an inequality in triangle

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an inequality in triangle

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Source: tuymaada 2006 - problem 6

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337 posts

maky

#1 h Jul 17, 2006, 12:21 PM • 2 Y

Y by Adventure10 and 1 other user

Let $ABC$ be a triangle, $G$ it`s centroid, $H$ it`s orthocenter, and $M$ the midpoint of the arc $\widehat{AC}$ (not containing $B$ ). It is known that $MG=R$ , where $R$ is the radius of the circumcircle. Prove that $BG\geq BH$ .

Proposed by F. Bakharev

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246 posts

#2 h Jul 17, 2006, 1:05 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Denote by $D$ midpoint of $AC$ . We have that $MO \mid \mid BH$ . From yhis is $\angle MOG= \angle GHB$ . It is also $\angle BGH= \angle DGO \leq MCO = \angle MOG$ . Hence $BH\leq BG$ . Equality holds if $\angle DGM=0$ , but then points $B,H,G$ lies on same line. SO there must be $H = G \Longleftrightarrow AB=AC=BC$ .
q.e.d.

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#3 h Jun 5, 2008, 9:34 PM • 2 Y

Y by Adventure10, Mango247

Ok, now realized my solution has some flaws in it.

This post has been edited 1 time. Last edited by omerftekin, Dec 5, 2011, 6:31 PM

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2 posts

#4 h Jun 6, 2008, 4:58 PM • 1 Y

Y by Adventure10

what is tagiyev's pisi lemma?can you send me this lemma?

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¬[ƒ(Gabriel)³²¹º]¼

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¬[ƒ(Gabriel)³²¹º]¼

#5 h Jun 6, 2008, 6:32 PM • 2 Y

Y by Adventure10 and 1 other user

another proof: Call O the circumcenter, K the projection of B on OH and $\theta$ the angle between the euler line and AC. So we have $\sin \theta = \frac{\overline{GO}}{2r}$ then $\overline{KH} = \overline{BH} \cdot \sin \theta = 2R \cos \beta \sin \theta = \overline{GO}\cos \beta \le \overline{GO} = \frac{\overline{GH}}{2}$ . Then $\overline{KH} \ge \overline{KG} \Longleftrightarrow \angle GHB \ge \angle HGB \Longleftrightarrow \overline{BG} \ge \overline{BH}$

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1345 posts

#6 h Jun 7, 2008, 7:32 PM • 2 Y

Y by Adventure10, Mango247

Let $K$ be the midpoint of segment $AC$ and $O$ be the circumcenter of triangle $ABC$ . $BGH$ similar to $KGO$ and $BG\geq BH$ iff $GK\geq OK$ . $MG=MO=R$ and by the triangle inequality $GK+MK\geq MG$ , $GK+R-OK\geq R$ and $GK\geq OK$ . The proof is completed.

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