$ABC$ üçgeni ya eşkenar ya da ikizkenar dik üçgen olmalıdır. Çözümü İngilizce olarak hazırlamıştım ama bu sorun olmayacaktır diye düşünüyorum. Aşağıdaki çizimimdeki harflendirmelerde küçük değişiklikler yaptım. Buna göre ilerleyelim:
Question: In triangle $ABC$, squares $BADE$, $CBFG$, and $ACHK$ are constructed on the sides of the triangle and extending outward. The points $ D, E, F, G, H, K $ are concyclic. Determine all triangles $ABC$ that satisfy these conditions.
Solution [Lokman Gökçe]: If the points $ D, E, F, G, H, K $ are concyclic, then the center of this circle coincides with the circumcenter of triangle $ABC $. Let's denote this center as $ O $. If the triangle $ABC$ is equilateral, then the points $ D, E, F, G, H, K $ are concyclic.
Now, let's consider the cases where at least two sides of triangle $ABC$ are different from each other. Let $|BC|=a$, $|CA|=b$, $|AB|=c$, $\angle A = \alpha$, $\angle B = \beta$, $\angle C= \gamma $, $|OA|=|OB|=|OC|=R$. For this, we can assume $ \beta \neq \gamma $. With angle chasing, we find $ \angle OBF = 180^\circ - \alpha $, $ \angle OAK = 180^\circ - \beta $ and $ \angle OAD = 180^\circ - \gamma $. Also, $|BF|=a$, $|AK|=b$, $|AD|=c$. Since point $O$ is the center, we have $|OF| = |OK| = |OD|$. We have the identity $\cos(180^\circ - \theta) = - \cos \theta$. By applying the cosine theorem in triangles $OBF $, $ OAK $, and $OAD $, we get $ R^2 + a^2 + 2Ra \cos \alpha = R^2 + b^2 + 2Rb \cos \beta = R^2 + c^2 + 2Rc \cos \gamma $. Therefore,
$$ a^2 + 2Ra \cos \alpha = b^2 + 2Rb \cos \beta = c^2 + 2Rc \cos \gamma .$$
By the sine theorem, $ a = 2R\sin\alpha$, $ b = 2R\sin\beta$, $ c = 2R\sin\gamma$. Thus,
\begin{equation*}
\begin{split}
\sin^2\beta + \sin\beta \cos\beta & = \sin^2\gamma + \sin\gamma \cos\gamma \\
\sin^2\beta -\sin^2\gamma & = \dfrac{1}{2}(\sin 2\gamma - \sin 2\beta) \\
(\sin\beta + \sin\gamma)(\sin\beta -\sin\gamma) & = \dfrac{1}{2}(\sin 2\gamma - \sin 2\beta) \\
2\sin\left(\frac{\beta + \gamma}{2}\right)\cos\left(\frac{\beta - \gamma}{2}\right)\cdot 2\cos\left(\frac{\beta + \gamma}{2}\right)\sin\left(\frac{\beta - \gamma}{2}\right) & = \cos(\beta + \gamma)\sin(\beta - \gamma) \\
\sin(\beta + \gamma)\sin(\beta - \gamma) & = - \cos(\beta + \gamma)\sin(\beta - \gamma) \\
\tan(\beta + \gamma) & = -1
\end{split}
\end{equation*}
and we get that $\beta + \gamma = 135^\circ $, $\alpha = 45^\circ$.
With similar reasoning, from $a^2 + 2Ra \cos \alpha = b^2 + 2Rb \cos \beta$, we find that $\alpha = \beta $ or $\alpha + \beta = 135^\circ$. Hence, we conclude that $\alpha = \beta = 45^\circ, \gamma = 90^\circ$ or $\alpha = \gamma = 45^\circ, \beta = 90^\circ$. By symmetry, another solution is $\beta = \gamma = 45^\circ, \alpha = 90^\circ$.
