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Given an equilateral triangle $ABC$ and a point $O$ on the plain of $\triangle ABC$, such that $\angle AOC=90^{\circ}, \angle BOC=75^{\circ}$. Find the angles of the triangle constructed from segments $AO,BO,CO$.

I have proof with trigonometry, but I am interested in a solution without it! Thanks in advance! (The answer is $135,30,15$.

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  • $\begingroup$ Do you have reason to believe that such a solution exists? $\endgroup$ – rogerl Jun 24 '18 at 22:44
  • $\begingroup$ @rogerl: of course it exists, $O$ can be found by intersecting two suitable circles. It is practical t consider the symmetric of $O$ with respect to the sides of $ABC$. Also: contests do not usually propose open problems. $\endgroup$ – Jack D'Aurizio Jun 25 '18 at 13:00
  • $\begingroup$ Yes, there is, I'll post solution in a moment :) $\endgroup$ – Oldboy Jun 25 '18 at 13:05
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So we have equilateral triangle $\triangle ABC$ with point $O$ such that $\angle AOC=90^{\circ}, \angle BOC=75^{\circ}$.

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Construct equilateral triangle $\triangle BOP$:

$$\angle CBA=\angle PBO\implies \angle CBP=\angle ABO,$$ $$BP=BO,\space BC=BA$$

The conclusion is:

$$\triangle BPC \cong \triangle BOA$$

$$\angle CPB=\angle AOB=90^{\circ}+75^{\circ}=165^{\circ}$$

$$OA=PC$$

So the triangle $\triangle OPC$ has sides $OP=OB$, $PC=OA$ and $OC$. If we find the angles of that triangle, we are done. And that is easy:

$$\angle COP = \angle COB-\angle POB=75^{\circ}-60^{\circ}=15^{\circ}$$

$$\angle CPO=360^{\circ}-\angle CPB-\angle BPO=360^{\circ}-165^{\circ}-60^{\circ}=135^{\circ}$$

The third angle is:

$$\angle OCP=180^{\circ}-\angle COP-\angle CPO=180^{\circ}-135^{\circ}-15^{\circ}=30^{\circ}$$

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  • $\begingroup$ (+1) Very nice solution. I reached the same conclusion by reflecting $O$ with respect to the sides of $ABC$, but that was pretty much unnecessary. $\endgroup$ – Jack D'Aurizio Jun 25 '18 at 14:56
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    $\begingroup$ @JackD'Aurizio Actually, that angle of $90^\circ$ almost killed my desire to solve this problem. I thought that the right angle had some special value so I started to think about circle over $AC$ passing through $O$, chased some central and inscribed angles, all sort of things...which turned out to be a (huge) waste of time. If the OP had used $91^\circ$ and $73^\circ$, I would have tried some simple angle chasing right from the start... "Fine tuned" input values can actually make problems harder, not simpler. :) $\endgroup$ – Oldboy Jun 25 '18 at 15:10

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