BdMO National 2012: Higher Secondary, Secondary 07

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Moon
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BdMO National 2012: Higher Secondary, Secondary 07

Unread post by Moon » Sat Feb 11, 2012 11:22 pm

Problem:
In an acute angled triangle $ABC$, $\angle A= 60^0$. Prove that the bisector of one of the angles formed by the altitudes drawn from $B$ and $C$ passes through the center of the circumcircle of the triangle $ABC$.
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Re: BdMO National 2012: Higher Secondary, Secondary 07

Unread post by photon » Wed Feb 15, 2012 7:38 am

Moon wrote:Problem:
In an acute angled triangle $ABC$, $\angle A= 60^0$. Prove that the bisector of one of the angles formed by the altitudes drawn from $B$ and $C$ passes through the center of the circumcircle of the triangle $ABC$.
what is that "angles formed by the altitudes drawn from $B$ and $C$" ?
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Re: BdMO National 2012: Higher Secondary, Secondary 07

Unread post by *Mahi* » Wed Feb 15, 2012 11:35 am

Let the feet of altitudes from $B,C$ be $E,F$. If $BE$ and $CF$ intersects at $H$ then the angles are $\angle BHC,\angle CHE,\angle EHF,\angle BHF$.
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Re: BdMO National 2012: Higher Secondary, Secondary 07

Unread post by photon » Fri Feb 17, 2012 10:09 pm

O is center.radius R
\[BC=\frac{2R}{sinA}=\sqrt{3}R\]
$\angle BOC=2x60^O=120^O$ AND $\angle BHC=120^O$. so, $B,H,O,C$ are cyclic.
in tri $BOC$, \[sin \angle OBC=\frac{sinA.R}{BC}=\frac{sin60^O.R}{\sqrt{3}R}=\frac{1}{2}\]
so, $\angle OBC=30^O$, :arrow: $\angle OHC=30^O$[$B,H,O,C$ are cyclic.]
so, $HO$ bisects $\angle CHE=60^O$.Done.
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Re: BdMO National 2012: Higher Secondary, Secondary 07

Unread post by SANZEED » Sun Jan 26, 2014 6:12 pm

When the altitudes intersect, 2 acute angles and 2 obtuse angles are formed. If my check is right, then $OH$ always divides the acute ones.
Do we need to prove this?
And also, there is an easy euclidean proof taking this fact into account.
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