107. XY parallel BC and X,Y are isogonal/izotomic points.

by Virgil Nicula, Sep 9, 2010, 3:57 PM

Lemma. Let $ABC$ be a triangle with circumradius $R$ . If $\{x,y,z\}\subset \mathbb{R}$ Then $\boxed{\ \begin{array}{c} x+y+z=0\implies yz\cdot a^2+zx\cdot b^2+xy\cdot c^2\le 0\\\\ M(x,y,z)\implies -p_w(M)=yza^2+zxb^2+xyc^2\\\\ yz\cdot a^2+zx\cdot b^2+xy\cdot c^2\ \le\ R^2\cdot\left(x+y+z\right)^2\end{array}\ }$ .
Quote:
Let $X(x,y,z)$ , $Y$ be two points in the plane of $\triangle ABC$ , where $(x,y,z)$ are the barycentrical coordinates of $X$ w.r.t. given triangle. Prove that :

$\blacktriangleright$ If $Y$ is the isogonal point of $X$ , then $Y\left(\frac {a^2}{x},\frac {b^2}{y},\frac {c^2}{z}\right)$ and $XY\parallel BC\ \iff$ exists the relation $yz(y+z)\cdot a^2=x^2\left(z\cdot b^2+y\cdot c^2\right)$ .

Particular case. If $X$ , $Y$ are the centroid and symmedian point of $\triangle ABC$ respectively, then $GS\parallel BC\ \iff\ b^2+c^2=2a^2\ \iff\ OH\perp AM$ .

Proof. Denote $T\in AS\cap BC$ and the second intersection $N$ of $AM$ with the circumcircle $w$ . Thus, $SG\parallel BC$ $\iff$ $\frac {SA}{ST}=\frac {GA}{GM}$ $\iff$ $\frac {b^2+c^2}{a^2}=2$ $\iff$ $b^2+c^2=2a^2$ . On the other hand, $AM\perp OH\iff$ $AM\perp OG$ $\iff$ $GA=GN$ $\iff$ $m_a=3\cdot MN$ . From power of $M$ w.r.t. $w$ obtain $MA\cdot MN=MB\cdot MC$ $\iff$ $m_a\cdot\frac {m_a}{3}=\frac {a^2}{4}$ $\iff$ $4m_a^2=3a^2$ $\iff$ $2a^2=b^2+c^2$ .

$\blacktriangleright$ If $Y$ is the isotomic point of $X$ , then $Y(yz,zx,xy)$ and $XY\parallel BC\ \iff$ exists the relation $x^2=yz$ .

Particular case. If $X$ , $Y$ are $N$ - Nagel's point and $\Gamma$ - Gergonne's point of $\triangle ABC$ , then $N\Gamma\parallel BC\ \iff\ b^2+c^2=a(b+c)$ .

Proof. Denote $T\in AN\cap BC$ and $L\in A\Gamma\cap BC$ . Thus, $N\Gamma \parallel BC$ $\iff$ $\frac {NA}{NT}=\frac {\Gamma A}{\Gamma L}$ $\iff$ $\frac {p-b}{p-a}+\frac {p-c}{p-a}=\frac {p-a}{p-b}+\frac {p-a}{p-c}$ $\iff$ $\frac {a}{p-a}=\frac {a(p-a)}{(p-b)(p-c)}$ $\iff$ $(p-b)(p-c)=(p-a)^2$ $\iff$ $b^2+c^2=a(b+c)$ .

Proposed problem. Prove that in any triangle $ABC$ exists the chain $\boxed{\ \begin{array}{c} \sqrt{abc\left(a^3+b^3+c^3\right)}\ \le\ R\cdot (ab+bc+ca)\ \le\ 4R(R+r)^2\end{array}\ }$ .

Proof. Denote the isotomic $J(bc,ca,ab)$ of incenter $I(a,b,c)$ w.r.t. the triangle $ABC$ . From the remarkable property - The power $p_w(M)$ of the point $M(x,y,z)$ w.r.t. the circumcircle $w$ of the triangle $ABC$ is given by the relation $-p_w(M)=\frac {a^2yz+b^2zx+c^2xy}{(x+y+z)^2}=R^2-OM^2\le R^2$ - obtain for $M:=J$ that $\frac {a^2\cdot ca\cdot ab+b^2\cdot ab\cdot bc+c^2\cdot bc\cdot ca}{(ab+bc+ca)^2}\le R^2$$\iff$ $abc\left(a^3+b^3+c^3\right)\le R^2\cdot (ab+bc+ca)^2$ , i.e. the conclusion of the proposed problem.
This post has been edited 57 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:30 AM

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    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

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