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 $135,30,15$.

  • $\begingroup$ 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

So we have equilateral triangle $\triangle ABC$ with point $O$ such that $\angle AOC=90^{\circ}, \angle BOC=75^{\circ}$.

enter image description here

Construct equilateral triangle $\triangle BOP$:

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

The conclusion is:

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

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


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:

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

| cite | improve this answer | |
  • $\begingroup$ (+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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.