an acute angled triangle with $\angle{BAC}=60^0$

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an acute angled triangle with $\angle{BAC}=60^0$

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Source: APMO 2007

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3595 posts

#1 h Mar 31, 2007, 5:19 AM • 5 Y

Y by Adventure10, Mango247, Rounak_iitr, DEKT, and 1 other user

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 $2\angle{AHI}= 3\angle{ABC}$ .

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#2 h Mar 31, 2007, 5:44 PM • 3 Y

Y by vsathiam, Adventure10, Mango247

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#3 h Mar 31, 2007, 5:54 PM • 1 Y

Y by Adventure10

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 .

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#4 h Dec 8, 2010, 9:53 AM • 3 Y

Y by aats411, Adventure10, Mango247

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, $3\angle ABC=6\beta=2\angle AHI$ .

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#5 h Mar 13, 2015, 3:52 PM • 1 Y

Y by Adventure10

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

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#6 h Jul 3, 2015, 1:30 PM • 3 Y

Y by igli.2001, Adventure10, Mango247

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

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AMN300

#7 h May 30, 2016, 10:37 PM • 2 Y

Y by Mai-san, Adventure10

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#9 h Jul 6, 2017, 9:55 AM • 2 Y

Y by Adventure10, Mango247

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

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#11 h Apr 15, 2020, 5:28 PM • 3 Y

Y by Mango247, Mango247, Mango247

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, $\angle AHI=180^\circ-\angle BHI-\angle FHB=60^\circ-\frac{\gamma}{2}+\beta=60^\circ-\frac{120^\circ-\beta}{2}+\beta=\frac{3\beta}{2} \implies 2\angle AHI=3\beta$

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dchenmathcounts

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dchenmathcounts

#13 h Jun 15, 2020, 5:43 PM

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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.

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star32

#14 h Jul 17, 2020, 5:29 PM • 1 Y

Y by TwilightZone

solution sketch

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Reason: typo:(

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#15 h Aug 13, 2020, 12:32 PM • 1 Y

Y by TwilightZone

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.

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#16 h Oct 17, 2020, 7:35 PM

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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 $2\angle AHI = 3 \angle B$ .

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#17 h Dec 20, 2020, 8:16 PM

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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

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#18 h Dec 21, 2020, 9:40 AM • 2 Y

Y by shalomrav, Mango247

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

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#19 h Mar 8, 2021, 12:49 AM • 1 Y

Y by JustinLee2017

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 $2\measuredangle AHI=6\measuredangle CBI=3\measuredangle CBA$ . $\blacksquare$

Remarks

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#20 h Apr 16, 2021, 8:37 AM

Y by

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:
$\angle AHI=180^{\circ}-\angle IHD=180^{\circ}-(\angle IHB+\angle BHD)=180^{\circ}-(\frac{\angle C}{2}+\angle C)=180^{\circ}-\frac{3\angle C}{2}$ $2\angle AHI=360^{\circ}-3\angle C=3(120^\circ-\angle C)=3\angle B.\blacksquare$

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#21 h Dec 17, 2021, 3:34 AM • 1 Y

Y by centslordm

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, $\angle AHI=(180-\angle IHC)+(180-AHC)=\tfrac{1}{2}\angle B+180-\angle A-\angle C=\tfrac{3}{2}\angle B.$ $\square$

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Mahdi_Mashayekhi

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#22 h Jan 9, 2022, 6:05 AM

Y by

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.

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#23 h Jan 17, 2023, 4:36 AM • 1 Y

Y by Mango247

$BHIC$ 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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lifeismathematics

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lifeismathematics

#24 h Apr 6, 2023, 9:41 AM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(42cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.982013289036544, xmax = 16.091971207087482, ymin = -7.555732004429678, ymax = 8.00945071982281; /* image dimensions */ /* draw figures */ draw((-0.2750542635658916,5.839758582502767)--(-9.51982945736434,-4.371944629014396), linewidth(2)); draw((-0.2750542635658916,5.839758582502767)--(2.932316722037652,-0.5513997785160583), linewidth(2)); draw((-9.51982945736434,-4.371944629014396)--(2.932316722037652,-0.5513997785160583), linewidth(2)); draw(circle((-0.9208974895600552,1.0958517060424493), 2.7050286774217995), linewidth(2)); draw((xmin, -3.25925925925926*xmin + 4.943285427176898)--(xmax, -3.25925925925926*xmax + 4.943285427176898), linewidth(2)); /* line */ draw((xmin, 0.5018450184501846*xmin + 0.405534360659222)--(xmax, 0.5018450184501846*xmax + 0.405534360659222), linewidth(2)); /* line */ draw((xmin, -0.9053117782909931*xmin + 2.103261087624264)--(xmax, -0.9053117782909931*xmax + 2.103261087624264), linewidth(2)); /* line */ draw((-0.2750542635658916,5.839758582502767)--(-0.9208974895600541,1.0958517060424489), linewidth(2)); draw((-0.9208974895600541,1.0958517060424489)--(1.2064943515156186,1.0110075407556212), linewidth(2)); draw((-0.9208974895600541,1.0958517060424489)--(-9.51982945736434,-4.371944629014396), linewidth(2)); draw((-0.9208974895600541,1.0958517060424489)--(2.932316722037652,-0.5513997785160583), linewidth(2)); /* dots and labels */ dot((-0.2750542635658916,5.839758582502767),dotstyle); label("$A$", (-0.1807198228128475,6.07559468438538), NE * labelscalefactor); dot((-9.51982945736434,-4.371944629014396),dotstyle); label("$B$", (-9.425495016611295,-4.136108527131783), NE * labelscalefactor); dot((2.932316722037652,-0.5513997785160583),dotstyle); label("$C$", (3.0266511627906953,-0.3155636766334448), NE * labelscalefactor); dot((-0.9208974895600541,1.0958517060424489),dotstyle); label("$I$", (-0.8174772978959038,1.3352890365448493), NE * labelscalefactor); dot((-0.12745211158699393,-1.490192488832711),dotstyle); label("$D$", (-0.03921816168327944,-1.2589080841638987), NE * labelscalefactor); dot((-2.9262243575179765,2.9112977389281434),dotstyle); label("$E$", (-2.822084163898118,3.151227021040973), NE * labelscalefactor); dot((1.4967668262111673,2.3091444991970103),dotstyle); label("$F$", (1.5880509413067534,2.538053156146178), NE * labelscalefactor); dot((1.2064943515156186,1.0110075407556212),dotstyle); label("$H$", (1.3050476190476172,1.240954595791804), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Claim:-points $B,I,H,C$ are concylic

Proof:- we notice that $\angle{BHC}=180^{\circ}-\angle{A}=120^{\circ}$ and $\angle{BIC}=90^{\circ}+\frac{A}{2}=120^{\circ}$ so we get $\angle{BHC}=\angle{BIC}$ hence points $B,I,H,C$ are concylic $\blacksquare$

now we have $\angle{AHI}+\angle{AHC}+\angle{CHI}=360^{\circ} \implies \angle{AHI}=360^{\circ}-\left(180^{\circ}-\frac{\angle{B}}{2}\right)-(180^{\circ}-\angle{B}) \implies \angle{AHI}=\frac{3\angle{B}}{2} \implies 2\angle{AHI}=3\angle{ABC}$ $\square$

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#25 h Jul 4, 2023, 7:38 AM

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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.

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#26 h Mar 12, 2024, 2:22 PM

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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. $\square$

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