98. Symmedian & median (metrical relations).

by Virgil Nicula, Sep 3, 2010, 4:54 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=364921
Quote:
In $\triangle ABC$, the $A$-median meets $BC$ at $M$ and the circumcircle again at $A_1$. The $A$-symmedian meets $BC$ at $S$ and the circumcircle again at $A_2$ . Prove that $AS=\frac {2bc}{b^2+c^2}\cdot AM$ and $O\in A_1A_2\iff (b^2+c^2)^2+4b^2c^2 =$ $ a^2(b^2+c^2)$ $\iff$ $\cos A=-\frac {2bc}{b^2+c^2}=$ $-\frac {AS}{AM}$ $\iff$ $\tan\frac A2=\frac {b+c}{|b-c|}$ .

Proof. Observe that $\frac {SB}{c^2}=\frac {SC}{b^2}=\frac {a}{b^2+c^2}$ , $A_1A_2\parallel BC$ , i.e. $BA_2=CA_1$ , $BA_1=CA_2$ and $MS=\frac {a|b^2-c^2|}{2(b^2+c^2)}$ . Using similarities $ABA_2\sim AMC$ , $ACA_2\sim AMB$ and $ABA_1\sim ASC$ $\implies$ $\left\{\begin{array}{c}
AA_1\cdot s_a=AA_2\cdot m_a=bc \\\\
BA_2=CA_1=\frac {ac}{2m_a}\\\\
BA_1=\frac {ab^2c}{(b^2+c^2)s_a}\ ,\ CA_2=\frac {ab}{2m_a}\end{array}\right\|$ . Observe that $BA_1=CA_2\implies$ $\boxed{s_a=\frac {2bc}{b^2+c^2}\cdot m_a}$ .

Observe that $\boxed{O\in A_1A_2}$ $\iff$ $AS\perp AM$ $\iff$ $SM^2=m_a^2+s_a^2$ $\iff$ $\frac {a^2(b^2-c^2)^2}{4(b^2+c^2)^2}=m_a^2\cdot\frac {(b^2+c^2)^2+4b^2c^2}{(b^2+c^2)^2}$ $\iff$

$\left[2(b^2+c^2)-a^2\right]\left[(b^2+c^2)+4b^2c^2\right]=a^2(b^2+c^2)^2-4a^2b^2c^2$ $\iff$ $\boxed{(b^2+c^2)^2+4b^2c^2=a^2(b^2+c^2)}$ $\iff$

$(b^2+c^2)(b^2+c^2-a^2)+4b^2c^2=0$ $\iff$ $(b^2+c^2)\cos A+2bc=0$ $\iff$ $\cos A=-\frac {2bc}{b^2+c^2}$ $\iff$ $\boxed{\cos A=-\frac {s_a}{m_a}}$ .

Quote:
Proposed problem. In $\triangle ABC$ with $c<b$ the $A$-median and $A$-symmedian meet $BC$ at $M$ and at $S$ . Prove that $AM\perp AS$ $\iff$ $A=\frac {\pi}{2}+2\cdot\arctan\frac cb$ .
Quote:
Generalization. Let $ABC$ be a triangle. Consider two points $\{M,S\}\subset [BC]$ of $\triangle ABC$ so that $\widehat {BAM}\equiv\widehat {CAS}$ .

The lines $AM$ , $AS$ meet again the circumcircle of $ABC$ in $A_1$ , $A_2$ respectively. Then $AS=\frac {(m+1)bc}{mb^2+c^2}\cdot AM$ and

$\boxed{O\in A_1A_2\iff (mb^2+c^2)^2+(m+1)^2b^2c^2 = a^2(mb^2+c^2)\iff\cos A=-\frac {2mbc}{m^2b^2+c^2}\iff\tan\frac A2=\frac {mb+c}{|mb-c|}}$ .
Quote:
Proposed problem. In $\triangle ABC$ denote $D\in (BC)$ for wich $\widehat{DAB}\equiv\widehat {DAC}$ . Consider two points $\{M,S\}\subset (BC)$
so that $M\in (DC)$ , $\frac {MB}{MC}=m$ and $\widehat {BAM}\equiv\widehat {CAS}$ . Prove that $AM\perp AS$ $\iff$ $A=\frac {\pi}{2}+2\cdot\arctan\frac {c}{mb}$ .
Quote:
Proposed problem. Fie triunghiul $ABC$ in care notam lungimea $m_a$ a medianei din varful $A$ . Sa se arate ca mediana
si simediana din varful $A$ trisecteaza unghiul $\widehat {BAC}$ daca si numai daca $\boxed{\ m_a=\frac {\left|b^2-c^2\right|}{2\cdot\min\ \{b,c\}}\ }$ .
Demonstratie. Notam $A$-mediana $[AM]$ si $A$-simediana $[AS]$ , unde $\{M,S\}\subset (BC)$ . Presupunem fara a restrange generalitatea ca $c<b$ . Se arata usor ca $BS=\frac {ac^2}{b^2+c^2}$ si $SM=\frac {a\left(b^2-c^2\right)}{2\left(b^2+c^2\right)}$ . Asadar $[AS$ este bisectoarea unghiului $\widehat{BAM}$ daca si numai daca $\frac {AB}{AM}=\frac {BS}{SM}$ $\Longleftrightarrow$ $\frac {c}{m_a}=$ $\frac {ac^2}{b^2+c^2}\cdot\frac {2\left(b^2+c^2\right)}{a\left(b^2-c^2\right)}$ $\Longleftrightarrow$ $m_a=\frac {\left|b^2-c^2\right|}{2\cdot\min\ \{b,c\}}$ .
Observatie. Notam proiectia $D$ a lui $A$ pe $BC$ . Se arata usor ca $m_a=\frac {\left|b^2-c^2\right|}{2\cdot\min\ \{b,c\}}\Longleftrightarrow $ $\sin \widehat{MAD}=\frac {\min\{b,c\}}{a}\ \Longleftrightarrow\ 4b^2c^2+c^4=b^4+a^2c^2$ .
This post has been edited 33 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:34 AM

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