Angle formed by orthocenter, incenter and circumcenter of a triangle $>135^\circ$?

Question

If $H$ is the orthocenter, $I$ the incenter and $O$ the circumcenter of a triangle , the I need to show that the angle $HIO>135^\circ$

With the assumptions of $OI^2=R^2-2Rr$, $OH^2=9R^2-(a^2+b^2+c^2)$, $HI^2=2r^2-4R^2\cos A\cos B \cos C$ and $R^2\cdot8(1+\cos A\cos B\cos C)= a^2+b^2+c^2$

Applying the $cosine$ rule I got to something like $$\cos (HIO)=\frac{2r^2-2Rr+R^2\cdot 4\cos A\cos B\cos C}{2(2r^2-4R^2\cos A\cos B\cos C)(R^2-2Rr)}$$

I need to show that the LHS is between $\left[-\frac{1}{\sqrt{2}},-1\right]$

How t0 proceed?

Is there any simpler way to prove the question (without applying cosine rule) ?

Answer 1

Too long for a comment.

Put $X=4R^2\cos A\cos B \cos C$.

We have to show that $$-\frac 1{\sqrt{2}}>\cos (HIO)=\frac{HI^2+OI^2-OH^2}{2\cdot HI\cdot OI}.$$

Since it is assmed that $2\cdot HI\cdot OI>0$, we have to show that

$$OH^2- HI^2-OI^2>\sqrt{2}\cdot HI\cdot OI$$

$$2Rr-2r^2-X>\sqrt{2(2r^2-X)(R^2-2Rr)}$$

Since $HI^2>0$ and $OI^2>0$, $2r^2>X$ and $R>2r$, so $2Rr-2r^2-X>0$. Thus we have to show that

$$(2Rr-2r^2-X)^2>2(2r^2-X)(R^2-2Rr)$$

$$X^2+2XR^2+4Xr^2+4r^4>8RrX$$

The respective quadratic equation for $X$ has a discriminant $D=(R-2r)^2R(R-4r)$ and roots $X_1=4Rr-2r^2-R^2-\sqrt{D}$ and $X_2=4Rr-2r^2-R^2+\sqrt{D}$. So if $R<4r$ then $D<0$ and the inequality is proved.

Otherwise I tried to evaluate $X$ in terms of $R$ and $r$. According to exercises at p.23 of “Topics in Inequalities - Theorems and Techniques” by Hojoo Lee (version February 25, 2006), we have $X=s^2-(2R+r)^2$. Moreover, in W. J. Blundon (see Problem E1935, Amer. Math. Monthly 73 (1966), 1122) found the best possible inequalities of the form $A(R,r)\le s^2\le B(R, r)$, where $A(x, y)$ and $B(x, y)$ are real quadratic forms $\alpha x^2+\beta xy+\gamma y^2$, namely $$16Rr-5r^2\le s^2\le 4R^2+4Rr+3r^2.$$

This implies $$12Rr-4R^2-6r^2\le X\le 2r^2.$$

Unfortunately, this doesn’t help because

$$12Rr-4R^2-6r^2\le X_1\le X_2\le 2r^2.$$

Answer 2

It's wrong.

Try $a=3$, $b=10$ and $c=11.$

Answer 3

It is not always true that $\angle{HIO}\gt 135^\circ$.

A counterexample is $(a,b,c)=(85,13,88)$.

$\qquad\qquad$

For $(a,b,c)=(85,13,88)$, we have $$\begin{align}R&=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac{2431}{4\sqrt{186}} \\\\r&=\frac{1}{2}\sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c)}{a+b+c}}=\frac{40\sqrt{186}}{93} \\\\OI^2&=R^2-2Rr=\frac{1451307}{992} \\\\OH^2&=9R^2-(a^2+b^2+c^2)=\frac{2712387}{992} \\\\HI^2&=2r^2+4R^2-\frac{a^2+b^2+c^2}{2}=\frac{109875}{248} \\\\\cos(\angle{HIO})&=\frac{HI^2+OI^2-OH^2}{2\cdot HI\cdot OI}=-\frac{13693}{\sqrt{708721585}}\approx -0.5143524 \\\\\angle{HIO}&\approx 120.95^\circ\end{align}$$

Answer 4

This formula calculates the $θ$ angle formed by Orthocenter(H),Incenter(I) and circumcenter(O) of ABC triangle (sides:a,b,c).

$θ(a,b,c)=acot\left [\left [ \frac{(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}{(-a+b+c)(a-b+c)(a+b-c)}-2abc+(-a+b+c)(a-b+c)(a+b-c) \right ] \sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c)}{4(a+b+c)^3(a-b)^2(a-c)^2(b-c)^2}} \right ] $

$θ(3,10,11)= π -acot(\frac{523}{1344}\sqrt{6})=133.627°$

defining $θ$ as a function in the interval of existence of the ABC triangle to use the Calculus tools

$\begin{array}{} |b-c|<a<b+c & a≠b & a≠c \end{array} $

$Range(θ)=(\frac{ π }{2}, π )$

$θ=\frac{ π }{2}$ in the right triangle, $θ(\sqrt{b^2+c^2-2bc·cos(\frac{ π }{2})},b,c)$ whith $c→∞$ or $c→0$. $θ=π$ for $a=|b-c|$, $a=b$, $a=c$ and $a=b+c$.

maximum and minimum

$\begin{array}{} \frac{d}{da}θ(a,b,c)=0 & solve&for&a & \end{array}$

$\begin{array}{} r1=2.43938 & r2=10.54162 & r3=12.39471 \end{array}$

$θ(r3,b,c)=132.292°$

$\begin{array}{} θ(a,10,11)=135°=\frac{3}{4}π & a1=1.82913 & a2=3.27918 \end{array}$

$\begin{array}{} \frac{R}{r}<4 & ? \end{array}$

$k=\frac{R}{r}=\frac{2abc}{(-a+b+c)(a-b+c)(a+b-c)}$