Hong Kong 2020 TST 2 Q1

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Source: 2020HKTST2 Q1

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#1 h Oct 26, 2019, 10:37 AM • 2 Y

Y by mathematicsy, Adventure10

Let $\Delta ABC$ be an acute triangle with incenter $I$ and orthocenter $H$ . $AI$ meets the circumcircle of $\Delta ABC$ again at $M$ . Suppose the length $IM$ is exactly the circumradius of $\Delta ABC$ . Show that $AH\geq AI$ .

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#2 h Oct 26, 2019, 10:24 PM • 3 Y

Y by mathematicsy, Adventure10, Mango247

Blastzit wrote:

Let $\Delta ABC$ be an acute triangle with incenter $I$ and orthocenter $H$ . $AI$ meets the circumcircle of $\Delta ABC$ again at $M$ . Suppose the length $IM$ is exactly the circumradius of $\Delta ABC$ . Show that $AH\geq AI$ .

Let $O$ be the circumcenter of $\odot{ABC}$ .By Fact 5 $MI=MB=MC=MO=OB=OC$ .Hence $OMC$ and $OMB$ are equilateral triangles.Hence $\angle{BAC}=2\cdot \angle{MAC}=60^{\circ}$ .Also $AI\cdot R=AI\cdot IM=R^2-OI^2=2Rr \implies AI=2r$ .Now $R=2R\cos{60^{\circ}}=AH\geq AI=2r$ which is just Euler's Theorm so we are done.

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zuss77

#3 h Oct 27, 2019, 9:49 AM • 2 Y

Y by AlastorMoody, Adventure10

$M$ -center of $\odot IBC$ which in this case is a reflection of $\odot ABC$ about $BC$ , so it goes through $H$ .
$AH+HM\geq AM \implies AH \geq AI$

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#5 h Jan 3, 2020, 1:29 PM • 1 Y

Y by Adventure10

Identity:. As $MO=MI=MB=MC$ we get that $,O,I,B,C$ lie on a circle $\omega$ .
So $\angle BOC=\angle BIC$ , using above identity we have $\angle BHC=\angle BIC$ and hence $H\in\omega$ . Now $AH+HM\geq AM \implies AH\geq AI$ .

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Mathematicsislovely

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Mathematicsislovely

#6 h Jan 21, 2020, 10:34 PM • 1 Y

Y by Adventure10

Let $I_A,I_B,I_C$ be the $A,B,C$ excenters of $\Delta ABC$ .We have $I_A$ is the reflection of $I$ in $M$ .So $II_A=2R$ .Again if $O_1$ is the centre of $\Delta I_AI_BI_C$ then $I_AO_1=2R$ as circumcircle of $\Delta ABC$ is nine point circle of $\Delta I_AI_BI_C$ .So $II_A=I_AO_1$ .As $I$ is the orthocentre of $\Delta I_AI_BI_C$ and $O_1$ is the circumcenter of $\Delta I_AI_BI_C$ and the are of equal length so $\angle{I_CI_AI_B}=60^{\circ}$ .So $\angle{I_CI_AI_B}=90^{\circ}-\angle{BAC}/2=60^{\circ}$ so $\angle{BAC}=60^{\circ}$ .So $\angle{BHC}=120^{\circ}$ .

Now note that $\angle{BHC}+\angle{BI_AC}=120^{\circ}+60^{\circ}=180^{\circ}$ so $BHICI_A$ is concyclic.By fact 5 $M$ is the centre of this circle.So $MH=MI$ .Now $AH+HM\geq AM \implies AH\geq AI$ .

This post has been edited 1 time. Last edited by Mathematicsislovely, Jan 21, 2020, 10:35 PM

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