377. Some remarkable properties of the symmedian.

by Virgil Nicula, Jun 8, 2013, 4:44 PM

PP1. Let $ABC$ be a triangle with the circumcircle $w=C(O,R)$ . Denote $XX$ - the tangent line to the circle $w$ at the point $X$ . The $A$-symmedian cut $BC$ in $S$ and

meet again $w$ in $E$. Denote the midpoint $M$ of $[BC]$ , $T\in AA\cap BC$ , $L\in BB\cap CC$ and $AM=m_a$ , $AS=s_a$ . Prove that the following relations:

$1\blacktriangleright\ \ \boxed{s_a=\frac {2bc}{b^2+c^2}\cdot m_a}$ . Generally, if $AS$ is the $A$-isogonal line of the line $AM$ for which $\frac {MB}{MC}=m$ , then $\boxed{AS=\frac {(m+1)bc}{mb^2+c^2}\cdot AM}$ .

$2\blacktriangleright\ \ AE\cdot m_a=bc$ and $\boxed{TA=\frac {abc}{b^2+c^2}}\ ,\ T\in EE$ .

$3\blacktriangleright\ \ \frac {EB}{c}=\frac {EC}{b}=\frac {a}{2m_a}$ (the quadrilateral $ABEC$ is harmonically) and $L\in \overline{ASE}$ .

$4\blacktriangleright\ \ \boxed{h_a+\left(\frac a2+m_a\right)\sin B\le \frac {3\sqrt 3}{2}m_a}$ . Particular case. If $A=90^{\circ}$ , then $h_a+\max\{b,c\}\le \frac {3\sqrt 3}{4}a$ .


Proof.

$1\blacktriangleright$ Is well-known that $\frac {SB}{SC}\cdot\frac {MB}{MC}=\left(\frac cb\right)^2$ , i.e. $\frac {SB}{SC}=\frac {c^2}{mb^2}\implies$ $\left\{\begin{array}{ccc}
\frac {MB}{m}=\frac {MC}{1}=\frac a{m+1} & \implies & MC=\frac a{m+1}\\\\
\frac {SB}{c^2}=\frac {SC}{mb^2}=\frac a{mb^2+c^2} & \implies & SB=\frac {ac^2}{mb^2+c^2}\end{array}\right|\ (*)$ . Denote $m\left(\widehat{BAS}\right)=m\left(\widehat{CAM}\right)=\phi$ .

Apply the theorem of Sines in the triangles: $\left\{\begin{array}{cc}
\triangle BAS\ : & \frac {AS}{\sin B}=\frac {SB}{\sin\phi}\\\\
\triangle AMC\ : & \frac {AM}{\sin C}=\frac {MC}{\sin\phi}\end{array}\right|\implies$ $\frac {AS}{AM}\cdot \frac {\sin C}{\sin B}=\frac {SB}{MC}\stackrel{(*)}{\implies}$ $\frac {AS}{AM}\cdot\frac cb=\frac {ac^2}{mb^2+c^2}\cdot\frac {m+1}{a}\implies$ $AS=\frac {(m+1)bc}{mb^2+c^2}\cdot AM$ .

$ 2\wedge 3\blacktriangleright\ \left\{\begin{array}{ccc}
\triangle ABE\sim\triangle AMC & \implies & \frac {AE}{b}=\frac {c}{AM}=\frac {EB}{CM}\\\\
\triangle ACE\sim\triangle AMB & \implies & \frac {AE}{c}=\frac {b}{AM}=\frac {EC}{BM}\\\\
\triangle TAB\sim\triangle TCA & \implies & \frac {TA}{TC}=\frac {TB}{TA}=\frac cb\end{array}\right|\implies$ $AE\cdot m_a=bc$ and $\frac {EB}{c}=\frac {EC}{b}=\frac {a}{2m_a}$ , i.e. the quadrilateral $ABEC$ is harmonically: $EB\cdot b=EC\cdot c$ .

Observe that $\left\{\begin{array}{c}
TB=TA\cdot\frac cb\\\\
TC=TA\cdot \frac bc\end{array}\right|\implies$ $TA^2=TB\cdot TC$ and $\frac {TB}{TC}=\left(\frac cb\right)^2$ , i.e. $\frac {TB}{c^2}=\frac {TC}{b^2}=\frac a{b^2+c^2}\implies$ $TA^2=TB\cdot TC=$ $\frac {a^2b^2c^2}{\left(b^2+c^2\right)^2}\implies$ $TA=\frac {abc}{b^2+c^2}$ .

Denote $T_1\in EE\cap BC$ . Prove similarly $\frac {T_1B}{T_2C}=\left(\frac {EB}{EC}\right)^2=$ $\left(\frac cb\right)^2=\frac {TB}{TC}\implies$ $\frac {T_1B}{T_1C}=\frac {TB}{TC}\implies$ $T_1\equiv T\implies$ $T\in EE$ .

Denote $\left\{\begin{array}{c}
L_1\in AE\cap BB\\\\
T_2\in AE\cap CC\end{array}\right|$ . Therefore, $\left\{\begin{array}{c}
\frac {L_1E}{L_1A}=\left(\frac {EB}c\right)^2\\\\
\frac {L_2E}{L_2A}=\left(\frac {CE}b\right)^2\end{array}\right|\stackrel{\frac {EB}c=\frac {CE}b}{\implies}$ $\frac {L_1E}{L_1A}=\frac {L_2E}{L_2A}\implies$ $L_1\equiv L_2\equiv L\implies L\in\overline{ASE}$ .

$4\blacktriangleright$ Apply the remarkable inequality $\boxed{2s\le 3R\sqrt 3}$ to $\triangle AEC\ :\ AE+EC+CA\le 3R\sqrt 3\iff$ $\frac {bc}{m_a}+\frac {ab}{2m_a}+b\le 3R\sqrt 3\iff$ $2bc+ab+2bm_a\le 6Rm_a\sqrt 3\iff$

$4Rh_a+ab+2bm_a\le 6Rm_a\sqrt 3\iff$ $h_a+\frac {ab}{4R}+\frac {bm_a}{2R}\le\frac {3\sqrt 3}{2}m_a\iff$ $h_a+\left(\frac a2+m_a\right)\sin B\le \frac {3\sqrt 3}{2}m_a$ . We have the equality if and only if

the triangle $ABM$ is equilateral $\iff$ $c=\frac a2=m_a\iff$ $a=2c$ and $4m_a^2=a^2\iff$ $a=2c$ and $b^2+c^2=a^2\iff$ $\frac a2=\frac b{\sqrt 3}=\frac c1$ .
This post has been edited 31 times. Last edited by Virgil Nicula, Nov 15, 2015, 1:51 PM

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