What Acute Angle You Have There!

Let M be the midpoint of BC. It is well-known (and easy to prove with Euler line or homothety) that AH = 2OM. Thus BO = CO = AO = AH = 2OM. Because OM is perpendicular to BC <MOB = <MOC = arccos(1/2) = 60 degrees, so <BOC = 60 degrees. Then it follows that <BAC = 60 degrees.

Let $D$ be the point on the circumcircle so that $BD$ is a diameter. It's easy to see $AHCD$ is a parallelogram. So $DC=AH = AO = R$ . Hence $DC = \frac{1}{2} BD$ . Therefore $\angle{BAC} = \angle{BDC} = 60^{\circ}$ . So $\angle{ABC} = 85^{\circ}$ .

Let $\angle OAC$ and $\angle OAB$ be $\alpha$ and $\beta$ respectively. Then $\angle BAC = \alpha + \beta$ , $\angle ABC = 90^\circ - \alpha$ . [Do you see why? It's just angle chasing. - Calvin]

Let $AC = 1$ . Then $AO = \frac {1}{2 \cdot \cos \alpha}$ , $AH= \frac {\cos (\alpha + \beta)}{\sin (90^\circ - \alpha)} = \frac {\cos (\alpha + \beta)}{\cos \alpha}$ . Hence, $\cos (\alpha + \beta) = \frac {1}{2}$ , so $\angle BAC = \alpha + \beta = 60^\circ$ [Since we are given that $ABC$ is an acute triangle - Calvin].

If we choose the origin to be the orthocentre $O$ , then the position vectors of $A,B,C$ are $\mathbf{a},\mathbf{b},\mathbf{c}$ , where $|\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = R$ , where $R$ is the outradius.

It is a standard calculation that the position vector of the orthocentre is $\overrightarrow{OH}= \mathbf{a}+\mathbf{b}+\mathbf{c}$ . Thus the condition of the question states that $|\mathbf{b}+\mathbf{c}| = R$ , and so $R^2 = 2R^2 + \mathbf{b}\cdot\mathbf{c}$ . Thus $\mathbf{b}\cdot\mathbf{c} = -\tfrac12R^2$ . The angle between $\mathbf{b}$ and $\mathbf{c}$ is $2A$ , so we deduce that $\cos2A = -\tfrac12$ . Thus $2A = 120^\circ$ , and hence $A = 60^\circ$ . Since $C = 35^\circ$ , it follows that $B = 85^\circ$ .

In this special case, let H be the same point as O, just to keep things simple. The altitudes of all the sides must be the same because the orthocenter is in the same spot of the circumcenter. And because of that, the 3 angles of the triangle must be all the same. Since a triangle has 180 degrees, you divide 180 by 3 and get 60 degrees.

Thats triangle make a same long.we can prove that with sin rule $a/sina=b/sib=c/sinc=2R$ we get $sin a=sqrt(2)/3$ so <BAC is 60 degree

We draw $OX \perp BC$ . We use the fact that $AH = 2OX$ ,i.e the distance of the orthocentre from vertex $A$ is twice the distance of the the circumcenter from the side opposite to vertex $A$ [ It has a proof, which i'm not going to give here]. Thus, in right angled triangle $OXC$ , we have $OX = \frac{1}{2} \times AH = \frac{1}{2} \times AO$ , $OC = AO$ (since they are the circumradii). Morover, O is the centre of the circumcircle of $\triangle ABC$ . Hence, we have $\angle BOC = 2 \angle A$ And as $\triangle OBX \cong \triangle OXC$ (can be easily verifired) , we have $\angle XOC = \angle A$ .

Now, $\cos \angle XOC = \dfrac{OX}{OC} = \dfrac{1}{2}$ . Hence $\angle XOC = \angle A = 60^{\circ}$ or $120^{\circ}$ . But $\angle A$ is acute and we conclude $\angle A = 60^{\circ}$ . Thus, $\angle B = 180^{\circ} - 60^{\circ} - 35^{\circ}$ = $\boxed{85^{\circ}}$