AIME II

Because $AL = CL = KL$, the circumcenter of $\triangle ACK$ is $L$. Similarly, the circumcenter of $\triangle ABL$ is $K$.
Because $\triangle AKL$ is equilateral, $\angle ALK = \angle AKL = 60^\circ$, and it follows that $\angle ACK = 30^\circ = \angle ABL$.
Because
$$\angle BAK = \angle BAC - \angle KAL - \angle LAC = 30^\circ - \angle LAC = 30^\circ - \angle LCA = \angle ACK - \angle LCA = \angle KCL,$$
it follows that the isosceles triangles $\triangle ABK$ and $\triangle CKL$ are congruent. Thus $AB = CK$. Because $\angle KBL = \angle ABL - \angle ABK = \angle ACK - \angle LCK$, it is also true that $\triangle BLK \cong \triangle ACL$ and $AC = BL$.
It follows by SAS that $\triangle ABL \cong \triangle KCA$.

Therefore
$$\begin{align*} \text{Area}(BKLC) &= \text{Area}(\triangle ABC) - \text{Area}(\triangle ALK) - \text{Area}(\triangle ACL) - \text{Area}(\triangle ABK) \\ &= \text{Area}(\triangle ABC) - \text{Area}(\triangle ALK) - \text{Area}(\triangle BLK) - \text{Area}(\triangle ABK) \\ &= \text{Area}(\triangle ABC) - \text{Area}(\triangle ABL). \end{align*}$$

Set $c = AB = CK$ and $b = AC = BL$. Then
$$\text{Area}(\triangle ABL) = \frac{bc \sin(\angle ABL)}{2} = \frac{bc}{4} \quad \text{and} \quad \text{Area}(\triangle ABC) = \frac{bc}{2},$$
from which $\text{Area}(BKLC) = \frac{bc}{4}$. Applying the Law of Cosines to $\triangle ABL$ gives
$$14^2 = AL^2 = b^2 + c^2 - bc\sqrt{3} = 38^2 - bc\sqrt{3}.$$
Thus $bc\sqrt{3} = 38^2 - 14^2 = 52 \cdot 24$, implying that $\text{Area}(BKLC) = 52 \cdot 2\sqrt{3} = 104\sqrt{3}$. The requested coefficient of $\sqrt{3}$ is 104.

As in the first solution, $AC = BL$. Observe that a 60° rotation around the center of $\triangle AKL$ moves $\triangle AKB$ onto $\triangle KLC$ and $\triangle KLB$ onto $\triangle LAC$. Hence there exists a point $P$ such that $\triangle AKL$ and $\triangle PBC$ are concentric equilateral triangles, as shown below.

Once $\triangle AKL$ is removed from $\triangle PBC$, the area of the quadrilateral $BKLC$ is $\frac{1}{3}$ of the area of the remaining shape. Thus
$$\text{Area}(BKLC) = \frac{1}{3}(\text{Area}(\triangle PBC) - \text{Area}(\triangle AKL)) = \frac{1}{12}(38^2 - 14^2)\sqrt{3} = 104\sqrt{3},$$
as in the first solution.