In triangle $A B C \angle C A B=75^{\circ}, \angle A B C=45^{\circ}$. On side $C A$ a point $K$ is taken, and on side $C B$ a point $M$, $C K: A K=3: 1$.

Find $K M: A B$, if this ratio is less than $3 / 4$, and line $M K$ cuts off from triangle $A B C$ a triangle similar to it.
Note that the line $KM$ cannot be parallel to $AB$, because in this case $KB: AB = CK: AC = 3: 4$, which contradicts the condition. Therefore,

$\angle CKM = \angle B = 45^{\circ}, \angle CMK = \angle CAB = 75^{\circ}$.

Let $AK = 1$. Then $CK = 3, AC = 4$. By the Law of Sines, $AC / \sin 45^{\circ} = AB / \sin 60^{\circ}$, from which $AB = 2\sqrt{6}$.

Similarly, from triangle $MKC$, we find that $KM = \frac{3 \sin 60^{\circ}}{\sin 75^{\circ}} = \frac{3 \sqrt{6}}{\sqrt{3} + 1} (\sin 75^{\circ}$ is calculated in problem

$\underline{61207)}$.

Therefore, $KM / AB = \frac{3(\sqrt{3} - 1)}{4}$.

## Answer

$\frac{3(\sqrt{3} - 1)}{4}$