44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

by Virgil Nicula, Jun 8, 2010, 1:48 AM

Quote:
Let $ABC$ be a triangle with the incircle $C(I,r)$ and the circumcircle $w=C(O,R)$ . Denote

midpoint $M$ of $[BC]$ . Prove that $\left\|\ \begin{array}{ccc} IA=IO & \Longleftrightarrow & 2r(b+c)=aR\\\\ IA=IM & \Longleftrightarrow & a=2b\ \vee\ a=2c\\\\ IO=IM & \Longleftrightarrow & A=90^{\circ}\ \vee\ \cos A=\frac {2r}{R}\end{array}\ \right\|$ (here).

Method 1 (metric). Let $2s=a+b+c$ , $D\in AI\cap BC$ and $\{A,S\}=AI\cap w$ .

$\blacktriangleright\ IA=IO\Longleftrightarrow IA^2=IO^2\Longleftrightarrow IA^2=R^2-IA\cdot IS$ $\Longleftrightarrow$ $AI\cdot AS=R^2$ $\Longleftrightarrow$ $\frac {b+c}{2s}\cdot AD\cdot AS=R^2$ $\Longleftrightarrow$

$bc(b+c)=2sR^2$ $\Longleftrightarrow$ $2r(b+c)=aR$ . In conclusion, $\boxed {\ IA=IO\ \Longleftrightarrow\ \frac {b+c}{a}=\frac {R}{2r}\ }$ .

$\blacktriangleright\ IA=IM\Longleftrightarrow$ $IA^2=IM^2$ $\Longleftrightarrow$ $\frac {bc(s-a)}{s}=$ $r^2+\left(\frac {b-c}{2}\right)^2$ $\Longleftrightarrow$ $4bc(s-a)=4sr^2+s(b-c)^2$ $\Longleftrightarrow$ $4(s-a)[bc-(s-b)(s-c)]=$

$s(b-c)^2$ $\Longleftrightarrow$ $4(s-a)^2=(b-c)^2$ $\Longleftrightarrow$ $2(s-a)=|b-c|$ . In conclusion, $\boxed {\ IA=IM\Longleftrightarrow a=2b\ \vee\ a=2c\ }$ .

$\blacktriangleright\ IO=IM\Longleftrightarrow$ $IO^2=IM^2$ $\Longleftrightarrow$ $R^2-2Rr=r^2+\left(\frac {b-c}{2}\right)^2$ $\Longleftrightarrow$ $ 4sR(R-2r)=4sr^2+s(b-c)^2$ $\Longleftrightarrow$ $4sR^2=2abc+s(b-c)^2+4(s-a)(s-b)(s-c)$ $\Longleftrightarrow$

$8sR^2=4abc+(b-c)^2(a+b+c)+(b+c-a)(c+a-b)(a+b-c)$ $\Longleftrightarrow$ $8sR^2=a^2(a+b+c)+2a(b^2+c^2-a^2)$ $\Longleftrightarrow$ $2a\left(b^2+c^2-a^2\right)=$

$(a+b+c)\left(4R^2-a^2\right)$ $\Longleftrightarrow$ $4aS\cot A=sa^2\cot^2A$ $\Longleftrightarrow$ $\cot A=0\ \vee\ 4S=as\cot A$ $\Longleftrightarrow$ $A=90^{\circ}\ \vee\ 4r\sin A=a\cos A$ . Insa $a=2R\sin A$ .

In conclusion, $\boxed {\ IO=IM\ \Longleftrightarrow\ A=90^{\circ}\ \vee\ \cos A=\frac {2r}{R}\ }$ .

Exemple. Triunghiuri in care $A\ne90^{\circ}$ si totusi $IO=IM$ , unde $M$ este mijlocul lui $[BC]$ .

EX.1 Triunghiul pitagoreic $B$-dreptunghic $ABC$ pentru care $a=3$ , $b=5$ , $c=4$ .

EX.2 Triunghiul $B$-isoscel $ABC$ pentru care $a=c=2$ , $b=3$ .

EX.3 Triunghiul $ABC$ cu $A=60^{\circ}$ si $R=4r$ .

EX.4. In triunghiul $ABC$ pentru care $\cos A=\frac {2r}{R}$ si $R=r\left(1+\sqrt 5\right)$ triunghiul $IOM$ este echilateral.
This post has been edited 29 times. Last edited by Virgil Nicula, Nov 27, 2015, 7:46 AM

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