Given a triangle ABC we construct a new triangle DEF such that DE parallel to AB and contains C, EF is parallel to BC and contains A and similarly, FD is parallel to AC and sontains B. Clearly the triangle ABC is similar to DEF with the enlargment soefficient equal to 2. Moreover each of the triangles AFB, CBD and EAC is congruent to ABC. Therefore FB=Bd, DC=CE and EA=AF. Since H is the orthocentre of ABC, BH is perpendicular to AC. Hence BH is perpendicular to FD as FD is parallel to AC. Thus BH is the perpendicular bisector to FD. Similarly HA and HC are the perpendicular bisectors to EF and ED respectively. Therefore H is the circumcentre os the perpendicular bisectors to EF and ED respectively. Therefore H is the circumcentre of the triangle DEF. Hence FH is the circumradius of DEF. Therefore FH=2R In the triangle BOH, OH<BH+BO Since FBH is a right triangle BH<FH. Hence BH<2R Besides BO=R Therefore OH < BH + BO < 2R + R = 3R Davron

Easy to see $G$ is in the interior of triangle $ABC$,so it's in the interior of circle $O$, $OG<R$,and $OH=3OG<3R$

let A(a),B(b),C(c) and |a|=|b|=|c|=R then OH=|a+b+c| we must prove that OH<3r <=> |a+b+c|<|a|+|b|+|c| and that is wellknown ineq

The problem is obvious if we use well-known formula: $|OH|^2 = 9R^2 - a^2 - b^2 - c^2$.

I can do stronger $OH \leq HA+HB+HC-3R < 3R$ thence $HA+HB+HC \geq 3R \geq GA+GB+GC$

TomciO wrote: The problem is obvious if we use well-known formula: $ |OH|^2 = 9R^2 - a^2 - b^2 - c^2$. Can you prove this formula?

This one is the basic formula, every senior student who studies in the Math Major class in my country can solve this easily . The main idea of proving this formula is using another well- known formula- The leibniz formula for the centroid point of triangle ABC: OA^2+OB^2+OC^2=GA^2+GB^2+GC^2+3OG^2 From the above, we easily get this result:OG=1/3sqrt( 9R^2-(a^2+b^2+c^2). But OH=3OG (From Euler line of triangle ABC). Then we get the result

shobber wrote: Given a nondegenerate triangle $ ABC$, with circumcentre $ O$, orthocentre $ H$, and circumradius $ R$, prove that $ |OH| < 3R$. In complex coordinates let $ a,b,c,o,h$ be $ A,B,C,O,H$. Then $ h = a + b + c - 2o$, and thus $ |OH| = |h - o| = |(a - o) + (b - o) + (c - o)|$. But $ 3R = |a - o| + |b - o| + |c - o|$, so it is left to prove: $ |(a - o) + (b - o) + (c - o)| < |a - o| + |b - o| + |c - o|$, which is just the triangle inequaltiy (There can't be equality since $ a - o,b - o,c - o$, never could be in line in a non-degenerated triangle)

dgreenb801 wrote: TomciO wrote: The problem is obvious if we use well-known formula: $ |OH|^2 = 9R^2 - a^2 - b^2 - c^2$. Can you prove this formula? You can Stewart's bash using triangle $ HAO$ and cevian $ AG$. It's not too hard to express the sides of $ \triangle HAO$ and $ AG$ in terms of $ OH$, $ R$, $ a$, $ b$, and $ c$.

$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\Rightarrow c^2 = |AB|^2 = \left(\overrightarrow{OB} - \overrightarrow{OA}\right)^2 = 2R^2 - 2.\overrightarrow{OA}.\overrightarrow{OB}\Rightarrow \\ \Rightarrow 2.\overrightarrow{OA}.\overrightarrow{OB} = 2R^2 - c^2\mbox{ and analogue: } \\ 2.\overrightarrow{OA}.\overrightarrow{OC} = 2R^2 - b^2; \\ 2.\overrightarrow{OB}.\overrightarrow{OC} = 2R^2 - a^2\Rightarrow \\ \Rightarrow \overrightarrow{OH} = 3.\overrightarrow{OG} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}\Rightarrow \\ \Rightarrow |OH|^2 = 9.|OG|^2 = 3R^2 + 2.(\overrightarrow{OA}.\overrightarrow{OB} + \overrightarrow{OA}.\overrightarrow{OC} + \overrightarrow{OB}.\overrightarrow{OC}) = \\ = 3R^2+2\left[(2R^2 - c^2) + (2.R^2 - b^2) + (2.R^2 - a^2)\right] = 9R^2 - (a^2 + b^2 + c^2).$

Even easier with complex numbers Assume wlog that $ \triangle ABC$ has circumcenter in $ o$. Then use the real product: $ OH^2 = |a + b + c|^2 = (a + b + c) \cdot (a + b + c) = 9R^2 - |a - b|^2 - |b - c|^2 - |c - a|^2$ $ = 9R^2 - \alpha^2 - \beta^2 - \gamma^2$

One liner (?)

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Same as my solution

Toss the problem on the complex plane with the circumcircle of $\triangle ABC$ as the unit circle. Then it suffices to show $$|h| < |a|+|b|+|c| \rightarrow |a+b+c| < |a|+|b|+|c|$$which is clearly true. $\blacksquare$