
1. Determine the angles of the triangles which satisfy the following property: There exist $K$ and $M$ inside $A B$ and $A C$ respectively, such that circumcircles of the quadrilaterals $A K L M$ and $K B C M$ are the same, where $L$ is the intersection of $M B$ and $K C$.

(Jaroslav Švrček)


Solution. The quadrilateral is cyclic iff $\angle C M B=\angle C K B$ or $\angle A K L=\angle A M L$ (see Fig.1). The quadrilateral $A K L M$ is cyclic iff $\angle A K L+\angle A M L=180^{\circ}$. In the sought situation all four angles above have to be right, consequently $K$ and $M$ are the foots of the altitudes in $A B C$. Thus $A B C$ has to be acute, and $L$ has to be its orthocenter. The circumcircle of $K B C M$ is the Thales' circle over the diameter $B C$ and the circumcircle of $A K L M$ is the Thales' circle over the diameter $A L$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f0f309b65082ec06ea04g-14.jpg?height=371&width=458&top_left_y=872&top_left_x=526)

Fig. 1

These circumcircles are the same iff their diameters $B C$ and $A L$ are the same. Let the angles in $A B C$ be $\alpha, \beta, \gamma$ in a usual way. The right triangles $C K B$ and $A K L$ are similar, namely the angles at $C$ and at $A$ are the same: $\angle B A L=\angle B C K=90^{\circ}-\beta$. That is why $|B C|=|A L|$ iff $|A K|=|C K|$, that is $A K C$ is right and isosceles.

All together, $A B C$ fulfills the condition of the problem if and only if is acute with $\alpha=45^{\circ}$. For acute angles $\beta$ and $\gamma$ we have then $\beta+\gamma=135^{\circ}$.

Conclusion. The solutions are the triples $(\alpha, \beta, \gamma)=\left(45^{\circ}, 45^{\circ}+\phi, 90^{\circ}-\phi\right)$, where $\phi \in\left(0^{\circ}, 45^{\circ}\right)$.
