac{3}{2})} = \sqrt{\frac{10}{3}}$, and
$$a = \frac{(10 / 3) - 1}{2\sqrt{10 / 3}} = \left[\frac{7}{2\sqrt{30}}\right]$$