O problema data la ONM in Germania.

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Virgil Nicula
Euler
Posts: 622
Joined: Fri Sep 28, 2007 11:23 pm

O problema data la ONM in Germania.

Post by Virgil Nicula »

In \( \triangle\ ABC \) notam ortocentrul \( H \) , centrul \( O \) al cercului circumscris si centrul \( I \)

al cercului inscris. Definim punctele \( \left|\ \begin{array}{c}
E\in BH\ \cap\ AC\\\\\\\\
F\in CH\ \cap\ AB\end{array}\ \right| \)
si \( \left|\ \begin{array}{c}
N\in BI\ \cap\ AC\\\\\\\\
P\in CI\ \cap\ AB\end{array}\ \right|\ . \)


Sa se arate ca \( I\in EF\ \Longleftrightarrow\ \cos B+\cos C=\cos A\ \Longleftrightarrow\ O\in NP\ . \)
Last edited by Virgil Nicula on Sun Feb 08, 2009 9:28 pm, edited 2 times in total.
Marius Mainea
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Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Post by Marius Mainea »

Daca I apartine lui EF, din teorema bisectoarei \( \frac{FI}{IE}=\frac{AF}{AE}=\frac{b}{c}=k \) de unde

\( \overline{AI}=\frac{1}{k+1}\overline{AF}+\frac{k}{k+1}\overline{AE}=\frac{b\cos A}{b+c}\overline{AB}+\frac{c\cos A}{b+c}\overline{AC} \)

Deasemenea vectorul de pozitie al lui I fata de reperul A este

\( \overline{AI}=\frac{b}{a+b+c}\overline{AB}+\frac{c}{a+b+c}\overline{AC} \)

si din cele doua relatii evident rezulta ca \( (a+b+c)\cos A=b+c \), relatie echivalenta cu \( \cos B+\cos C=\cos A \).

Reciproc, presupunand ca \( \cos B+\cos C=\cos A \) obtinem ca \( \overline{AI}=\frac{1}{k+1}\overline{AF}+\frac{k}{k+1}\overline{AE} \), ceea ce inseamna ca I apartine lui EF.
Virgil Nicula
Euler
Posts: 622
Joined: Fri Sep 28, 2007 11:23 pm

Post by Virgil Nicula »

In \( \triangle\ ABC \) notam ortocentrul \( H \) , centrul \( O \) al cercului circumscris si centrul \( I \)

al cercului inscris. Definim punctele \( \left|\ \begin{array}{c}
E\in BH\ \cap\ AC\\\\\\\\
F\in CH\ \cap\ AB\end{array}\ \right| \)
si \( \left|\ \begin{array}{c}
N\in BI\ \cap\ AC\\\\\\\\
P\in CI\ \cap\ AB\end{array}\ \right|\ . \)


Sa se arate ca \( I\in EF\ \Longleftrightarrow\ \cos B+\cos C=\cos A\ \Longleftrightarrow\ O\in NP\ . \)
Demonstratie.

Se stie ca \( I(a,b,c) \) si \( \left|\ \begin{array}{c}
H(\tan A,\tan B,\tan C)\\\\\\
O(\sin 2A,\sin 2B,\sin 2C)\end{array}\ \right| \)
- coordonate baricentrice in raport cu \( \triangle ABC \) .

Asadar, \( E(\tan A\ ,\ 0\ ,\ \tan C) \) , \( F(\tan A,\ \tan B\ ,\ 0) \) si \( N(a\ ,\ 0\ ,\ c) \) , \( P(a\ ,\ b\ ,\ 0)\ . \) In concluzie,

\( \begin{array}{ccc}
I\in EF & \Longleftrightarrow & \left|\begin{array}{ccc}
a & b & c \\\\\\\\
\tan A & 0 & \tan C\\\\\\\\
\tan A & \tan B & 0\end{array}\right|=0\\\\\\\\
O\in NP & \Longleftrightarrow & \left|\begin{array}{ccc}
\sin 2A & \sin 2B & \sin 2C \\\\\\\\
a & 0 & c\\\\\\\\
a & b & 0\end{array}\right|=0\end{array}\ \begin{array}{c}
\searrow\\\\
\nearrow\end{array}\Longleftrightarrow\left|\begin{array}{ccc}
\cos A & \cos B & \cos C \\\\\\\\
1 & 0 & 1\\\\\\\\
1 & 1 & 0\end{array}\right|\ \Longleftrightarrow\ \cos B+\cos C=\cos A\ . \)
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