Answer is $4:3$ which can be found by similarity. Stats- 16 students solved it during the contest.

This is quite an interesting result My solution was by barycentric coordinates, noting that $D$ is the symmedian point. In the end, we get that \[ F = \left( -b^2 : -a^2 : 4c^2 \right), \]where $(u:v:w)$ denotes unhomogeneized coordinates. I would appreciate seeing a synthetic solution.

@above, using the fact that $D$ is the symmedian point, the angle bisector theorem along with ceva and Menelaus gives the result.

Answer: $\boxed{\frac{4}{3}}$. Let $CL$ be the altitude. Note that $(CL)$ is the incircle of triangle $FAB$. Let $a=BF, b=AF, c=AB$, $r, s$ be the inradius and semi-perimeter of $\triangle AFB$ and $h$ the length of $F$-altitude. Then, $$\triangle ALC \sim \triangle CLB \implies LC^2=LB \cdot LA.$$Hence, $$\ 4r^2=(s-a)(s-b) \implies \frac{4(s-c)}{s}=1,$$hence $a+b=3c$. Also, $$\frac{S_{ABF}}{S_{ABC}}=\frac{h}{2r}=\frac{\frac{2\Delta}{c}}{\frac{2\Delta}{s}}=\frac{s}{c}=\frac{4}{3}.$$

Let $X \in AF$ and $Y \in BF$ such that $C \in XY$ and $XY \parallel AB$, and let $E \equiv CD \cap AB$. Note that $D$ is the incenter of $XYAB$. Also, $\triangle CXD \sim \triangle EDA$ and $\triangle CEA \sim \triangle BEC$. Thus, \[\frac{CX}{\frac12 CE} = \frac{CX}{DE} = \frac{CD}{AE} = \frac{\frac12 CE}{AE} \iff CX = \frac{CE^2}{4AE} = \frac{BE}{4}\]Also, $CY = \frac{AE}{4}$. Thus, \[d(F, XY) : d(F, AB) = 1 : 4 \iff d(C, AB) : d(F, AB) = 3 : 4 \iff S_{ABF}:S_{ABC} = 4: 3.\] absurdist wrote: In the end, we get that \[ F = \left( -b^2 : -a^2 : 4c^2 \right), \]where $(u:v:w)$ denotes unhomogeneized coordinates. How did you get this?

Very triggy problem. Let $E$ be the foot of the $C$-altitude and $a, b, c$ have their usual meaning. Note that $\triangle CEA \sim \triangle BCA \sim \triangle BEC$. So, we have $AE = \frac{b^2}{c}$, $BE = \frac{a^2}{c}$ and $CE = \frac{ab}{c}$. Now, \[\cot \angle DAE = \frac{AE}{DE} = \frac{2 AE}{CE} = \frac{2b}{a}\]\[\cot \angle FAB = \cot 2 \angle DAE = \frac{\cot^2 \angle DAE - 1}{2\cot \angle DAE} = \frac{4b^2 - a^2}{4ab} \]Similarly, $\cot \angle FBA = \frac{4a^2 - b^2}{4ab}$. By Sine Rule and the formula of area in terms of two sides and a vertex angle, we have \[S_{ABF} = \frac{c^2 \sin \angle FAB \sin \angle FBA}{\sin( \angle FAB + \angle FBA)} = \frac12\cdot\frac{c^2}{\cot \angle FAB + \cot \angle FBA} = \frac12\cdot\frac{c^2}{3\cdot\frac{a^2+b^2}{4 ab}} = \frac43 S_{ABC}.\]

Suppose $AB = 1$, so that $S_{ABC} = \frac{1}{2}\cos A \sin A$. Let $\alpha = \angle BAD$ and $\beta = \angle ABD$ so that $\tan \alpha = \frac{1}{2}\tan A$ and $\tan \beta = \frac{1}{2} \tan B$. Notice that $D$ is the center of $FAB$, and let the incircle of $FAB$ touch $FA$ at $P$. To compute the area $S_{FAB}$, we find the semiperimeter of $FAB$ and multiply that by the inradius $r = \frac{1}{2}\cos A \sin A$, which is half the $C$-altitude in $ABC$. If we let $FP = x$, we get $$\tan \angle DFP = \tan (90 - \alpha - \beta) = \frac{r}{x} = \frac{\cos A \sin A}{2x}.$$ We see that $$\tan (90 - \alpha - \beta) = \frac{1}{\tan(\alpha + \beta)} = \frac{1-\tan \alpha \tan \beta}{\tan \alpha + \tan \beta} = \frac{1 - \frac{1}{4}\tan A \tan B}{\frac{1}{2}(\tan A + \tan B)} = \frac{3}{2(\tan A + \tan B)}.$$ Plugging this back into our previous equation gives $x = \frac{1}{3}\cos A \sin A(\tan A + \tan B) = \frac{1}{3} (\sin^2 A + \sin^2 B) = \frac{1}{3}$. Thus, we see that $FAB$ has semiperimeter $x + AB = \frac{1}{3} + 1 = \frac{4}{3}$. The ratio $\frac{S_{FAB}}{S_{ABC}} = \frac{\frac{4}{3}r}{\frac{1}{2}\cos A \sin A} = \frac{4}{3}$.