It problem is given by Nairi Sedrakyan and, my opinion, this is fundamental fact So let's prove that !

by using $OI^{2}=R^{2}-2Rr$,$OH^{2}=9R^{2}-(a^{2}+b^{2}+c^{2})$, and $IH^{2}=4R^{2}+2r^{2}-0.5(a^{2}+b^{2}+c^{2})$ and the famous lemma $R\ge\ 2r$,we can get $OI^{2}+IH^{2}<OH^{2}$ easily. but I can't solve the second part

Okay, with horizon's hint, a) becomes $4r^2 + a^2 + b^2 + c^2 < 8R^2 + 4rR$ and I can't prove it. Would someone help me?

$4r^2 < 8R^2$ is obviously true from $2r < R$ $a^2 +b^2 + c^2 < 8R^2$ is true because $9R^2 - (a^2+b^2+c^2) = OH^2 \ge 0$. It can also be proved by using that $\sin^2 A + \sin^2 B + \sin^2 C \le \frac{9}{4}$ which is well known.

hatchguy wrote: $a^2 +b^2 + c^2 < 8R^2$ is true because $9R^2 - (a^2+b^2+c^2) = OH^2 \ge 0$.. This is false. The fact that $a^2+b^2+c^2\le 9R^2$ does not necessarily imply that $a^2+b^2+c^2<8R^2$ ... However, the inequality $4r^2+a^2+b^2+c^2< 8R^2+4Rr$ (which is equivalent to $\widehat{OIH}> 90^{\circ}$) follows immediately from Gerretsen's old result i.e. $\boxed{\, a^2+b^2+c^2\, \le\, 8R^2+4Rr\, }$ and Euler's inequality i.e. $\boxed{\, R\, \ge\, 2r\, }$ . On the other hand, the second part of the problem asking to show that $\widehat{OIH}>135^{\circ}$ is true only if $\triangle\, ABC$ is acute-angled. I don't know why the OP left out this important condition.

I'm stupid, please delete my comment(post)

Mateescu Constantin wrote: On the other hand, the second part of the problem asking to show that $\widehat{OIH}>135^{\circ}$ is true only if $\triangle\, ABC$ is acute-angled. I don't know why the OP left out this important condition. Could you explain how does it follow in case $\triangle ABC$ is acute-angled ?

samirka259 wrote: Mateescu Constantin wrote: On the other hand, the second part of the problem asking to show that $\widehat{OIH}>135^{\circ}$ is true only if $\triangle\, ABC$ is acute-angled. I don't know why the OP left out this important condition. Could you explain how does it follow in case $\triangle ABC$ is acute-angled ? Well... I don't remember if I had a solution for this claim back in 2012, but I don't see anything else other than $Rrs$ bashing . Note that: \[\widehat{OIH}>135^{\circ} \iff \frac {OI^2 + IH^2 - OH^2}{2 \cdot OI \cdot IH} < -\frac {\sqrt{2}}2,\] which expressed in terms of the classical $R$, $r$ and $s$ elements reduces to: \[\begin{array}{c}\displaystyle \left(R^2 - 2Rr\right) + \left(4R^2 + 4Rr + 3r^2 - s^2\right) - \left(9R^2 + 8Rr + 2r^2 - 2s^2\right) < \\\\ \displaystyle -\sqrt{2}\cdot \sqrt{\left(R^2 - 2Rr\right) \cdot \left(4R^2 + 4Rr + 3r^2 - s^2\right)}\end{array} \] \[\iff \sqrt{2\left(R^2 - 2Rr\right) \cdot \left(4R^2 + 4Rr + 3r^2 - s^2\right)} < \underbrace{4R^2 + 6Rr - r^2 - s^2}_{\ge 0}\] \[\iff 2\left(R^2 - 2Rr\right) \cdot \left[\left(4R^2 + 6Rr - r^2 - s^2\right) - \left(2Rr - 4r^2\right)\right] < \left(4R^2 + 6Rr - r^2 - s^2\right)^2\] \[\iff -4Rr\left(R - 2r\right)^2 < \left(4R^2 + 6Rr - r^2 - s^2\right)^2 - 2R\left(R - 2r\right)\left(4R^2 + 6Rr - r^2 - s^2\right)\] \[\iff R^2\left(R - 2r\right)^2 -4Rr\left(R - 2r\right)^2 < \left[\left(4R^2 + 6Rr - r^2 - s^2\right) - R\left(R - 2r\right)\right]^2\] \[\iff R\left(R - 2r\right)^2 \left(R - 4r\right) < \left(3R^2 + 8Rr - r^2 - s^2\right)^2.\] If $R < 4r$ then we are clearly done. Otherwise, we are left to prove that: \[\boxed{\left(R - 2r\right)\sqrt{R\left(R - 4r\right)} < \left| 3R^2 + 8Rr - r^2 - s^2 \right|}\ ,\] which we will do in two cases. Case 1. If $3R^2 + 8Rr - r^2 \ge s^2$ then by taking into account the well-known inequality for non-obtuse triangles, i.e. $s\ge 2R + r$, we see that $(2R + r)^2 \le 3R^2 + 8Rr - r^2$ holds true for $\tfrac Rr \in\left[2, 2 + \sqrt{2}\right]$ , which contradicts the assumption that $\tfrac Rr \ge 4$. So this case is clearly impossible. Case 2. It remains to study the only valid case when $3R^2 + 8Rr - r^2 < s^2$. Here we are left to show that: \[\left(R - 2r\right)\sqrt{R\left(R - 4r\right)} + \left(3R^2 + 8Rr - r^2\right) < s^2,\] which can be proven via the same inequality $s\ge 2R + r$. Indeed, \[\left(R - 2r\right)\sqrt{R\left(R - 4r\right)} + \left(3R^2 + 8Rr - r^2\right) < \left(2 R + r\right)^2\] \[\begin{array}{c}\displaystyle \stackrel{t = \frac Rr}{\iff} (t - 2)\sqrt{t (t - 4)} < t^2 - 4 t + 2\end{array}\] \[\iff t (t - 4) (t - 2)^2 < (t^2 - 4 t + 2)^2 \iff 4 > 0,\] which is obviously true and this completes the proof. It is interesting to note the following case when triangle $ABC$ is obtuse having two angles $\to 90^{\circ}$. In this example we have $\widehat{OIH}\approx 91.12^{\circ}$ which means that the condition for $\triangle ABC$ to be acute was essential in establishing the fact that $\widehat{OIH} > 135^{\circ}$. Finally, by computer we can find that $\approx 135.33^{\circ}$ is the best upper bound for $\widehat{OIH}$ in an acute-angled triangle, so the proposed inequality (missing the constraint for $\triangle ABC$) was almost optimal .

Thank you!

@Mateescu Constantin, Good job for staying on AoPS for 6 years! That is extremely impressive!

Mateescu Constantin wrote: samirka259 wrote: Mateescu Constantin wrote: On the other hand, the second part of the problem asking to show that $\widehat{OIH}>135^{\circ}$ is true only if $\triangle\, ABC$ is acute-angled. I don't know why the OP left out this important condition. Could you explain how does it follow in case $\triangle ABC$ is acute-angled ? Well... I don't remember if I had a solution for this claim back in 2012, but I don't see anything else other than $Rrs$ bashing . Note that: \[\widehat{OIH}>135^{\circ} \iff \frac {OI^2 + IH^2 - OH^2}{2 \cdot OI \cdot IH} < -\frac {\sqrt{2}}2,\] which expressed in terms of the classical $R$, $r$ and $s$ elements reduces to: \[\begin{array}{c}\displaystyle \left(R^2 - 2Rr\right) + \left(4R^2 + 4Rr + 3r^2 - s^2\right) - \left(9R^2 + 8Rr + 2r^2 - 2s^2\right) < \\\\ \displaystyle -\sqrt{2}\cdot \sqrt{\left(R^2 - 2Rr\right) \cdot \left(4R^2 + 4Rr + 3r^2 - s^2\right)}\end{array} \] \[\iff \sqrt{2\left(R^2 - 2Rr\right) \cdot \left(4R^2 + 4Rr + 3r^2 - s^2\right)} < \underbrace{4R^2 + 6Rr - r^2 - s^2}_{\ge 0}\] \[\iff 2\left(R^2 - 2Rr\right) \cdot \left[\left(4R^2 + 6Rr - r^2 - s^2\right) - \left(2Rr - 4r^2\right)\right] < \left(4R^2 + 6Rr - r^2 - s^2\right)^2\] \[\iff -4Rr\left(R - 2r\right)^2 < \left(4R^2 + 6Rr - r^2 - s^2\right)^2 - 2R\left(R - 2r\right)\left(4R^2 + 6Rr - r^2 - s^2\right)\] \[\iff R^2\left(R - 2r\right)^2 -4Rr\left(R - 2r\right)^2 < \left[\left(4R^2 + 6Rr - r^2 - s^2\right) - R\left(R - 2r\right)\right]^2\] \[\iff R\left(R - 2r\right)^2 \left(R - 4r\right) < \left(3R^2 + 8Rr - r^2 - s^2\right)^2.\] If $R < 4r$ then we are clearly done. Otherwise, we are left to prove that: \[\boxed{\left(R - 2r\right)\sqrt{R\left(R - 4r\right)} < \left| 3R^2 + 8Rr - r^2 - s^2 \right|}\ ,\] which we will do in two cases. Case 1. If $3R^2 + 8Rr - r^2 \ge s^2$ then by taking into account the well-known inequality for non-obtuse triangles, i.e. $s\ge 2R + r$, we see that $(2R + r)^2 \le 3R^2 + 8Rr - r^2$ holds true for $\tfrac Rr \in\left[2, 2 + \sqrt{2}\right]$ , which contradicts the assumption that $\tfrac Rr \ge 4$. So this case is clearly impossible. Case 2. It remains to study the only valid case when $3R^2 + 8Rr - r^2 < s^2$. Here we are left to show that: \[\left(R - 2r\right)\sqrt{R\left(R - 4r\right)} + \left(3R^2 + 8Rr - r^2\right) < s^2,\] which can be proven via the same inequality $s\ge 2R + r$. Indeed, \[\left(R - 2r\right)\sqrt{R\left(R - 4r\right)} + \left(3R^2 + 8Rr - r^2\right) < \left(2 R + r\right)^2\] \[\begin{array}{c}\displaystyle \stackrel{t = \frac Rr}{\iff} (t - 2)\sqrt{t (t - 4)} < t^2 - 4 t + 2\end{array}\] \[\iff t (t - 4) (t - 2)^2 < (t^2 - 4 t + 2)^2 \iff 4 > 0,\] which is obviously true and this completes the proof. It is interesting to note the following case when triangle $ABC$ is obtuse having two angles $\to 90^{\circ}$. In this example we have $\widehat{OIH}\approx 91.12^{\circ}$ which means that the condition for $\triangle ABC$ to be acute was essential in establishing the fact that $\widehat{OIH} > 135^{\circ}$. Finally, by computer we can find that $\approx 135.33^{\circ}$ is the best upper bound for $\widehat{OIH}$ in an acute-angled triangle, so the proposed inequality (missing the constraint for $\triangle ABC$) was almost optimal . TelvCohl's solution on $OIH>135$: Let $ \mu =\cos \angle A \cos \angle B \cos \angle C >0 $ We have to prove $$ \cos \angle OIH=\frac{{OI}^2+{HI}^2-{OH}^2}{2 \cdot OI \cdot HI}< \frac{-1}{\sqrt{2}} $$$$ \Longleftrightarrow ({OI}^2+{HI}^2-{OH}^2)^2>2 \cdot {OI}^2 \cdot {HI}^2 $$$$\Longleftrightarrow (\text{R}\text{r}-{\text{r}}^2-2 {\text{R}}^2 \mu)^2>({\text{R}}^2-2\text{R}\text{r})({\text{r}}^2-2 {\text{R}}^2 \mu) $$$$ \Longleftrightarrow ({\text{r}}^2-2 {\text{R}}^2 \mu)^2 + 2 \mu ({\text{R}}^2-2 \text{R} \text{r} )^2 >0 $$

Ovchinnikov Denis wrote: Let $ABC$ arbitrary triangle ($AB \neq BC \neq AC \neq AB$) And O,I,H it's circum-center, incenter and ortocenter (point of intersection altitudes). Prove, that 1) $\angle OIH > 90^0$(2 points) 2)$\angle OIH >135^0$(7 points) balls for 1) and 2) not additive. I think i found some contradiction for second part I dont know how to insert what i found but it is for sure a contradiction

Mathmatin wrote: I think i found some contradiction for second part I dont know how to insert what i found but it is for sure a contradiction Note that the original problem statement didn't specify that $\triangle ABC$ must be acute-angled in the second part. There is a counterexample in my post #10 above which shows that the second inequality fails for an obtuse-angled triangle with angles $(\approx 90.57^{\circ}, \approx 89.42^{\circ}, \approx 0.01).$

Does there exist complex solutions?