Please write full! I didn't understand. And please write Day2 problems [Mod edit: Edited problem statement]

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Lemma. The orthocenter of a beautiful triangle is the same as the circumcenter $O$ of triangle $ABC$. Proof. We show that this point is also the Miquel point of the configuration. The Miquel point is the same as the circumcenter and the orthocenter because of angle chasing. Now assume there are two beautiful triangles belonging to the same point $K$. If $O\neq K$, the two beautiful triangles have parallel sides $MN$ and $M'N'$. Moving from $N$ to $N'$, the height from these two points is the perpendicular from $O$ to $MN$, then $M'N'$, yet it is also the line $NO$, then $N'O$. The direction of these two lines change in the opposite direction, hence the contradiction. So $O=K$ and by Thales we are done.

I'll show a sketch of the proof: 1. Let the triangles be $MNK$ and $MXY$. A spirall symetry of center $M$ sends $N$ to $X$ and $K$ to $Y$. 2. Quadrilaterals $MXAY$ and $MKAN$ are cyclic. 3. $\angle MKN=\angle MAN=\gamma$ and $\angle MNK=\angle MAK=\beta$ so $\alpha=90^\circ$

@randomusername - are MN and M'N' parallel? I think that they intersect, once you draw the figure. Maybe you meant something else? @vlad1m1r - please explain how did you prove Step 2. I believe that I got a simple angle chasing solution.

Konigsberg wrote: @vlad1m1r - please explain how did you prove Step 2. I believe that I got a simple angle chasing solution. It just follows from the fact that $MNK$ and $MXY$ are similar. $\Rightarrow MNX$ and $MKY$ are also similar $\Rightarrow \angle MYA=\angle MXB$.There is nothing complicated in this solution .It is a nice proof!!!!!!!

Any elementary solution please???

Let $\Delta MNK$ and $\Delta MN'K'$ be the two triangles.Let $X$ the point on $AB$ such that $MX=MB$ .We easily get that $MNXK$ and $MN'XK'$ are cyclic. $\Rightarrow \angle NXA=\angle N'XA=\angle BAC$ which forces us to conclude that $X/equiv A$

My solution : Let another triangle be $DEK$. Since the triangles $KDE$ and $KMN$ are similar. So there is a Spiral Similarity centered at $K$ and sending $D\to M$ and $E\to N$. So this Similarity also sends $D\to E$ $M\to N$ which means that triangles $KDM$ and $KEN$ are similar. So $\measuredangle KEN=\measuredangle KDM$ $(1)$ Let's say that $\measuredangle C=\beta$ $\measuredangle CDE=\gamma $ $\measuredangle EDK=\theta$ So from $(1)$ we get $\beta=90^{\circ}$

very easy problem for izho p1

Let $\triangle MNK$ and $\triangle MN'K'$ are two beautiful triangles with same angle orientation. Notice that they are spirally similar. Let $J=KN\cap K'N'$. Therefore $M$ is the miequel point of quadrilateral $AK'JN$. So $AK'MN$ and $AKMN$ is cyclic. So \[ \angle BAC=\angle ABC+\angle ACB. \]So $\angle BAC=90^\circ$.