Fie \( \triangle ABC \) si \( I \) centrul cercului inscris. \( R_1,\ R_2,\ R_3 \) sunt razele cercurilor circumscrise triunghiurilor \( BIC,\ AIC,\ AIB \).
Sa se arate ca daca \( R_1R_2+R_2R_3+R_3R_1=3R\sqrt[3]{4Rr^{2}}, \) atunci \( \triangle ABC \) este echilateral.
Conditie ca un triunghi sa fie echilateral
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Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
\( R_1=\frac{IB\cdot IC\cdot BC}{4\cdot \frac{IB\cdot IC\cdot sin(\angle BIC)}{2}}=\frac{a}{2\cos \frac{A}{2}}=2R\ \sin\frac{A}{2} \)
Analog \( R_2=2R\ \sin\frac{B}{2},\ R_3=2R\ \sin\frac{C}{2}. \)
Aplicam \( AM-GM \):
\( R_1R_2+R_2R_3+R_3R_1\geq 3\sqrt[3]{R_1^{2}R_2^{2}R_3^{2}}=3\sqrt[3]{64\cdot R^{6}\cdot \prod \sin^{2}\frac{A}{2}}=3\sqrt[3]{64\cdot R^{6}\cdot \frac{r^{2}}{16R^{2}}}=3R\sqrt[3]{4Rr^{2}}. \)
Suntem in cazul de egalitate deci \( R_1=R_2=R_3 \). De aici totul este clar.
Analog \( R_2=2R\ \sin\frac{B}{2},\ R_3=2R\ \sin\frac{C}{2}. \)
Aplicam \( AM-GM \):
\( R_1R_2+R_2R_3+R_3R_1\geq 3\sqrt[3]{R_1^{2}R_2^{2}R_3^{2}}=3\sqrt[3]{64\cdot R^{6}\cdot \prod \sin^{2}\frac{A}{2}}=3\sqrt[3]{64\cdot R^{6}\cdot \frac{r^{2}}{16R^{2}}}=3R\sqrt[3]{4Rr^{2}}. \)
Suntem in cazul de egalitate deci \( R_1=R_2=R_3 \). De aici totul este clar.