in fact $B,C,H,O,I$ are concyclic and that leads to an easy solution. by the way i was not a contestent at apmo .

Here is my solution: Easy angle chasing yields $\angle BIC=120^\circ$, while more angle chasing yields $\angle BHC=120^\circ$. Hence $BIHC$ is cyclic, and if we denote the foot of the perpendicular from $C$ to $AB$ as $D$, then $\angle IHD=\angle IBC$. Now, let $\angle ABC=2\beta$. It is easy to find $\angle DCB=90-2\beta$, hence $\angle AHD=2\beta$. Therefore, $3\angle ABC=6\beta=2\angle AHI$.

we first prove a lemma lemma = let $ABC$ be a triangle with $D$ as midpoint of minor arc $BC$ of circumcircle of $ABC$ also let $I$ be incentre of triangle $ABC$ than we prove that $D$ is centre of circumcircle of $BIC$. proof = it is evident that $A$ excentre of triangle $ABC$ lie on circumcircle of $BIC$ also $BD=DC$ and thus $\angle CBD=A/2 = \angle CAD=\angle BAI$ thus $A,I,D$ are collinear. now $\angle IBD = 90-C/2=\angle JCB=\angle BID$ where $J$ is $A$ excentre. and thus $BD=DI$ thus $D$ is circumcentre of $BIC$. main proof = by this lemma and $\angle A = 60$ we get that $\angle BOC=\angle BIC=\angle BHC = 120$ and hence $B,H,O,I,C$ are concyclic. by concyclicity of $HIJB$ we have $\angle HBJ = \angle AIH = 120-C-B/2=60+B/2$ in triangle $AHI$ we thus get $\angle AHI=3B/2$ since $\angle HAI=60-B$ and thus $2\angle AHI = 3\angle B$. we are done

A nice problem . We will show that points B,I, H and C are concyclic wich is too easy since angle<BIC = 90°+A/2 = 120°=180°-A=<BHC. Let (w) the cercle ( BIHC). Let Y be the intersection point of line AH with (w);then angle <YBC = <ABC . Prof By simple angle chasing <BAY = <HCB =<HYB = <AYB and (AY) perpendicular to (BC) so triangle BAY is isoceles then the equality follows. Now denote by X the center of (w) ,then : 2<AHI=360°-2<IHY = <IXY = 2<IBY =<ABC+ 2<CBY = <ABC + 2<ABC = 3<ABC , QED

This problem is easy. First we should know $B,I,H,C$ are concyclic. $<BHC=<CIB=120$

Let altitudes intersect sides $a,b,c$ in points $F,D,E$. Now, $\angle IBH=\angle ABD-\angle ABI=30^\circ-\frac{\beta}{2}$. $\angle ICH=\angle ACI-\angle ACH=30^\circ-\frac{\beta}{2} \implies \angle IBH=\angle ICH$. Using that quadrilateral $BCHI$ is cyclic we obtain: $\angle BCI=\angle BHI=\frac{\gamma}{2}$ and $\angle FHB=90^\circ-\angle FBH=90^\circ-\beta+\angle ABH=120^\circ-\beta$ Finally, $\angle AHI=180^\circ-\angle BHI-\angle FHB=60^\circ-\frac{\gamma}{2}+\beta=60^\circ-\frac{120^\circ-\beta}{2}+\beta=\frac{3\beta}{2} \implies 2\angle AHI=3\beta$

Note $\angle BHC=\angle BIC=\angle BOC=120^{\circ},$ so $H,O$ lie on $(BIC).$ We claim that $AHO$ is isosceles with $AH=AO.$ We use complex numbers. Put $ABC$ on the unit circle and note that $H=A+B+C=A+D,$ so $AH=OD=AO.$ Now note that $\angle CAH=\angle BAO=90^{\circ}-\angle C,$ so $\angle HAO=\angle A-(180^{\circ}-\angle 2C)=\angle C-\angle B=2\angle C-120^{\circ}.$ Thus $\angle AHO=150^{\circ}-\angle C=\angle B+30^{\circ}.$ Also note $\angle HIO=180^{\circ}-\angle CHO=180^{\circ}-\angle A+\angle B=120^{\circ}+\angle B,$ so $\angle IHO = 30^{\circ}-\frac{1}{2}\angle B.$ Thus $\angle AHI=\frac{3}{2}\angle B,$ as desired.

Let $M$ be the midpoint of minor arc $BC$ of $(ABC)$ and $O$ be its center. We have that $\triangle COM$ is equilateral and hence $OM=AH$. Also evidently $AH || OM$, hence $AOMH$ is a parallelogram. But since $OM=OA$ it's a rhombus, whence $OM=MH$. Now from the excenter/incenter lemma we have that $CHIOB$ is cyclic. We have $\angle AHI=\angle AOI=\angle AOC-\angle COH=\angle AOC-\angle IBC=\angle AOC-\frac{1}{2}\angle ABC=\frac{3}{2}\angle ABC$, as desired.

Note that we have $$\angle BIC = 180^{\circ}-\angle IBC - \angle ICB = 180^{\circ} - \frac{\angle ABC + \angle ACB}{2} = 180^{\circ} - \frac{180^{\circ}-\angle A}{2}$$and since $\angle A = 60^{\circ}$, then $\angle BIC = 120^{\circ}$. Now, it is well-known that $$\angle BHC = 180^{\circ}-\angle A = 180^{\circ}-60^{\circ}=120^{\circ}$$so as a result we have $\angle BIC = \angle BHC$ so $BIHC$ is cyclic. Now, reflecting $A$ over $BC$ and letting that point be $A'$ we know $A'$ is on the circumcircle of $BIHC$, so $$\angle AHI = 180^{\circ}- \angle A'HI = \angle A'BI = \angle B + \angle CBI = \frac32\angle B$$so $2\angle AHI = 3 \angle B$.

First note that $BIC$ are concyclic. I claim that $H$ lies on this circle. Let $\Gamma_1$ denote the circumcircle of $\triangle ABC$ and $\Gamma_2$ denote $(BIC)$. Then, arc $\overarc {BAC}$ has measure $240^{\circ}$. Then, since $\angle BIC=90^{\circ}+\frac{\angle A}{2}=120^{\circ}$, looking at $\Gamma_2$, $\overarc{BC}$ not containing $I$ has measure $240^{\circ}$ as well. Hence, $\Gamma_1$ and $\Gamma_2$ are reflections of each other across $BC$. Since the reflection of $H$ across $BC$ lies on $\Gamma_1$, $H$ must lie on $\Gamma_2$. Now, \[3\angle ABC=6\angle IBC=6(180-\angle IHC)=1080-6(180-\angle AHI+\angle ABC)=6\angle AHI-6\angle ABC,\]which implies the desired result

Points $I, H, O, B, C$ Will lie on same circle.. And then by some angle chase we will get the desired results

We start with a lemma. Lemma: In a triangle $ABC$ with orthocenter $H$, $AH=2R\cos A$. Proof. Consult problem Exercise 4.4.7 in the AoPS Precalculus Book. $\square$ Now note that $AH=2R\cos A=2R\cdot \frac12=R=AO$. Also, $AI=AI$ and $\measuredangle HAI=\measuredangle IAO$ since $H$ and $O$ are isogonal conjugates. Thus, $\triangle AHI\cong \triangle AOI$. Now observe the following angle equalities: \begin{align*} \angle BHC&=180^\circ-\angle A=120^\circ\\ \angle BIC&=90^\circ+\frac{\angle A}2=120^\circ\\ \angle BOC&=2\angle A=120^\circ \end{align*} so hexagon $CHIOBI_A$ is cyclic. The rest is just angle chasing: $$\measuredangle AHI=\measuredangle IOA=\measuredangle IOC+\measuredangle COA=\measuredangle IBC+2\measuredangle CBA=4\measuredangle CBI- \measuredangle CBI=3\measuredangle CBI$$and multiplying both sides by $2$ gives the desired $2\measuredangle AHI=6\measuredangle CBI=3\measuredangle CBA$. $\blacksquare$

APMO 2007 P2 wrote: Let $ABC$ be an acute angled triangle with $\angle{BAC}=60^\circ$ and $AB > AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$ . Prove that $2\angle{AHI}= 3\angle{ABC}$. Solution. Since $\angle A=60^{\circ}$, we have that $BIHC$ is cyclic. Now just by angle chasing, we get that: $$\angle AHI=180^{\circ}-\angle IHD=180^{\circ}-(\angle IHB+\angle BHD)=180^{\circ}-(\frac{\angle C}{2}+\angle C)=180^{\circ}-\frac{3\angle C}{2}$$$$2\angle AHI=360^{\circ}-3\angle C=3(120^\circ-\angle C)=3\angle B.\blacksquare$$

Claim: $BIHC$ is cyclic. Proof. Notice that \begin{align*}\angle HCI&=\tfrac{1}{2}\angle C-(90-60)\\&=\tfrac{1}{2}\angle C-30+\tfrac{1}{2}(\angle B+\angle C-120)\\&=\tfrac{1}{2}\angle B-(90-\angle C)\\&=\angle HBI.\end{align*}$\blacksquare$ Hence, $$\angle AHI=(180-\angle IHC)+(180-AHC)=\tfrac{1}{2}\angle B+180-\angle A-\angle C=\tfrac{3}{2}\angle B.$$$\square$

Let AD,BE and CF be altitudes of triangle. ∠EHF = 180 - 60 = 120 = ∠BHC and ∠BIC = ∠A + ∠B/2 + ∠C/2 = 120 so BIHC is cyclic. ∠AHI = ∠IHF + ∠FHA = ∠IBC + ∠DHC = ∠IBC + ∠ABC ---> ∠AHI = 3/2 ∠ABC. we're Done.

$BHIC$ is cyclic since $\angle BHC = 180^{\circ} - \angle A = 120^{\circ} = 90^{\circ} + \frac{\angle A}{2}$. $D$ be the foot from $A$ to $BC$. Consider $DHIC$ where $\angle HDC = 90^{\circ}$, $\angle IBC = \frac{\angle B}{2}$, $\angle HIC = 180^{\circ} - 90^{\circ} + \angle B = 90^{\circ} + \angle B$. Now, trivial calculation gives, $\angle AHI = \frac{3}{2}\angle B$ and check that, this is the result we wanted.

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Let $\angle ABC=2\theta$ and $\angle ACB=120-2\theta$. We wish to show that $\angle AHI=3\theta$. Claim: $HIBC$ is cyclic. This is because $$\angle BHC=180-\angle A=120$$and $$\angle BIC=90+(\angle A/2)=120.$$ Then, $$\angle IHB=\angle ICB=60-\theta.$$Let $E$ be the foot from $B$ to $AC$. Then, $$\angle AHE=\angle C=120-2\theta,$$so $$\angle AHI=180-\angle AHE-\angle IHB=180-(120-2\theta)-(60-\theta)=3\theta,$$hence done.

Observe that $B, C, H, I$ are concyclic, as $\angle BHC = 120^\circ = \angle BIC$ from some simple angle chasing. Now we have $\angle AHC = 180^\circ - B, \angle CHI = 180^\circ - \frac{B}{2} \implies \angle AHI = \frac{3B}{2}$, and we are done. $\square$