HMMT Spring, 2021

Solution:
Rotate $\triangle ABC$ around $A$ to $\triangle AB' C'$, such that $B'$ is on segment $AF$. Note that as $BD \parallel EF$, $AB=AD$. From this, $AB'=AB=AD$, and $B'=D$. Note that
$$\angle ADC' = \angle ABC = 180 - \angle ADC$$
because $ABCD$ is cyclic. Therefore, $C$, $D$, and $C'$ are collinear. Also, $AC' = AC$, and
$$\angle CAC' = \angle DAC + \angle C'AD = \angle DAC + \angle CAB = \angle EAF.$$
Thus, since $AE=AF$, $\triangle ACC' \sim \triangle AEF$. Now, we have
$$[ACC'] = [ACD] + [ADC'] = [ACD] + [ABC] = [ABCD]$$
But, $[ACC'] = \frac{CC'^2}{EF^2} \cdot [AEF]$, and we know that $CC' = CD + DC' = 4 + 3 = 7$. Thus,
$$\frac{[ABCD]}{[AEF]} = \frac{[ACC']}{[AEF]} = \frac{7^2}{20^2} = \frac{49}{400}.$$
The answer is $100 \cdot 49 + 400 = 5300$.

Solution:

Since $BD$ is parallel to $EF$ and $AE=AF$, we have $AB=AD$. Since $ABCD$ is cyclic, $\angle ABC + \angle ADC = 180^\circ$. Thus we can glue $\triangle ABC$ and $\triangle ADC$ as shown in the diagram above to create a triangle that is similar to $\triangle AEF$ and has the same area as $ABCD$. The base of this triangle has length $BC + CD = 3 + 4 = 7$, so the desired ratio is
$$\frac{7^2}{20^2} = \frac{49}{400}$$