104. Concursul Revistei de Matematica din Galati (RMG).

by Virgil Nicula, Sep 8, 2010, 12:52 AM

Quote:
Aratati ca într-un triunghi $ABC$ dreapta $N\Gamma\ \parallel\ BC\ \Longleftrightarrow\ F\in (AM)$ , unde $M$ - mijlocul laturii $[BC]$ , $N$ - punctul lui Nagel ,

$\Gamma$ - punctul lui Gergonne si $F$ - punctul lui Feuerbach (din lista scurta a concursului R.M.G. - Revista de Matematica din Galati din 2007).

Demonstratie. Vom arata ca $\underline{\overline{\left\|\ N\Gamma\ \parallel\ BC\ \Longleftrightarrow\ a(b+c)=b^2+c^2\ \Longleftrightarrow\ F\in (AM)\ \right\|}}$ .

Notam a doua intersectie $L$ a cercului Euler cu mediana $AM$ , ortocentrul $H$ , $D\in AH\cap BC$ si mijlocul $E$ al segmentului $[AH]$ .

Se stie ca punctele $M$ , $D$ , $E$ apartin cercului Euler si din puterea punctului $A$ fata de acest cerc se obtine : $AE\cdot AD=AL\cdot AM$ $\Longleftrightarrow$

$R\cos A\cdot h_a=AL\cdot m_a$ $\Longleftrightarrow$ $4Rh_a\cdot\cos A=4m_a\cdot AL$ $\Longleftrightarrow$ $2bc\cdot\cos A=4m_a\cdot AL$ $\Longleftrightarrow$ $\boxed{\ AL=\frac {b^2+c^2-a^2}{4m_a}\ }$ .

Deci $LM=AM-AL=\frac {4m_a^2-\left(b^2+c^2-a^2\right)}{4m_a}=$ $\frac {2\left(b^2+c^2\right)-a^2-\left(b^2+c^2\right)+a^2}{4m_a}$ , adica $\boxed{\ LM=\frac {b^2+c^2}{4m_a}\ }$ .

$\odot\ \ \boxed{\ N\Gamma\ \parallel\ BC\ }$ $\Longleftrightarrow$ $\frac {(p-b)+(p-c)}{p-a}=\frac {(p-a)(p-b)+(p-a)(p-c)}{(p-b)(p-c)}$ $\Longleftrightarrow$

$\frac {a}{(p-a)}=\frac {a(p-a)}{(p-b)(p-c)}$ $\Longleftrightarrow$ $ \boxed{\ (p-a)^2=(p-b)(p-c)\ }$ $\Longleftrightarrow$ $ \boxed{\ a(b+c)=b^2+c^2\ }$ .

$\odot\ \ \boxed{\ F\in AM\ }\ \Longleftrightarrow$ $L\in AM\cap w$ , unde $w=C(I,r)$ . Notam $\{L,S\}=AM\cap w$ , unde $L\in (AS)$ . Din $AL=\frac {b^2+c^2-a^2}{4m_a}$

si $LM=\frac {b^2+c^2}{4m_a}$ obtinem $p_w(A)=(p-a)^2=$ $AL\cdot AS$ $\Longrightarrow$ $AS=\frac {4m_a(p-a)^2}{b^2+c^2-a^2}$ . Asadar, $MS=MA-AS=$ $m_a-\frac {4m_a(p-a)^2}{b^2+c^2-a^2}$ ,

adica $MS=\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}$ . De asemenea, $p_w(M)=\frac {(b-c)^2}{4}=$ $MS\cdot ML=\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\cdot \frac {b^2+c^2}{4m_a}$ ,

adica $\left(b^2+c^2-a^2\right)(b-c)^2=$ $2\left(b^2+c^2\right)\left[a(b+c)-a^2-bc\right]\ \Longleftrightarrow\ \left[a(b+c)-\left(b^2+c^2\right)\right]^2=0$ $\Longleftrightarrow$ $\boxed{\ a(b+c)=b^2+c^2\ }$ .

Observatie. Apropos de subiectul topicului "Puncte importante intr-un triunghi". Pentru cei care nu cunosc semnificatiile punctelor remarcabile mentionate in enuntul problemei propuse le recomand sa consulte Google unde, spre surprinderea lor, vor intalni cel putin 100 de asemenea puncte ceea ce inseamna ca triunghiul a fost si ramane o preocupare de secole a multor matematicieni, multi dintre ei cu rezultate remarcabile la nivel inalt.

In demonstratie am folosit urmatoarele relatii metrice dintr-un triunghi care se pot dovedi fara dificultate : $ab+bc+ca=p^2+\sum(p-b)(p-c)=p^2+r(4R+r)\ \ ;\ \ IA^2=\frac {bc(p-a)}{p}$ .
This post has been edited 20 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:18 AM

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