45. \cos B+\cos C=1 <==> nice property.

by Virgil Nicula, Jun 15, 2010, 2:50 PM

Very very nice problem !
seifi-seifi wrote:
Let $ABC$ be a triangle with $AB\ne AC$ . The circle $\omega $ is tangent to $AB$ and $AC$ and circumcircle of $ABC$. The circle $\omega $ is tangent to circumcircle of $ABC$ at $G$. Prove that $AG \perp BC\ \Longleftrightarrow\ \cos B+\cos C=1$ .

Proof (metrical !). Denote $w=C(S,\rho )$ and $T\in AB\cap w$ . Is well-known or prove easly that $IT\perp IA$ , $ST\perp AB$ , $AT=\frac {bc}{s}$

and $\rho=\frac {r}{\cos^2\frac A2}=\frac {bcr}{s(s-a)}$ , where $2s=a+b+c$ . Therefore, $\boxed {\ AG\perp BC\ }$ $\Longleftrightarrow$ the ray $[AS$ is the bisector of $\widehat {GAO}$ $\Longleftrightarrow$

$m(\widehat{AGS})=2\cdot m(\widehat{GAS})$ $\stackrel{(*)}{\Longleftrightarrow}$ $AS^2=SG\cdot(SG+AG)$ $\Longleftrightarrow$ $AS^2-\rho^2=\rho\cdot GA$ $\Longleftrightarrow$ $AT^2=\rho\cdot GA$ $\Longleftrightarrow$

$\frac {b^2c^2}{s^2}=\frac {bcr}{s(s-a)}\cdot AG$ $\Longleftrightarrow$ $AG=\frac {bc(s-a)}{sr}$ $\Longleftrightarrow$ $2R\cos (B-C)=\frac {bc(s-a)}{sr}$ $\Longleftrightarrow$ $2R\cos (B-C)=\frac {2(s-a)}{\sin A}$ $\Longleftrightarrow$

$2(s-a)=2R\sin A\cos (B-C)$ $\Longleftrightarrow$ $2(s-a)=R(\sin 2B+\sin 2C)$ $\Longleftrightarrow$ $2(s-a)=b\cdot\cos B+c\cdot\cos C$ $\Longleftrightarrow$


$b+c-a=(b+c)(\cos B+\cos C)-a$ $\Longleftrightarrow$ $b+c=(b+c)(\cos B+\cos C)$ $\Longleftrightarrow$ $\boxed {\ \cos B+\cos C=1\ }$ .

====================================================================================

$(*)$ Here I used the well-known property : "In a triangle $ABC$ exists the equivalence $B=2C\Longleftrightarrow b=c(c+a)\ \ \vee\ \ b=c$ .
This post has been edited 4 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:47 AM

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