Angle AIO is less than 90 degrees iff 2a =<b+c

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Angle AIO is less than 90 degrees iff 2a =<b+c

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WakeUp

#1 h Nov 7, 2010, 7:40 PM • 2 Y

Y by Adventure10, Mango247

Let $I$ be the incentre and $O$ the circumcentre of a non-equilateral triangle $ABC$ . Prove that $\angle AIO \le 90^{\circ}$ if and only if $2BC\le AB+AC$ .

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jgnr

#2 h Nov 8, 2010, 12:03 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$\angle AIO\le90^{\circ}$

$\Leftrightarrow AI^2+IO^2\ge AO^2$

$\Leftrightarrow \frac{r^2}{\sin^2\frac{A}2}+R(R-2r)\ge R^2$

$\Leftrightarrow r^2\ge (1-\cos A)Rr$

$\Leftrightarrow r\ge (1-\cos A)R$

$\Leftrightarrow \frac{A}{s}\ge(1-\frac{b^2+c^2-a^2}{2bc})\frac{abc}{4A}$

$\Leftrightarrow \frac{8A^2}{a+b+c}\ge\frac{a^2-(b-c)^2}{2bc}\cdot abc$

$\Leftrightarrow 16A^2\ge(a-b+c)(a+b-c)a(a+b+c)$

$\Leftrightarrow (a+b+c)(a+b-c)(a-b+c)(-a+b+c)\ge (a-b+c)(a+b-c)a(a+b+c)$

$\Leftrightarrow -a+b+c\ge a$

$\Leftrightarrow 2a\le b+c$

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#3 h Jan 9, 2014, 2:54 PM • 4 Y

Y by SMOJ, tchebytchev, Adventure10, Mango247

jgnr wrote:

$\angle AIO\le90^{\circ}$

$\Leftrightarrow AI^2+IO^2\ge AO^2$

$\Leftrightarrow \frac{r^2}{\sin^2\frac{A}2}+R(R-2r)\ge R^2$

$\Leftrightarrow r^2\ge (1-\cos A)Rr$

$\Leftrightarrow r\ge (1-\cos A)R$

$\Leftrightarrow \frac{A}{s}\ge(1-\frac{b^2+c^2-a^2}{2bc})\frac{abc}{4A}$

$\Leftrightarrow \frac{8A^2}{a+b+c}\ge\frac{a^2-(b-c)^2}{2bc}\cdot abc$

$\Leftrightarrow 16A^2\ge(a-b+c)(a+b-c)a(a+b+c)$

$\Leftrightarrow (a+b+c)(a+b-c)(a-b+c)(-a+b+c)\ge (a-b+c)(a+b-c)a(a+b+c)$

$\Leftrightarrow -a+b+c\ge a$

$\Leftrightarrow 2a\le b+c$

So complicated..
Let $AI\cap k_{ABC}=D$
We know that $DC=DB=DI$
By Ptolemy, we have $AD\cdot BC=(AB+AC)\cdot DI \rightarrow DI=\dfrac{AD\cdot BC}{AB+AC}$
Now we have:
$AO=OD \rightarrow \measuredangle AIO\leq 90^{o} \Leftrightarrow DI\leq\dfrac{AD}{2} \Leftrightarrow 2BC\leq AB+AC$

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