In $\triangle IME$, $O$ and $H$ are circumcenter and orthocenter. $\overline{IO} = \overline{IH}, ~\angle EIH = 2\cdot \angle HIO$, find $\angle HEO$.

Question

Let $\triangle IME$ be an acute triangle and points $O$ and $H$ be the circumcenter and orthocenter, respectively, such that $\overline{IO} = \overline{IH}$. If $\angle EIH = 2\cdot \angle HIO$, find the value of angle $\angle HEO$.

I know $\overline{EO} = \overline{IO} = \overline{IH}$, since $O$ is the circumcenter, but I do not know what to do with this information. I also thought about using trigonometry since I know $\sin{2\alpha} = 2\sin{\alpha}\cos{\alpha}$, and one angle is two times the other, but I don't know the value of any of the sides or angles, so I don't see how this can be useful. Then, I thought about extending line $IH$ since there would be a right angle and then use trig, but this isn't working. Also, I'm struggling to see the connection between the given information, since I can't see any useful relation between the orthocenter and the circumcenter, and I don't know why the triangle needs to be acute (maybe it's just so that there's only one answer?).

I am a high school student, and this problem is adapted from an entrance exam I am studying for. Nothing I tried so far worked, so I have no idea of what to do. Any help would be really appreciated.

Answer 1

Lemma: $H$ and $O$ are isogonal conjugates $\implies\angle EIH=\angle MIO=2\angle HIO=2\alpha.$

Lemma: $IH=2R\cos(5\alpha)\implies \cos(5\alpha)=0.5\implies \alpha=12^{\circ}.$

Angle chasing leads to $\angle HEO=8\alpha-90^{\circ}=6^{\circ}$ .

Answer 2

The radius of circle (I, O) is equal to that of circumcircle. We use this fact that the circle passing orthocenter and two vertices E and M has equal radius to that of circumcircle. that is we have $\angle HEO=\alpha$ and HIO opposite to equal arcs HO belonging to circles (I, O) and (EMO), which gives:

$\alpha = \frac {\widehat{HIO}}2$

J is where the bisectors of angles of triangle pass. IJ also bisects angle HIO, so we may write:

$\angle EIG=2\alpha +2(4\alpha)=10\alpha$

$\angle OIE=\angle OEI$

$\angle OEI=\angle HEI+\alpha\Rightarrow \angle HEI=(\angle OEI=\angle OIE=6\alpha)-\alpha=5\alpha$

Finally:

$\angle HEI+\angle EIG=5\alpha +10\alpha=15\alpha=90^o\Rightarrow \alpha = 6^o $