IMO Hk TST

The ratio can be $\frac{5}{3}$ or $\frac{5}{4}$.
Let $a = BC$, $b = CA$, $c = AB$, $d = a+b$, $s = \frac{a+b+c}{2}$, $x = MA$, $y = MB$ and $z = MC$. Also, let $r$ be the inradius of $\triangle ABC$, and let $r'$ be the inradius of $\triangle AMC$.
Firstly, by considering $\triangle AMC$, we obtain $r' = \frac{2[AMC]}{b+x+z}$. Similarly, by considering $\triangle BMC$, we obtain $r' = \frac{2[BMC]}{a+y+z}$. Equating these, and using $\frac{[AMC]}{[BMC]} = \frac{x}{y}$, we obtain
$$\frac{b+x+z}{a+y+z} = \frac{x}{y}.$$
This implies $\frac{b+z}{a+z} = \frac{x}{y}$. Since $x+y=c$, we obtain
$$x = \frac{(b+z)c}{a+b+2z} \quad \text{and} \quad y = \frac{(a+z)c}{a+b+2z}. \qquad (1)$$

Secondly, we have
$$O_1P = AP \tan \frac{A}{2} = \frac{b+x-z}{2} \cdot \frac{r}{s-a} = \frac{(b+x-z)r}{b+c-a}$$
and similarly $O_2Q = \frac{(a+y-z)r}{a+c-b}$. Since these are equal, we have
$$\frac{b+x-z}{b+c-a} = \frac{a+y-z}{a+c-b}.$$
Using (1), this becomes
$$\frac{(a+b+2z)(b-z) + (b+z)c}{b+c-a} = \frac{(a+b+2z)(a-z) + (a+z)c}{a+c-b}.$$

Clearing denominators and simplifying, we get
$$4(a-b)z^2 = (a-b)(a+b+c)(a+b-c).$$
This implies $a = b$ or $z = \sqrt{s(s-c)}$. If $a = b$, then we have $x = y = \frac{c}{2}$ by (1), and hence
$$z = \sqrt{a^2 - x^2} = \sqrt{\left(a + \frac{c}{2}\right)\left(a - \frac{c}{2}\right)} = \sqrt{s(s-c)}.$$
Therefore,
$$z = \sqrt{s(s-c)} \qquad (2)$$
holds in any case.
Thirdly, we use the given condition. Note that $PQO_2O_1$ is a rectangle, and so
$$[PQO_2O_1] = PQ \cdot r' = \left(\frac{x+z-b}{2} + \frac{y+z-a}{2}\right) r' = \frac{(c+2z-d)r'}{2}.$$
Next, we have
$$[ABC] = [AMC] + [BMC] = \frac{b+x+z}{2} \cdot r' + \frac{a+y+z}{2} \cdot r' = \frac{(c+2z+d)r'}{2}.$$
Since $6[PQO_2O_1] = [ABC]$, we get $6(c+2z-d) = c+2z+d$. This implies
$$z = \frac{7d - 5c}{10}. \qquad (3)$$
Comparing (2) and (3), we solve
$$\left(\frac{7d - 5c}{10}\right)^2 = s(s-c) = \frac{(d+c)(d-c)}{4}.$$
This is the same as $(3d - 5c)(4d - 5c) = 0$. Therefore, we have
$$\frac{d}{c} = \frac{5}{3}, \frac{5}{4}.$$
Both values are possible. For example, when $(a, b, c) = (5, 5, 6)$, $(5, 5, 8)$, both $\triangle AMC$ and $\triangle BMC$ are right-angled triangles with side lengths 3, 4, 5. It is easy to find $r' = 1$ and verify $6[PQO_2O_1] = 12 = [ABC]$.